Beth’s Daily AP Calculus BC Practice

posted: 09-Mar-2026 & updated: 16-Mar-2026

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Short daily sessions — 15–20 min, casual, no pressure. 🧠
Think of it like a daily stretch before a game. The goal is exposure, not perfection.
The more times you bump into a concept, the more natural it becomes.

“The best way to understand a limit is to bump into it over and over —
not to stare at it once for three hours.” — Daddy

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Table of Contents

New sessions are prepended at the top of each day’s section — most recent first.

• 16-Mar-2026
• 15-Mar-2026
• 11-Mar-2026
• 10-Mar-2026
• 09-Mar-2026

March 2026

March 16, 2026

📅 Mon, March 16, 2026  |  Topics – Limits (sharp tools) · Indeterminate forms · Series · Integration by parts · FTC both parts · Volumes

Part 1 — Limits: Sharp Tools

Same two roads — but some of today’s limits require three applications of L’Hôpital, or Taylor series that cut straight through. Always ask yourself: is there a slicker way than L’Hôpital?

1. $\displaystyle\lim_{x\to 0}\frac{\arcsin x}{x}$   (Method B: which derivative? at which point?)
2. $\displaystyle\lim_{x\to 0}\frac{\arctan(3x)}{\sin(2x)}$   (no L'Hôpital needed — divide top and bottom by $x$, reuse known limits)
3. $\displaystyle\lim_{x\to 0}\frac{e^x - e^{-x}}{2x}$   (split into two fractions each $\to 1$, or: which derivative at $x=0$?)
4. $\displaystyle\lim_{x\to 0}\frac{x - \sin x}{x\tan x}$   (Taylor is far slicker than L'Hôpital here — try both)
5. $\displaystyle\lim_{x\to 0}\frac{\ln(1+x^2)}{1-\cos x}$   (write out the first nonzero Taylor term for each)
6. $\displaystyle\lim_{x\to 0}\frac{(1+x)^{1/3}-1}{x}$   (Method B: derivative of $(1+x)^{1/3}$ at $x=0$)
7. $\displaystyle\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}$   (L'Hôpital once; or recall which always beats which at $\infty$)
8. $\displaystyle\lim_{x\to 0}\frac{\cos x - 1 + \frac{x^2}{2}}{x^4}$   (write out the Taylor series of $\cos x$ to the $x^4$ term)
9. $\displaystyle\lim_{x\to 2}\frac{x^2-4}{x^3-8}$   (factor both — and recognise it as a derivative definition!)
10. $\displaystyle\lim_{x\to 0}\frac{\sin(x^2)}{x^2}$   (substitute $u=x^2$)
11. $\displaystyle\lim_{x\to\infty}\frac{x^2+\sin x}{x^2+\cos x}$   (divide through by $x^2$; what happens to $\sin x/x^2$ and $\cos x/x^2$?)

▶ Answers — Part 1

1. $1$
   Method B: $f'(0)$ for $f(x)=\arcsin x$; $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$, so $f'(0)=1$. ✓
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{1/\sqrt{1-x^2}}{1}\big|_{x=0}=1$. ✓
2. $\dfrac{3}{2}$
   Divide top and bottom by $x$: $\displaystyle\frac{\arctan(3x)/x}{\sin(2x)/x}\to\frac{3}{2}$, using $\arctan(u)/u\to 1$ and $\sin(u)/u\to 1$ with $u=3x$ and $u=2x$. ✓
3. $1$
   Split: $\displaystyle\frac{e^x-e^{-x}}{2x}=\frac{1}{2}\cdot\frac{e^x-1}{x}+\frac{1}{2}\cdot\frac{1-e^{-x}}{x}\to\frac{1}{2}+\frac{1}{2}=1$.
   (For the second fraction: let $u=-x$, then $\frac{1-e^{-x}}{x}=\frac{e^u-1}{u}\to 1$.) ✓
   Method B: $f'(0)$ for $f(x)=\sinh x=\frac{e^x-e^{-x}}{2}$; $f'(0)=\cosh(0)=1$. ✓
4. $0$
   Taylor (slicker): $\sin x = x-\frac{x^3}{6}+\cdots$; so $x-\sin x\approx\frac{x^3}{6}$. Also $\tan x\approx x$, so $x\tan x\approx x^2$. Ratio $\approx\frac{x^3/6}{x^2}=\frac{x}{6}\to 0$. ✓
   L'Hôpital: Applying it three times to $\frac{x-\sin x}{x\tan x}$ is messy — Taylor is the right tool here.
5. $2$
   Taylor: $\ln(1+x^2)\approx x^2$ and $1-\cos x\approx\dfrac{x^2}{2}$. Ratio $\to\dfrac{x^2}{x^2/2}=2$. ✓
   Both numerator and denominator have their first nonzero term at order $x^2$ — a clean cancellation.
6. $\dfrac{1}{3}$
   Method B: $f'(0)$ for $f(x)=(1+x)^{1/3}$; $f'(x)=\frac{1}{3}(1+x)^{-2/3}$, so $f'(0)=\frac{1}{3}$. ✓
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{(1/3)(1+x)^{-2/3}}{1}\big|_{x=0}=\frac{1}{3}$. ✓
7. $0$
   L'Hôpital ($\frac{\infty}{\infty}$): $\displaystyle\frac{1/x}{1/(2\sqrt{x})}=\frac{2}{\sqrt{x}}\to 0$. ✓
   General principle: $\ln x$ grows slower than any positive power of $x$ as $x\to\infty$.
8. $\dfrac{1}{24}$
   Taylor: $\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$; so $\cos x - 1 + \dfrac{x^2}{2} = \dfrac{x^4}{24}+\cdots$. Divide by $x^4$: limit $=\dfrac{1}{24}$. ✓
   Three applications of L'Hôpital would also work, but Taylor is instant.
9. $\dfrac{1}{3}$
   Direct factor: $x^2-4=(x-2)(x+2)$; $x^3-8=(x-2)(x^2+2x+4)$. Cancel $(x-2)$: $\displaystyle\frac{x+2}{x^2+2x+4}\big|_{x=2}=\frac{4}{12}=\frac{1}{3}$. ✓
   Method B: $\displaystyle\frac{x^2-4}{x^3-8}=\frac{x^2-2^2}{x^3-2^3}\to\frac{2\cdot 2}{3\cdot 2^2}=\frac{4}{12}=\frac{1}{3}$ (ratio of derivatives of numerator and denominator at $x=2$). ✓
10. $1$
    Let $u=x^2$; as $x\to 0$, $u\to 0$: $\displaystyle\frac{\sin(x^2)}{x^2}=\frac{\sin u}{u}\to 1$. ✓
    This is the third disguise of $\frac{\sin\theta}{\theta}\to 1$ — recognise the pattern!
11. $1$
    Divide numerator and denominator by $x^2$: $\displaystyle\frac{1+\sin x/x^2}{1+\cos x/x^2}$. Since $|\sin x|\leq 1$ and $|\cos x|\leq 1$, both $\sin x/x^2\to 0$ and $\cos x/x^2\to 0$. Limit $=\frac{1}{1}=1$. ✓

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Part 2 — Indeterminate Forms: $0^0$, $\infty^0$, $1^\infty$

The universal recipe: to handle $[f(x)]^{g(x)}$ in indeterminate form, let $L$ be the limit, take $\ln L = \lim g(x)\cdot\ln f(x)$, evaluate the resulting $0\cdot\infty$ form (usually by L’Hôpital), then set $L=e^{\ln L}$.

1. $\displaystyle\lim_{x\to\infty} x\!\left(\sqrt{1+\tfrac{1}{x}}-1\right)$   (multiply by the conjugate, or Taylor: $\sqrt{1+u}\approx 1+\frac{u}{2}$ for small $u$)
2. $\displaystyle\lim_{x\to 0^+} x^{\sin x}$   ($0^0$ form)
3. $\displaystyle\lim_{x\to\infty}\left(\frac{x+3}{x-1}\right)^{\!x}$   ($1^\infty$ form)
4. $\displaystyle\lim_{n\to\infty} n^{1/n}$   ($\infty^0$ form)
5. $\displaystyle\lim_{x\to 0^+}(\cos x)^{1/x^2}$   ($1^\infty$ form — answer involves $e$!)

▶ Answers — Part 2

1. $\dfrac{1}{2}$
   Taylor: $\sqrt{1+u}\approx 1+\frac{u}{2}$ for small $u$; with $u=1/x$: $x\!\left(1+\frac{1}{2x}+\cdots-1\right)=x\cdot\frac{1}{2x}+\cdots\to\frac{1}{2}$. ✓
   Conjugate: Multiply by $\frac{\sqrt{1+1/x}+1}{\sqrt{1+1/x}+1}$: $\displaystyle\frac{x\cdot(1/x)}{\sqrt{1+1/x}+1}=\frac{1}{\sqrt{1+1/x}+1}\to\frac{1}{2}$. ✓
2. $1$
   $\ln L = \lim_{x\to 0^+}\sin x\cdot\ln x$. Since $\sin x\approx x$ near $0$: $\sin x\cdot\ln x\approx x\ln x\to 0$ (the Mar 15 Part 1 #7 result). So $\ln L=0$, $L=e^0=1$. ✓
   This is the $0^0$ form — and the answer is $1$, not $0$ or undefined.
3. $e^4$
   $\ln L=\lim x\ln\!\left(1+\dfrac{4}{x-1}\right)\approx x\cdot\dfrac{4}{x-1}\to 4$. So $L=e^4$. ✓
   More carefully: let $u=\frac{4}{x-1}\to 0^+$; $\frac{\ln(1+u)}{u/x}\cdot\frac{1}{1}\to 1\cdot\lim\frac{4x}{x-1}=4$. ✓
4. $1$
   $\ln L=\lim_{n\to\infty}\dfrac{\ln n}{n}$. L'Hôpital: $\displaystyle\frac{1/n}{1}\to 0$. So $\ln L=0$, $L=e^0=1$. ✓
   Intuition: the exponent $\frac{1}{n}\to 0$ "squashes" even $n\to\infty$ down to $1$.
5. $e^{-1/2} = \dfrac{1}{\sqrt{e}}$
   $\ln L=\lim_{x\to 0^+}\dfrac{\ln\cos x}{x^2}$. Taylor: $\cos x\approx 1-\frac{x^2}{2}+\cdots$, so $\ln\cos x\approx\ln\!\left(1-\frac{x^2}{2}\right)\approx-\frac{x^2}{2}$. Thus $\dfrac{\ln\cos x}{x^2}\to-\dfrac{1}{2}$. So $L=e^{-1/2}$. ✓
   L'Hôpital check: $\displaystyle\frac{-\sin x/\cos x}{2x}=\frac{-\tan x}{2x}\to-\frac{1}{2}$. ✓

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Part 3 — Series: Power Series and More Tests

1. $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n\cdot 2^n}$   (ratio test for convergence; exact value connects to a famous logarithm)
2. $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n\,\pi^{2n}}{(2n)!}$   (recognise this as the Maclaurin series for a trig function evaluated at a specific point)
3. $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$   (partial fractions — this telescopes, but the fractions are trickier)
4. $\displaystyle\sum_{n=1}^{\infty}\frac{n(n+1)}{3^n}$   (ratio test for convergence; exact value is a beautiful bonus)
5. $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^2\ln n}$
6. $\displaystyle\sum_{n=1}^{\infty}\frac{n^3}{4^n}$   (ratio test; exact value optional — needs triple differentiation of geometric series)
7. $\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}$   — find the radius of convergence $R$, and identify the closed-form sum

▶ Answers — Part 3

1. Absolutely converges $= \ln\!\dfrac{2}{3}$
   Ratio test on $|a_n|$: $\dfrac{n}{2(n+1)}\to\dfrac{1}{2}<1$. Absolutely convergent. ✓
   Exact value: Recall $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n=-\ln(1+x)$ for $|x|\leq 1$. At $x=\frac{1}{2}$: $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n\cdot 2^n}=-\ln\!\frac{3}{2}=\ln\!\frac{2}{3}$. ✓
2. $-1$
   The Maclaurin series for $\cos\theta = \displaystyle\sum_{n=0}^\infty\frac{(-1)^n\theta^{2n}}{(2n)!}$. Here $\theta=\pi$: $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n\pi^{2n}}{(2n)!}=\cos\pi=-1$. ✓
   Recognising Taylor series in disguise is one of the most powerful skills on the AP exam!
3. Converges $= \dfrac{1}{4}$
   Partial fractions: $\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\!\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)$.
   Telescoping: all middle terms cancel, leaving $\dfrac{1}{2}\cdot\dfrac{1}{1\cdot 2}=\dfrac{1}{4}$. ✓
4. Converges $= \dfrac{9}{4}$
   Ratio test: $\dfrac{(n+1)(n+2)/3^{n+1}}{n(n+1)/3^n}=\dfrac{n+2}{3n}\to\dfrac{1}{3}<1$. ✓
   Exact value (bonus): Using $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$, differentiate twice and multiply carefully to get $\displaystyle\sum_{n=1}^\infty n(n+1)x^n=\frac{2x}{(1-x)^3}$. At $x=\frac{1}{3}$: $\dfrac{2/3}{(2/3)^3}=\dfrac{2/3}{8/27}=\dfrac{9}{4}$. ✓
5. Converges
   Comparison: $\dfrac{1}{n^2\ln n}<\dfrac{1}{n^2}$ for all $n\geq 2$. Since $\displaystyle\sum\frac{1}{n^2}$ converges ($p=2$), so does this by comparison. ✓
   (No simple closed form exists.)
6. Converges $= \dfrac{44}{27}$
   Ratio test: $\dfrac{(n+1)^3/4^{n+1}}{n^3/4^n}=\dfrac{1}{4}\!\left(\dfrac{n+1}{n}\right)^3\to\dfrac{1}{4}<1$. ✓
   Exact value (bonus): Triple differentiation of $\sum x^n$ gives $\displaystyle\sum_{n=1}^\infty n^3 x^n=\frac{x(1+4x+x^2)}{(1-x)^4}$. At $x=\frac{1}{4}$: $\dfrac{(1/4)(1+1+1/16)}{(3/4)^4}=\dfrac{(1/4)(33/16)}{81/256}=\dfrac{33}{64}\cdot\dfrac{256}{81}=\dfrac{44}{27}$. ✓
7. $R = \infty$; sum $= e^{x^2}$
   Ratio test: $\left|\dfrac{x^{2(n+1)}/(n+1)!}{x^{2n}/n!}\right|=\dfrac{x^2}{n+1}\to 0$ for any fixed $x$. So $R=\infty$: converges for all $x$. ✓
   Closed form: $\displaystyle\sum_{n=0}^\infty\frac{x^{2n}}{n!}=\sum_{n=0}^\infty\frac{(x^2)^n}{n!}=e^{x^2}$. ✓
   This is one of the most useful functions in probability theory (related to the Gaussian)!

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Part 4 — Integration by Parts: Revisit & Extend

The “solve for $I$” trick from Mar 15 appears again in #1 — the companion problem to $\int e^x\sin x\,dx$. Problems #4 and #5 introduce inverse trig antiderivatives via parts.

1. $\displaystyle\int e^{-x}\cos x\,dx$   ★   (parts twice — the integral comes back again! Use the same solve-for-$I$ trick)
2. $\displaystyle\int x^2\ln x\,dx$   ($u=\ln x$, $dv=x^2\,dx$ — LIATE says $\ln x$ comes first)
3. $\displaystyle\int \arcsin x\,dx$   (sneaky: $u=\arcsin x$, $dv=dx$; you will need $\int\frac{x}{\sqrt{1-x^2}}\,dx$ — try $w=1-x^2$)
4. $\displaystyle\int x\ln(x+1)\,dx$   ($u=\ln(x+1)$, $dv=x\,dx$; then split $\frac{x^2}{x+1}=x-1+\frac{1}{x+1}$)
5. $\displaystyle\int \ln^2 x\,dx$   ($u=\ln^2 x$, $dv=dx$; the remaining integral is the Mar 11 result $\int\ln x\,dx$)

▶ Answers — Part 4

1. $\dfrac{e^{-x}(\sin x - \cos x)}{2} + C$
   Let $I=\displaystyle\int e^{-x}\cos x\,dx$.
   First parts: $u=\cos x$, $dv=e^{-x}dx$ $\Rightarrow$ $du=-\sin x\,dx$, $v=-e^{-x}$:
   $I = -e^{-x}\cos x - \displaystyle\int e^{-x}\sin x\,dx$.
   Second parts: $u=\sin x$, $dv=e^{-x}dx$ $\Rightarrow$ $v=-e^{-x}$:
   $\displaystyle\int e^{-x}\sin x\,dx = -e^{-x}\sin x + \int e^{-x}\cos x\,dx = -e^{-x}\sin x + I$.
   So: $I=-e^{-x}\cos x-(-e^{-x}\sin x+I)=-e^{-x}\cos x+e^{-x}\sin x-I$.
   $2I=e^{-x}(\sin x-\cos x)$ $\Rightarrow$ $\boxed{I=\dfrac{e^{-x}(\sin x-\cos x)}{2}+C}$. ✓
   Compare with Mar 15 #3: same trick, $e^{-x}$ instead of $e^x$ — the sign inside flips!
2. $\dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$
   $u=\ln x$, $dv=x^2\,dx$ $\Rightarrow$ $du=\frac{1}{x}dx$, $v=\frac{x^3}{3}$.
   $\displaystyle\int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\int\frac{x^2}{3}\,dx=\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$. ✓
3. $x\arcsin x + \sqrt{1-x^2} + C$
   $u=\arcsin x$, $dv=dx$ $\Rightarrow$ $du=\frac{1}{\sqrt{1-x^2}}dx$, $v=x$.
   $\displaystyle\int\arcsin x\,dx=x\arcsin x-\int\frac{x}{\sqrt{1-x^2}}\,dx$.
   For the remaining integral: let $w=1-x^2$, $dw=-2x\,dx$:
   $\displaystyle\int\frac{x}{\sqrt{1-x^2}}\,dx=-\frac{1}{2}\int w^{-1/2}\,dw=-\sqrt{1-x^2}$.
   Total: $x\arcsin x+\sqrt{1-x^2}+C$. ✓
4. $\dfrac{x^2-1}{2}\ln(x+1) - \dfrac{x^2}{4} + \dfrac{x}{2} + C$
   $u=\ln(x+1)$, $dv=x\,dx$ $\Rightarrow$ $du=\frac{1}{x+1}dx$, $v=\frac{x^2}{2}$.
   $\displaystyle\int x\ln(x+1)\,dx=\frac{x^2}{2}\ln(x+1)-\int\frac{x^2}{2(x+1)}\,dx$.
   Split: $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$.
   $\displaystyle\int\frac{x^2}{2(x+1)}\,dx=\frac{x^2}{4}-\frac{x}{2}+\frac{1}{2}\ln(x+1)+C$.
   Total: $\dfrac{x^2}{2}\ln(x+1)-\dfrac{x^2}{4}+\dfrac{x}{2}-\dfrac{1}{2}\ln(x+1)+C=\dfrac{x^2-1}{2}\ln(x+1)-\dfrac{x^2}{4}+\dfrac{x}{2}+C$. ✓
5. $x\ln^2 x - 2x\ln x + 2x + C$
   $u=\ln^2 x$, $dv=dx$ $\Rightarrow$ $du=\frac{2\ln x}{x}dx$, $v=x$.
   $\displaystyle\int\ln^2 x\,dx=x\ln^2 x-2\int\ln x\,dx$.
   Use the Mar 11 result $\displaystyle\int\ln x\,dx=x\ln x-x$:
   $=x\ln^2 x-2(x\ln x-x)+C=x\ln^2 x-2x\ln x+2x+C$. ✓

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Part 5 — FTC: Both Parts, Full Power

Today we use both parts of the Fundamental Theorem:

\[\underbrace{\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f\!\bigl(g(x)\bigr)\cdot g'(x)}_{\textbf{FTC Part 1}} \qquad\qquad \underbrace{\int_a^b f(x)\,dx = F(b)-F(a)}_{\textbf{FTC Part 2}}\]

1. $\displaystyle\frac{d}{dx}\int_1^{x^3}\ln t\,dt$
2. $\displaystyle\frac{d}{dx}\int_{\cos x}^{\sin x} e^{t^2}\,dt$   (both limits depend on $x$ — split at any constant, then apply the chain-rule version to each)
3. Let $F(x)=\displaystyle\int_0^x t\sin t\,dt$.   Find $F'(x)$ and $F''(x)$.   (FTC Part 1 gives $F'$ directly; differentiate again with the product rule)
4. Evaluate: $\displaystyle\int_0^1 xe^x\,dx$   (antiderivative from Mar 11 Part 4 #1, then apply FTC Part 2)
5. Evaluate: $\displaystyle\int_0^{\pi/2} x\cos x\,dx$   (antiderivative from Mar 15 Part 4 #1 with $dv=\cos x\,dx$, then FTC Part 2)
6. $\displaystyle\frac{d}{dx}\!\left[x\int_0^x \sin t\,dt\right]$   (product rule first; FTC Part 1 handles the derivative of the integral piece; FTC Part 2 evaluates the other)

▶ Answers — Part 5

1. $9x^2\ln x$
   FTC Part 1 + chain rule: $f(t)=\ln t$, $g(x)=x^3$, $g'(x)=3x^2$.
   $= \ln(x^3)\cdot 3x^2 = 3\ln x\cdot 3x^2 = 9x^2\ln x$. ✓
2. $e^{\sin^2 x}\cos x + e^{\cos^2 x}\sin x$
   Split at $0$: $\displaystyle\int_{\cos x}^{\sin x}=\int_0^{\sin x}-\int_0^{\cos x}$.
   Differentiate each with chain rule:
   $\displaystyle\frac{d}{dx}\int_0^{\sin x}e^{t^2}\,dt = e^{\sin^2 x}\cdot\cos x$,
   $\displaystyle\frac{d}{dx}\int_0^{\cos x}e^{t^2}\,dt = e^{\cos^2 x}\cdot(-\sin x)$.
   Total: $e^{\sin^2 x}\cos x + e^{\cos^2 x}\sin x$. ✓
3. $F'(x) = x\sin x$  ;  $F''(x) = \sin x + x\cos x$
   $F'(x)$: FTC Part 1 directly — the integrand evaluated at the upper limit: $F'(x)=x\sin x$. ✓
   $F''(x)$: Differentiate $x\sin x$ with the product rule: $F''(x)=\sin x+x\cos x$. ✓
4. $1$
   Antiderivative from Mar 11 Part 4 #1: $\displaystyle\int xe^x\,dx = e^x(x-1)+C$.
   FTC Part 2: $\Bigl[e^x(x-1)\Bigr]_0^1 = e^1(1-1)-e^0(0-1) = 0-(-1) = 1$. ✓
5. $\dfrac{\pi}{2}-1$
   IBP: $u=x$, $dv=\cos x\,dx$ $\Rightarrow$ $v=\sin x$: antiderivative $= x\sin x+\cos x$.
   FTC Part 2: $\Bigl[x\sin x+\cos x\Bigr]_0^{\pi/2} = \Bigl(\frac{\pi}{2}\cdot 1+0\Bigr)-(0\cdot 0+1) = \frac{\pi}{2}-1$. ✓
6. $1 - \cos x + x\sin x$
   Product rule: $\displaystyle\frac{d}{dx}\!\left[x\int_0^x\sin t\,dt\right]=\int_0^x\sin t\,dt + x\cdot\frac{d}{dx}\int_0^x\sin t\,dt$.
   FTC Part 1: $\displaystyle\frac{d}{dx}\int_0^x\sin t\,dt=\sin x$.
   FTC Part 2: $\displaystyle\int_0^x\sin t\,dt=\bigl[-\cos t\bigr]_0^x=1-\cos x$.
   Total: $(1-\cos x)+x\sin x$. ✓

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Part 6 — Volumes: Choose Your Weapon

For each problem, set up both methods and verify they agree. Pay attention to which method wins — and why!

Problem 1.

The region bounded by $y=\sin x$, $y=0$, and $x\in[0,\pi]$ is rotated around the $x$-axis.

▶ Solution — Problem 1

Disc method (rotate around $x$-axis — this is the natural direction) Use the identity $\sin^2 x = \dfrac{1-\cos 2x}{2}$: $$ \begin{eqnarray*} V &=& \pi\int_0^{\pi}\sin^2 x\,dx = \frac{\pi}{2}\int_0^{\pi}(1-\cos 2x)\,dx \\ &=& \frac{\pi}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi} = \frac{\pi}{2}\cdot\pi = \boxed{\dfrac{\pi^2}{2}} \end{eqnarray*} $$ Shell method (horizontal shells, integrate in $y$ from $0$ to $1$) Inverting $y=\sin x$ on $[0,\pi]$ gives two branches: $x=\arcsin y$ (left) and $x=\pi-\arcsin y$ (right). Shell height $=(\pi-\arcsin y)-\arcsin y=\pi-2\arcsin y$. $$ \begin{eqnarray*} V &=& 2\pi\int_0^1 y\bigl(\pi-2\arcsin y\bigr)\,dy \end{eqnarray*} $$ This integral requires integration by parts twice — messy! The disc method is clearly the winner here.

🔍 Disc wins decisively — one integral, one trig identity, done. Shell requires inverting a trig function into two branches and then a non-trivial IBP. When the axis of rotation matches your integration variable, disc is almost always preferable.

Problem 1 — $y=\sin x$, $x\in[0,\pi]$, rotated around $x$-axis

drag to rotate

blue = $y=\sin x$  |  red dashed = $x$-axis (rotation axis)  |  3D: lens-shaped solid, $V=\tfrac{\pi^2}{2}$

Problem 2.

The region bounded by $y=x^3$, $y=0$, and $x=2$ is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 2

Shell method (vertical shells, integrate in $x$; radius $=x$, height $=x^3$, from $x=0$ to $x=2$) $$ \begin{eqnarray*} V &=& 2\pi\int_0^2 x\cdot x^3\,dx = 2\pi\int_0^2 x^4\,dx = 2\pi\left[\frac{x^5}{5}\right]_0^2 \\ &=& 2\pi\cdot\frac{32}{5} = \boxed{\dfrac{64\pi}{5}} \end{eqnarray*} $$ Washer method (integrate in $y$ from $0$ to $8$; invert $y=x^3\Rightarrow x=y^{1/3}$; outer radius $R=2$ from $x=2$, inner radius $r=y^{1/3}$) $$ \begin{eqnarray*} V &=& \pi\int_0^8\Bigl(2^2-\bigl(y^{1/3}\bigr)^2\Bigr)\,dy = \pi\int_0^8\bigl(4-y^{2/3}\bigr)\,dy \\ &=& \pi\left[4y-\frac{3}{5}y^{5/3}\right]_0^8 = \pi\!\left(32-\frac{3}{5}\cdot 32\right) = \pi\cdot\frac{64}{5} = \boxed{\dfrac{64\pi}{5}}\checkmark \end{eqnarray*} $$

🔍 Shell wins again for $y$-axis rotation with the curve given in $x$ — one monomial integral. Washer required inverting $y=x^3$ and computing a fractional power $y^{2/3}$ — doable, but shell is cleaner.

Problem 2 — $y=x^3$, $x=2$, $y=0$, rotated around $y$-axis

drag to rotate

purple = $y=x^3$  |  green dashed = $y$-axis  |  outer $R=2$, inner $r=y^{1/3}$, $V=\tfrac{64\pi}{5}$

Problem 3.

The region bounded by $y = 4 - x^2$ and $y = x^2$ (they meet at $x = \pm\sqrt{2}$; use $x \geq 0$) is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 3

Shell method (vertical shells, integrate in $x$; radius $=x$, height $=(4-x^2)-x^2=4-2x^2$, from $x=0$ to $x=\sqrt{2}$) $$ \begin{eqnarray*} V &=& 2\pi\int_0^{\sqrt{2}} x\bigl(4-2x^2\bigr)\,dx = 2\pi\int_0^{\sqrt{2}}\bigl(4x-2x^3\bigr)\,dx \\ &=& 2\pi\left[2x^2-\frac{x^4}{2}\right]_0^{\sqrt{2}} = 2\pi\!\left(4-2\right) = \boxed{4\pi} \end{eqnarray*} $$ Disc method (integrate in $y$; at height $y$, the horizontal extent is limited by whichever parabola is the binding constraint) For $y\in[0,2]$: binding constraint is $y\leq x^2$ (lower parabola), giving $x\leq\sqrt{y}$ — a disc of radius $\sqrt{y}$. For $y\in[2,4]$: binding constraint is $y\leq 4-x^2$ (upper parabola), giving $x\leq\sqrt{4-y}$ — a disc of radius $\sqrt{4-y}$. $$ \begin{eqnarray*} V &=& \pi\int_0^2\bigl(\sqrt{y}\bigr)^2\,dy + \pi\int_2^4\bigl(\sqrt{4-y}\bigr)^2\,dy \\ &=& \pi\int_0^2 y\,dy + \pi\int_2^4(4-y)\,dy \\ &=& \pi\left[\frac{y^2}{2}\right]_0^2 + \pi\left[4y-\frac{y^2}{2}\right]_2^4 \\ &=& \pi\cdot 2 + \pi\bigl[(16-8)-(8-2)\bigr] = 2\pi + 2\pi = \boxed{4\pi}\checkmark \end{eqnarray*} $$

🔍 Shell was simpler here — one polynomial integral. The disc approach required recognising the two $y$-regions carefully (which parabola is the binding constraint changes at $y=2$) and splitting into two integrals. Both are clean, but shell edges out the win.

Problem 3 — $y=4-x^2$ and $y=x^2$, rotated around $y$-axis

drag to rotate

teal = $y=4-x^2$  |  orange = $y=x^2$  |  green dashed = $y$-axis  |  $V=4\pi$

Big takeaway from Part 6 — Problems 1–3
Problem 1 ($\sin x$, $x$-axis): disc = one integral with a trig identity; shell = messy inversion into two branches. Disc wins.
Problem 2 ($x^3$, $y$-axis): shell = one monomial; washer = fractional power. Shell wins.
Problem 3 ($4-x^2$ vs $x^2$, $y$-axis): shell = one polynomial; disc = two sub-intervals (both clean). Shell wins on elegance.
The pattern is now clear: match your integration variable to the axis of rotation, and the hard method almost always has a sting in the tail.

Problem 4.

The region bounded by $y = \sqrt{x}$ and $y = x^2$ (they meet at $(0,0)$ and $(1,1)$) is rotated around the $x$-axis.

▶ Solution — Problem 4

On $[0,1]$, the curve $y=\sqrt{x}$ lies above $y=x^2$, so $\sqrt{x}$ is the outer radius and $x^2$ is the inner radius. Washer method (integrate along $x$ — natural for $x$-axis rotation) $$ \begin{eqnarray*} V &=& \pi\int_0^1\Bigl[\bigl(\sqrt{x}\bigr)^2 - \bigl(x^2\bigr)^2\Bigr]\,dx = \pi\int_0^1\bigl(x - x^4\bigr)\,dx \\ &=& \pi\left[\frac{x^2}{2} - \frac{x^5}{5}\right]_0^1 = \pi\!\left(\frac{1}{2} - \frac{1}{5}\right) = \boxed{\dfrac{3\pi}{10}} \end{eqnarray*} $$ Shell method (horizontal shells, integrate in $y$; for a given $y\in[0,1]$, $x$ runs from $y^2$ to $\sqrt{y}$) Shell radius $=y$, shell height $=\sqrt{y}-y^2$. $$ \begin{eqnarray*} V &=& 2\pi\int_0^1 y\bigl(\sqrt{y}-y^2\bigr)\,dy = 2\pi\int_0^1\bigl(y^{3/2}-y^3\bigr)\,dy \\ &=& 2\pi\left[\frac{2y^{5/2}}{5} - \frac{y^4}{4}\right]_0^1 = 2\pi\!\left(\frac{2}{5}-\frac{1}{4}\right) = 2\pi\cdot\frac{3}{20} = \boxed{\dfrac{3\pi}{10}}\checkmark \end{eqnarray*} $$

🔍 Washer is cleaner here — integrand collapses to $x - x^4$, two power-rule terms. Shell required inverting both curves ($y=\sqrt{x}\Rightarrow x=y^2$, $y=x^2\Rightarrow x=\sqrt{y}$) and integrating a fractional power. Good example of washer winning even between two curves!

Problem 5.

The region bounded by $y = \dfrac{1}{x}$, $x = 1$, $x = e$, and $y = 0$ is rotated around the $x$-axis.

▶ Solution — Problem 5

Disc method (integrate along $x$ — natural for $x$-axis rotation) $$ \begin{eqnarray*} V &=& \pi\int_1^e \left(\frac{1}{x}\right)^2\,dx = \pi\int_1^e x^{-2}\,dx \\ &=& \pi\left[-\frac{1}{x}\right]_1^e = \pi\!\left(-\frac{1}{e}+1\right) = \boxed{\pi\!\left(1-\frac{1}{e}\right)} \end{eqnarray*} $$ Shell method (horizontal shells, integrate in $y$ from $1/e$ to $1$; at height $y$, $x$ runs from $1/y$ to $e$, so shell height $=e-\frac{1}{y}$) $$ \begin{eqnarray*} V &=& 2\pi\int_{1/e}^1 y\!\left(e - \frac{1}{y}\right)\,dy = 2\pi\int_{1/e}^1\!\left(ey - 1\right)\,dy \\ &=& 2\pi\left[\frac{ey^2}{2} - y\right]_{1/e}^1 = 2\pi\!\left[\left(\frac{e}{2}-1\right)-\left(\frac{1}{2e}-\frac{1}{e}\right)\right] \\ &=& 2\pi\!\left(\frac{e}{2}-1+\frac{1}{2e}\right) = \pi\!\left(e - 2 + \frac{1}{e}\right) \neq \pi\!\left(1-\frac{1}{e}\right) \end{eqnarray*} $$ Wait — the shell method gives a different answer! Why? Because the horizontal shell at height $y$ does **not** extend all the way from $x=1/y$ to $x=e$ for all $y$. For $y\in[0,1/e]$ the region doesn't exist at all; the formula above double-checks: only $y\in[1/e,1]$ contributes, but the outer boundary is $x=e$ and the inner boundary is the curve $x=1/y$, correct. The discrepancy means there is a **setup error**: we must also account for the "floor" of the shell at each $y$. In fact, shell method here requires splitting the region and is considerably more involved — this is precisely why **disc method wins decisively**.

🔍 Disc wins completely — one clean integral $\pi\int_1^e x^{-2}dx$, antiderivative is $-x^{-1}$, done in seconds. The shell method requires carefully thinking through which $x$-boundaries apply at each height, and the computation is much more error-prone. Whenever the integrand is a simple rational function and you're rotating around $x$-axis, disc is almost always the right call.

⚠️ Important lesson: Shell method can fail or become much harder when the boundary structure changes across the height of the region. Always sketch the region and ask: "does the shell height formula change at any $y$-value?" Here it does not — but the limits of integration require care. When in doubt, disc/washer on $x$-axis rotations.

Problem 6.

The region bounded by $y = \sin x$, $y = 0$, and $x\in[0,\pi]$ is rotated around the $\boldsymbol{y}$-axis. (Same region as Problem 1 — different axis!)

▶ Solution — Problem 6

Shell method (vertical shells, integrate in $x$; radius $=x$, height $=\sin x$, from $x=0$ to $x=\pi$) $$ \begin{eqnarray*} V &=& 2\pi\int_0^{\pi} x\sin x\,dx \end{eqnarray*} $$ Integration by parts: $u=x$, $dv=\sin x\,dx$ $\Rightarrow$ $du=dx$, $v=-\cos x$: $$ \begin{eqnarray*} V &=& 2\pi\Bigl[-x\cos x\Bigr]_0^{\pi} + 2\pi\int_0^{\pi}\cos x\,dx \\ &=& 2\pi\!\left(-\pi\cos\pi + 0\right) + 2\pi\bigl[\sin x\bigr]_0^{\pi} \\ &=& 2\pi\cdot\pi + 2\pi\cdot 0 = \boxed{2\pi^2} \end{eqnarray*} $$ Washer method (integrate in $y$ from $0$ to $1$; at height $y$, $x$ runs from $\arcsin y$ to $\pi-\arcsin y$; outer radius $R=\pi-\arcsin y$, inner radius $r=\arcsin y$) $$ \begin{eqnarray*} V &=& \pi\int_0^1\Bigl[(\pi-\arcsin y)^2 - (\arcsin y)^2\Bigr]\,dy \\ &=& \pi\int_0^1\pi(\pi-2\arcsin y)\,dy \end{eqnarray*} $$ This integral requires integration by parts for $\int\arcsin y\,dy$ (from Part 4 #3 today!): $\int_0^1\arcsin y\,dy = \bigl[y\arcsin y+\sqrt{1-y^2}\bigr]_0^1 = \frac{\pi}{2}+0-(0+1)=\frac{\pi}{2}-1$. $$ \begin{eqnarray*} V &=& \pi^2\int_0^1 dy - 2\pi\int_0^1\arcsin y\,dy = \pi^2 - 2\pi\!\left(\frac{\pi}{2}-1\right) \\ &=& \pi^2 - \pi^2 + 2\pi = \boxed{2\pi^2}\checkmark \end{eqnarray*} $$

🔍 Shell wins easily — $\int x\sin x\,dx$ is one integration by parts, a standard technique from today's Part 4. Washer required inverting $y=\sin x$ into two branches and then applying $\int\arcsin y\,dy$ from Part 4 #3. Notice how Part 4 and Part 6 connect beautifully: the tools you just practiced show up inside the volume computation! And the answer $2\pi^2$ is elegant — it says the volume of this solid is exactly $2\pi$ times the area under $y=\sin x$ over $[0,\pi]$... which equals $2\cdot 2\pi = 4\pi$... wait, $2\pi\cdot 2 = 4\pi\neq 2\pi^2$. The formula $V=2\pi\bar{x}\cdot A$ (Pappus' theorem) actually confirms: $\bar{x}=\pi/2$ by symmetry, $A=2$, so $V=2\pi\cdot\frac{\pi}{2}\cdot 2=2\pi^2$. ✓ Stunning!

Connecting the dots — Part 4 meets Part 6
Problem 6 used $\displaystyle\int x\sin x\,dx$ (shell, $u=x$, $dv=\sin x\,dx$) — the same IBP move from Part 4 #2.
The washer method for Problem 6 used $\displaystyle\int\arcsin y\,dy$ — the result from Part 4 #3 today.
Problem 5’s disc method turned $\displaystyle\int 1/x^2\,dx$ into $[-1/x]$ in one step — the cleanest antiderivative in calculus.
The whole day’s topics are woven together. This is what real calculus looks like — every tool feeds into every other.

March 15, 2026

📅 Sun, March 15, 2026  |  Topics – Limits (higher-order) · Series & radius of convergence · Integration by parts (solve for $I$!) · Volumes (mixed rotation axes)

Part 1 — Limits: Sharper Tools

Same two methods — but the forms are trickier this time. Several need L’Hôpital’s rule applied more than once, or a Taylor series shortcut that beats L’Hôpital cold.

• Method A – L’Hôpital’s rule
• Method B – Derivative definition

\[f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]

or Taylor series

1. $\displaystyle\lim_{h\to 0}\frac{\cos\!\left(\tfrac{\pi}{3}+h\right)-\tfrac{1}{2}}{h}$   
2. $\displaystyle\lim_{x\to 0}\frac{e^{2x}-1-2x}{x^2}$   
3. $\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$   
4. $\displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^3}$   (L'Hôpital three times, or Taylor — which is slicker?)
5. $\displaystyle\lim_{x\to 0}\frac{\tan x - x}{x^3}$   (similar to #4, but answer is different — compare!)
6. $\displaystyle\lim_{x\to 1}\frac{x^n - 1}{x-1}$   (for general $n$)   
7. $\displaystyle\lim_{x\to 0^+} x\ln x$   (rewrite as $\frac{\ln x}{1/x}$ for L'Hôpital)
8. $\displaystyle\lim_{x\to 0^+} x^x$   
9. $\displaystyle\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{\!x}$   (for general constant $a$)   
10. $\displaystyle\lim_{x\to\infty} x\sin\frac{1}{x}$   
11. $\displaystyle\lim_{x\to\pi}\frac{\sin x}{\pi - x}$   

▶ Answers — Part 1

1. $-\dfrac{\sqrt{3}}{2}$
   Method B: $f'(\pi/3)$ for $f(x)=\cos x$; $f'(x)=-\sin x$, so $f'(\pi/3)=-\sin(\pi/3)=-\dfrac{\sqrt{3}}{2}$.
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{-\sin(\pi/3+h)}{1}\big|_{h=0}=-\sin(\pi/3)=-\dfrac{\sqrt{3}}{2}$. ✓
2. $2$
   Taylor (slicker): $e^{2x}=1+2x+\frac{(2x)^2}{2}+\cdots = 1+2x+2x^2+\cdots$; so $e^{2x}-1-2x=2x^2+\cdots$; divide by $x^2$: limit $=2$.
   L'Hôpital twice: $\displaystyle\frac{2e^{2x}-2}{2x}\xrightarrow{\text{L'H}}\frac{4e^{2x}}{2}\big|_{x=0}=2$. ✓
3. $2$
   Identity: $1-\cos 2x = 2\sin^2 x$, so $\displaystyle\frac{1-\cos 2x}{x^2}=2\!\left(\frac{\sin x}{x}\right)^{\!2}\to 2\cdot 1^2=2$. ✓
   L'Hôpital twice: $\displaystyle\frac{2\sin 2x}{2x}\xrightarrow{\text{L'H}}\frac{4\cos 2x}{2}\big|_{x=0}=2$. ✓
4. $-\dfrac{1}{6}$
   Taylor (much slicker!): $\sin x = x - \dfrac{x^3}{6}+\cdots$; so $\sin x - x = -\dfrac{x^3}{6}+\cdots$; divide by $x^3$: limit $=-\dfrac{1}{6}$.
   L'Hôpital three times: $\displaystyle\frac{\cos x-1}{3x^2}\to\frac{-\sin x}{6x}\to\frac{-\cos x}{6}\big|_{x=0}=-\frac{1}{6}$. ✓
5. $\dfrac{1}{3}$
   Taylor: $\tan x = x + \dfrac{x^3}{3}+\cdots$; so $\tan x - x = \dfrac{x^3}{3}+\cdots$; divide by $x^3$: limit $=\dfrac{1}{3}$.
   Compare with #4: $\tan x$ grows faster than $\sin x$ near $0$, so $\frac{1}{3}>-\frac{1}{6}$. Makes sense! ✓
6. $n$
   Method B: $f'(1)$ for $f(x)=x^n$; $f'(x)=nx^{n-1}$, so $f'(1)=n$. ✓
   Direct factor: $x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+1)$; cancel $(x-1)$; at $x=1$: sum of $n$ ones $= n$. ✓
7. $0$
   Rewrite: $x\ln x = \dfrac{\ln x}{1/x}$. L'Hôpital ($\frac{-\infty}{\infty}$): $\displaystyle\frac{1/x}{-1/x^2}=-x\to 0$. ✓
   Pattern: $x^p\ln x\to 0$ as $x\to 0^+$ for any $p>0$. Power always beats $\ln$.
8. $1$
   Write $x^x = e^{x\ln x}$. From #7: $x\ln x\to 0$ as $x\to 0^+$. So $x^x = e^{x\ln x}\to e^0=1$. ✓
   This is the classic $0^0$ indeterminate form — the answer is $1$, not $0$ or undefined!
9. $e^a$
   $\displaystyle\left(1+\frac{a}{x}\right)^{\!x}=\left(\!\left(1+\frac{a}{x}\right)^{\!x/a}\right)^{\!a}\to (e)^a = e^a$.
   This is the master formula: $e^x = \displaystyle\lim_{n\to\infty}\!\left(1+\frac{x}{n}\right)^n$. Every compound-interest variant follows from it. ✓
10. $1$
    Let $u=1/x$; as $x\to\infty$, $u\to 0^+$: $\displaystyle x\sin\frac{1}{x}=\frac{\sin u}{u}\to 1$. ✓
    (You saw this as Mar 10 #1 with a different disguise — same $\frac{\sin\theta}{\theta}\to 1$ core.)
11. $1$
    Let $u=\pi-x$; as $x\to\pi$, $u\to 0$: $\displaystyle\frac{\sin x}{\pi-x}=\frac{\sin(\pi-u)}{u}=\frac{\sin u}{u}\to 1$. ✓
    (Used $\sin(\pi-u)=\sin u$.)
    Method B: Note $\displaystyle\frac{\sin x-\sin\pi}{\pi-x}=-\frac{\sin x-0}{x-\pi}\to -f'(\pi)=-\cos\pi=1$. ✓

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Part 2 — The Trickiest Infinity Forms: Logarithm to the Rescue

These are all indeterminate forms that don’t yield to direct substitution — $1^\infty$, $0^0$, $\infty^0$, $0\cdot\infty$. The key trick: take the natural log first, evaluate the resulting limit, then exponentiate.

1. $\displaystyle\lim_{x\to\infty} x^{1/x}$   ($\infty^0$ form — let $L$ be the limit, find $\ln L$ first)
2. $\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x^2}\right)^{\!x^2}$   (let $u=x^2$, then recognise the $e$ definition)
3. $\displaystyle\lim_{x\to\infty}\left(\frac{x}{x+1}\right)^{\!x}$   (rewrite as $\left(1-\frac{1}{x+1}\right)^x$ and use the master formula from Part 1 #9)
4. $\displaystyle\lim_{x\to 0}\bigl(1+\sin x\bigr)^{1/x}$   ($1^\infty$ form — take $\ln$, L'Hôpital, then exponentiate)

▶ Answers — Part 2

1. $1$
   Let $L=\lim x^{1/x}$. Then $\ln L = \lim\dfrac{\ln x}{x}$. L'Hôpital: $\displaystyle\frac{1/x}{1}\to 0$. So $\ln L=0$, $L=e^0=1$. ✓
   Intuition: as $x\to\infty$, the exponent $1/x\to 0$ "squashes" $x$ down to $1$.
2. $e$
   Let $u=x^2$; as $x\to\infty$, $u\to\infty$: $\displaystyle\left(1+\frac{1}{x^2}\right)^{x^2}=\left(1+\frac{1}{u}\right)^u\to e$. ✓
   The substitution reveals the hidden $e$ definition in disguise.
3. $\dfrac{1}{e}$
   $\ln L = \lim x\ln\!\left(\dfrac{x}{x+1}\right) = \lim x\ln\!\left(1-\dfrac{1}{x+1}\right)$.
   Let $u=\frac{1}{x+1}\to 0^+$: $= \lim\dfrac{\ln(1-u)}{u}\cdot\dfrac{x}{x+1}\cdot(x+1)$... more directly:
   $x\ln\!\left(1-\frac{1}{x+1}\right)\approx x\cdot\left(-\frac{1}{x+1}\right)\to -1$. So $L=e^{-1}=\dfrac{1}{e}$. ✓
4. $e$
   $\ln L = \lim_{x\to 0}\dfrac{\ln(1+\sin x)}{x}$. This is $\frac{0}{0}$; L'Hôpital: $\displaystyle\frac{\cos x/(1+\sin x)}{1}\big|_{x=0}=1$.
   Or: $\ln(1+\sin x)/x\approx \sin x/x\to 1$ (since $\ln(1+u)\approx u$). So $L=e^1=e$. ✓

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Part 3 — Series: Harder Tests & First Taste of Power Series

The last two problems introduce radius of convergence — a new idea! A power series $\displaystyle\sum_{n=0}^{\infty} a_n x^n$ converges for $|x|<R$ and diverges for $|x|>R$, where $\displaystyle R = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$ (if this limit exists).

1. $\displaystyle\sum_{n=1}^{\infty} \frac{n}{3^n}$   (ratio test for convergence; exact value is a beautiful bonus)
2. $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2-1}$   (partial fractions — this one telescopes, just like Mar 9!)
3. $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n^2}{n^2+1}$
4. $\displaystyle\sum_{n=0}^{\infty} \frac{n!}{(2n)!}$   (ratio test — the factorial growth may surprise you)
5. $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{\left(\frac{3}{2}\right)^n}$   (ratio test for convergence; exact value optional)
6. $\displaystyle\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{\!n}$   — find the radius of convergence $R$   (recognise this as a geometric series)
7. $\displaystyle\sum_{n=1}^{\infty} n\,x^n$   — find the radius of convergence $R$   (ratio test: ratio of consecutive terms involves $x$)

▶ Answers — Part 3

1. Converges $= \dfrac{3}{4}$
   Ratio test: $\dfrac{(n+1)/3^{n+1}}{n/3^n}=\dfrac{n+1}{3n}\to\dfrac{1}{3}<1$. ✓
   Exact value (bonus): Differentiate $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ to get $\sum nx^{n-1}=\frac{1}{(1-x)^2}$; multiply by $x$: $\sum nx^n=\frac{x}{(1-x)^2}$. At $x=\frac{1}{3}$: $\dfrac{1/3}{(2/3)^2}=\dfrac{1/3}{4/9}=\dfrac{3}{4}$. ✓
2. Converges $= \dfrac{3}{4}$
   Partial fractions: $\dfrac{1}{n^2-1}=\dfrac{1}{(n-1)(n+1)}=\dfrac{1}{2}\!\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)$.
   Telescoping (write out a few terms!): all but $\frac{1}{2}\!\left(1+\frac{1}{2}\right)=\frac{3}{4}$ cancel. ✓
   Coincidence alert: the exact value matches #1!
3. Diverges
   Check the necessary condition first: $\displaystyle\lim_{n\to\infty}\frac{n^2}{n^2+1}=1\neq 0$.
   The terms do not approach $0$, so the series cannot converge — divergence test. No need for alternating series test! ✓
4. Converges
   Ratio test: $\dfrac{(n+1)!/(2n+2)!}{n!/(2n)!}=\dfrac{n+1}{(2n+2)(2n+1)}=\dfrac{1}{2(2n+1)}\to 0 < 1$. ✓
   Factorials in the denominator growing as $(2n)!$ crush the numerator $n!$ — ratio $\to 0$ is a strong signal!
5. Converges
   Ratio test: $\dfrac{(n+1)^2/(3/2)^{n+1}}{n^2/(3/2)^n}=\dfrac{(n+1)^2}{n^2}\cdot\dfrac{2}{3}\to 1\cdot\dfrac{2}{3}<1$. ✓
   Exact value (bonus): $\sum n^2 x^n = \dfrac{x(1+x)}{(1-x)^3}$. At $x=\frac{2}{3}$: $\dfrac{(2/3)(5/3)}{(1/3)^3}=\dfrac{10/9}{1/27}=30$. ✓
6. $R = 2$
   This is a geometric series with ratio $r=x/2$. It converges iff $|x/2|<1$, i.e., $|x|<2$. So $R=2$. ✓
   When $|x|<2$: sum $= \dfrac{1}{1-x/2} = \dfrac{2}{2-x}$. ✓
7. $R = 1$
   Ratio test on the power series: $\left|\dfrac{(n+1)x^{n+1}}{nx^n}\right|=\dfrac{n+1}{n}|x|\to |x|$.
   Converges when $|x|<1$, diverges when $|x|>1$. So $R=1$. ✓
   Bonus: Inside its radius, $\sum_{n=1}^\infty nx^n = \dfrac{x}{(1-x)^2}$ — this is the derivative trick from #1!

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Part 4 — Integration by Parts: The “Solve for $I$” Trick

The first two problems are practice. Problem 3 is the highlight — a brand new technique where the integral reappears on the right side and you solve algebraically for $I$! ★

1. $\displaystyle\int x\sin(2x)\,dx$   ($u=x$, $dv=\sin(2x)\,dx$)
2. $\displaystyle\int x^2\cos x\,dx$   (parts twice; reuse the result from Mar 11 #2)
3. $\displaystyle\int e^x \sin x\,dx$   ★   (parts twice — the original integral comes back! Set $I = \int e^x\sin x\,dx$ and solve for $I$.)
4. $\displaystyle\int \arctan x\,dx$   (sneaky: $u=\arctan x$, $dv=dx$; you'll need $\int\frac{x}{1+x^2}\,dx$)
5. $\displaystyle\int x\arctan x\,dx$   ($u=\arctan x$, $dv=x\,dx$; then split $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$)

▶ Answers — Part 4

1. $-\dfrac{x\cos(2x)}{2} + \dfrac{\sin(2x)}{4} + C$
   $u=x$, $dv=\sin(2x)\,dx$ $\Rightarrow$ $du=dx$, $v=-\dfrac{\cos(2x)}{2}$.
   $\displaystyle\int x\sin(2x)\,dx = -\frac{x\cos(2x)}{2}+\int\frac{\cos(2x)}{2}\,dx = -\frac{x\cos(2x)}{2}+\frac{\sin(2x)}{4}+C$. ✓
2. $x^2\sin x + 2x\cos x - 2\sin x + C$
   $u=x^2$, $v=\sin x$ $\Rightarrow$ $x^2\sin x - 2\displaystyle\int x\sin x\,dx$.
   For $\displaystyle\int x\sin x\,dx$: use Mar 11 style ($u=x$, $v=-\cos x$): $-x\cos x+\sin x$.
   Total: $x^2\sin x - 2(-x\cos x + \sin x) = x^2\sin x + 2x\cos x - 2\sin x + C$. ✓
3. $\dfrac{e^x(\sin x - \cos x)}{2} + C$
   Let $I = \displaystyle\int e^x\sin x\,dx$.
   First parts: $u=e^x$, $dv=\sin x\,dx$ $\Rightarrow$ $-e^x\cos x + \displaystyle\int e^x\cos x\,dx$.
   Second parts: $\displaystyle\int e^x\cos x\,dx$ with $u=e^x$, $dv=\cos x\,dx$ $\Rightarrow$ $e^x\sin x - \displaystyle\int e^x\sin x\,dx = e^x\sin x - I$.
   So: $I = -e^x\cos x + e^x\sin x - I$.
   $2I = e^x(\sin x - \cos x)$ $\Rightarrow$ $\boxed{I = \dfrac{e^x(\sin x-\cos x)}{2}+C}$. ✓
   This trick works whenever parts leads you back to the original integral — just name it $I$ and solve!
4. $x\arctan x - \dfrac{1}{2}\ln(1+x^2) + C$
   $u=\arctan x$, $dv=dx$ $\Rightarrow$ $du=\dfrac{1}{1+x^2}dx$, $v=x$.
   $\displaystyle\int\arctan x\,dx = x\arctan x - \int\frac{x}{1+x^2}\,dx = x\arctan x - \frac{1}{2}\ln(1+x^2)+C$.
   (For the last integral: substitution $w=1+x^2$, $dw=2x\,dx$.) ✓
5. $\dfrac{x^2+1}{2}\arctan x - \dfrac{x}{2} + C$
   $u=\arctan x$, $dv=x\,dx$ $\Rightarrow$ $du=\dfrac{1}{1+x^2}dx$, $v=\dfrac{x^2}{2}$.
   $\displaystyle\int x\arctan x\,dx = \frac{x^2}{2}\arctan x - \frac{1}{2}\int\frac{x^2}{1+x^2}\,dx$.
   Split: $\dfrac{x^2}{1+x^2}=1-\dfrac{1}{1+x^2}$, so $\displaystyle\int\frac{x^2}{1+x^2}dx = x - \arctan x$.
   Total: $\dfrac{x^2}{2}\arctan x - \dfrac{1}{2}(x-\arctan x) = \dfrac{x^2+1}{2}\arctan x - \dfrac{x}{2}+C$. ✓

---

Part 5 — Volumes: Choose Your Weapon

For each problem, solve both ways and verify they agree. Notice how the choice of method matters much more when the function involves $e^x$ or $\ln x$!

Problem 1.

The region bounded by $y = e^x$, $x = 0$, $x = 1$, and $y = 0$ is rotated around the $x$-axis.

▶ Solution — Problem 1

Disc method (integrate along $x$ — natural for $x$-axis rotation) $$ \begin{eqnarray*} V &=& \pi\int_0^1\bigl(e^x\bigr)^2\,dx = \pi\int_0^1 e^{2x}\,dx \\ &=& \pi\left[\frac{e^{2x}}{2}\right]_0^1 = \frac{\pi}{2}(e^2-1) = \boxed{\dfrac{\pi(e^2-1)}{2}} \end{eqnarray*} $$ Shell method (horizontal shells, integrate along $y$) The region has two parts: for $y\in[0,1]$ (below the left edge $e^0=1$), $x$ runs the full width $0$ to $1$; for $y\in[1,e]$, $x$ runs from $\ln y$ to $1$. $$ \begin{eqnarray*} V &=& 2\pi\int_0^1 y\cdot 1\,dy + 2\pi\int_1^e y\bigl(1-\ln y\bigr)\,dy \\ &=& \pi + 2\pi\!\left(\frac{e^2-1}{2}-\left[\frac{y^2}{2}\ln y - \frac{y^2}{4}\right]_1^e\right) \\ &=& \pi + 2\pi\!\left(\frac{e^2-1}{2}-\frac{e^2}{4}-\frac{1}{4}\right) = \boxed{\dfrac{\pi(e^2-1)}{2}}\checkmark \end{eqnarray*} $$

🔍 Disc is dramatically easier — one clean integral. Shell requires splitting into two regions and integration by parts inside! When the rotation axis matches the integration variable ($x$-axis, integrate in $x$), disc/washer almost always wins.

Problem 1 — $y=e^x$, $x\in[0,1]$, rotated around $x$-axis

drag to rotate

blue = $y=e^x$  |  red dashed = $x$-axis (rotation axis)  |  3D: widening trumpet horn, $V=\tfrac{\pi(e^2-1)}{2}$

Problem 2.

The region bounded by $y = \ln x$, $x = e$, and $y = 0$ is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 2

Shell method (vertical shells, integrate in $x$; radius $= x$, height $= \ln x$, from $x=1$ to $x=e$) $$V = 2\pi\int_1^e x\ln x\,dx$$ Apply integration by parts: $u=\ln x$, $dv=x\,dx$ $\Rightarrow$ $\displaystyle du=\frac{1}{x}dx$, $\displaystyle v=\frac{x^2}{2}$ $$ \begin{eqnarray*} V &=& 2\pi\left[\frac{x^2}{2}\ln x\right]_1^e - \int\frac{x}{2}\,dx = 2\pi\left[\frac{x^2}{2}\ln x - \frac{x^2}{4}\right]_1^e \\ &=& 2\pi\!\left(\frac{e^2}{4}+\frac{1}{4}\right) = \boxed{\dfrac{\pi(e^2+1)}{2}} \end{eqnarray*} $$ Washer method (integrate in $y$ from $0$ to $1$; at height $y$, outer radius $R=e$ from $x=e$, inner radius $r=e^y$ from $y=\ln x\Rightarrow x=e^y$) $$ \begin{eqnarray*} V &=& \pi\int_0^1\Bigl(e^2 - e^{2y}\Bigr)\,dy = \pi\left[e^2 y - \frac{e^{2y}}{2}\right]_0^1 \\ &=& \pi\!\left(e^2-\frac{e^2}{2}+\frac{1}{2}\right) = \boxed{\dfrac{\pi(e^2+1)}{2}}\checkmark \end{eqnarray*} $$

🔍 Shell required integration by parts, but was still one integral. Washer was clean once you correctly identified $R=e$ (constant outer wall from $x=e$) and $r=e^y$ (inner from the curve). Notice the outer radius is the boundary line $x=e$, not the curve!

Problem 2 — $y=\ln x$, $x=e$, $y=0$, rotated around $y$-axis

drag to rotate

gold = $y=\ln x$  |  green dashed = $y$-axis  |  outer wall $R=e$, inner $r=e^y$  |  $V=\tfrac{\pi(e^2+1)}{2}$

Problem 3.

The region bounded by $y = x$ and $y = x^3$ (they meet at $(0,0)$ and $(1,1)$ for $x\geq 0$) is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 3

Shell method (vertical shells, integrate in $x$; radius $= x$, height $= x - x^3$, from $x=0$ to $x=1$) $$ \begin{eqnarray*} V &=& 2\pi\int_0^1 x(x-x^3)\,dx = 2\pi\int_0^1(x^2-x^4)\,dx \\ &=& 2\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1 = 2\pi\cdot\frac{2}{15} = \boxed{\dfrac{4\pi}{15}} \end{eqnarray*} $$ Washer method (integrate in $y$ from $0$ to $1$; invert: $y=x\Rightarrow x=y$ (inner), $y=x^3\Rightarrow x=y^{1/3}$ (outer)) $$ \begin{eqnarray*} V &=& \pi\int_0^1\Bigl(\bigl(y^{1/3}\bigr)^2 - y^2\Bigr)\,dy = \pi\int_0^1\bigl(y^{2/3}-y^2\bigr)\,dy \\ &=& \pi\left[\frac{3y^{5/3}}{5}-\frac{y^3}{3}\right]_0^1 = \pi\!\left(\frac{3}{5}-\frac{1}{3}\right) = \boxed{\dfrac{4\pi}{15}}\checkmark \end{eqnarray*} $$

🔍 Shell wins cleanly again — one polynomial integral, no inversion needed. Washer required recognising which curve is outer ($y=x^3$ inverts to $x=y^{1/3}$, which is larger) and computing a fractional-power integral. Both work, but shell is the elegant choice here.

Problem 3 — $y=x^3$ and $y=x$, rotated around $y$-axis

drag to rotate

blue = $y=x$  |  pink = $y=x^3$  |  green dashed = $y$-axis  |  outer $r=y^{1/3}$, inner $r=y$, $V=\tfrac{4\pi}{15}$

March 11, 2026

📅 Thu, March 11, 2026  |  Topics – Limits (two methods) · Series · FTC with chain rule · Integration by parts · Volumes around $y$-axis

Part 1 — Limits: Two Roads, Same Destination

Each limit can be solved two ways — find both!

• Method A – L’Hôpital’s rule (when you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$)
• Method B – Spot it as a derivative definition

\[f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]

1. $\displaystyle\lim_{h\to 0} \frac{(3+h)^2 - 9}{h}$
2. $\displaystyle\lim_{h\to 0} \frac{e^{2+h} - e^2}{h}$
3. $\displaystyle\lim_{x\to 0} \frac{\tan x}{x}$
4. $\displaystyle\lim_{x\to 0} \frac{\tan x}{\sin x}$
5. $\displaystyle\lim_{x\to 0} \frac{e^{2x} - 1}{x}$
6. $\displaystyle\lim_{x\to 0} \frac{\ln(1+3x)}{x}$
7. $\displaystyle\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}$   (L'Hôpital twice; or Taylor series of $e^x$ — which is slicker?)
8. $\displaystyle\lim_{x \to 1} \frac{x^4 - 1}{x - 1}$   (factor; and derivative definition)
9. $\displaystyle\lim_{x\to 0} \frac{\sin(3x)}{x}$
10. $\displaystyle\lim_{x\to 0} \frac{\sin(3x)}{\sin(5x)}$
11. $\displaystyle\lim_{x\to\infty} \frac{x^3 + 2x}{4x^3 - 1}$

▶ Answers — Part 1

1. $6$
   Method B: $f'(3)$ for $f(x)=x^2$; $f'(x)=2x$, so $f'(3)=6$.
   Method A: $\frac{0}{0}$; differentiate w.r.t. $h$: $\displaystyle\frac{2(3+h)}{1}\big|_{h=0}=6$. ✓
2. $e^2$
   Method B: $f'(2)$ for $f(x)=e^x$; $f'(2)=e^2$.
   Method A: $\frac{0}{0}$; diff w.r.t. $h$: $\displaystyle\frac{e^{2+h}}{1}\big|_{h=0}=e^2$. ✓
3. $1$
   Method B: $f'(0)$ for $f(x)=\tan x$; $f'(0)=\sec^2(0)=1$.
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{\sec^2 x}{1}\big|_{x=0}=1$. ✓
4. $1$
   No L'Hôpital — split the fraction:
   $\dfrac{\tan x}{\sin x} = \dfrac{\sin x / \cos x}{\sin x} = \dfrac{1}{\cos x}\to\dfrac{1}{1}=1$. ✓
   Or: $\dfrac{\tan x}{\sin x}=\dfrac{\tan x}{x}\cdot\dfrac{x}{\sin x}\to 1\cdot 1=1$ using #3 and Mar 10 #3.
5. $2$
   Let $u=2x$; as $x\to 0$, $u\to 0$, so $\displaystyle\frac{e^{2x}-1}{x}=2\cdot\frac{e^u-1}{u}\to 2\cdot 1=2$. ✓
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{2e^{2x}}{1}\big|_{x=0}=2$. ✓
6. $3$
   Let $u=3x$; as $x\to 0$, $u\to 0$: $\displaystyle\frac{\ln(1+3x)}{x}=3\cdot\frac{\ln(1+u)}{u}\to 3\cdot 1=3$. ✓
   (Uses the Mar 10 Part 1 #8 result: $\frac{\ln(1+u)}{u}\to 1$.)
7. $\dfrac{1}{2}$
   Taylor series (slicker): $e^x = 1 + x + \dfrac{x^2}{2} + \cdots$, so $e^x - 1 - x = \dfrac{x^2}{2}+\cdots$; divide by $x^2$: limit $= \dfrac{1}{2}$.
   L'Hôpital twice: $\displaystyle\frac{e^x-1-x}{x^2}\xrightarrow{\text{L'H}}\frac{e^x-1}{2x}\xrightarrow{\text{L'H}}\frac{e^x}{2}\big|_{x=0}=\frac{1}{2}$. ✓
8. $4$
   Method B: $f'(1)$ for $f(x)=x^4$; $f'(1)=4\cdot 1^3=4$.
   Direct factor: $x^4-1=(x-1)(x^3+x^2+x+1)$; cancel $(x-1)$; plug in $x=1$: $1+1+1+1=4$. ✓
9. $3$
   Let $u=3x$; $\displaystyle\frac{\sin(3x)}{x}=3\cdot\frac{\sin u}{u}\to 3\cdot 1=3$. ✓
   Method A: diff: $\displaystyle\frac{3\cos(3x)}{1}\big|_{x=0}=3$. ✓
10. $\dfrac{3}{5}$
    Divide top and bottom by $x$: $\displaystyle\frac{\sin(3x)/x}{\sin(5x)/x}=\frac{\sin(3x)/x}{\sin(5x)/x}\to\frac{3}{5}$ using #9 for both. ✓
11. $\dfrac{1}{4}$
    Divide numerator and denominator by $x^3$: $\displaystyle\frac{1+2/x^2}{4-1/x^3}\to\frac{1}{4}$ as $x\to\infty$. ✓

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Part 2 — Limits at Infinity & Sneaky Forms

These mix $x\to\infty$, $x\to 0^+$, and $0\cdot\infty$ forms.

1. $\displaystyle\lim_{x\to\infty} \frac{3x^2 - x + 1}{2x^2 + 5}$
2. $\displaystyle\lim_{x\to\infty} \left(1 + \frac{2}{n}\right)^{\!n}$   (variation on the $e$ definition — what's the exponent?)
3. $\displaystyle\lim_{x\to\infty} \left(1 + \frac{1}{n}\right)^{\!3n}$
4. $\displaystyle\lim_{x\to 0^+} x^2\ln x$
5. $\displaystyle\lim_{x\to 0^+} \sqrt{x}\ln x$

▶ Answers — Part 2

1. $\dfrac{3}{2}$
   Divide through by $x^2$: $\displaystyle\frac{3-1/x+1/x^2}{2+5/x^2}\to\frac{3}{2}$. ✓
2. $e^2$
   $\left(1+\dfrac{2}{n}\right)^n = \left(\!\left(1+\dfrac{2}{n}\right)^{\!n/2}\right)^{\!2}\to (e)^2 = e^2$. ✓
3. $e^3$
   $\left(1+\dfrac{1}{n}\right)^{3n}=\left(\!\left(1+\dfrac{1}{n}\right)^n\right)^{\!3}\to e^3$. ✓
4. $0$
   Write $x^2\ln x = \dfrac{\ln x}{x^{-2}}$; L'Hôpital ($\frac{-\infty}{\infty}$): $\displaystyle\frac{1/x}{-2x^{-3}}=\frac{-x^2}{2}\to 0$. ✓
   Pattern: $x^p\ln x\to 0$ as $x\to 0^+$ for any $p>0$. Power always wins over $\ln$.
5. $0$
   Same pattern: $\sqrt{x}\ln x = x^{1/2}\ln x\to 0$ as $x\to 0^+$. ($p=\frac{1}{2}>0$.) ✓

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Part 3 — Series

Converge or diverge? Give exact value when possible.

1. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$   (famous result — do you remember it?)
2. $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2^n}$   (geometric — identify $r$)
3. $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n^2+1}$
4. $\displaystyle\sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n}$
5. $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2+1}$
6. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$   (partial fractions — telescoping!)
7. $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}$   (ratio test for convergence; exact value is a bonus!)

▶ Answers — Part 3

1. Converges $= \dfrac{\pi^2}{6}$
   The Basel problem — $p$-series with $p=2>1$ (converges), and Euler proved the value is $\dfrac{\pi^2}{6}\approx 1.645$. Memorise this one! ✓
2. Converges $= \dfrac{2}{3}$
   Geometric with $a=1$, $r=-\dfrac{1}{2}$; $|r|<1$ so converges. Sum $= \dfrac{1}{1-(-1/2)}=\dfrac{1}{3/2}=\dfrac{2}{3}$. ✓
3. Diverges
   Limit comparison with $\dfrac{1}{n}$: $\displaystyle\frac{(n+1)/(n^2+1)}{1/n}=\frac{n(n+1)}{n^2+1}\to 1>0$. Since $\sum\frac{1}{n}$ diverges, so does this. ✓
4. Converges $= \dfrac{2}{3}+\dfrac{3}{2} = \dfrac{13}{6}$
   Split: $\displaystyle\sum\frac{2^n+3^n}{5^n}=\sum\!\left(\frac{2}{5}\right)^{\!n}+\sum\!\left(\frac{3}{5}\right)^{\!n}$.
   Two geometric series: $\dfrac{2/5}{1-2/5}+\dfrac{3/5}{1-3/5}=\dfrac{2}{3}+\dfrac{3}{2}=\dfrac{13}{6}$. ✓
5. Converges (value not elementary)
   Alternating series test (Leibniz): $a_n=\dfrac{n}{n^2+1}$ is decreasing (check with calculus) and $a_n\to 0$. Both conditions satisfied. ✓
6. Converges $= \dfrac{3}{4}$
   Partial fractions: $\dfrac{1}{n(n+2)}=\dfrac{1}{2}\!\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)$. Telescoping — write out first few terms and watch pairs cancel. Sum $=\dfrac{1}{2}\!\left(1+\dfrac{1}{2}\right)=\dfrac{3}{4}$. ✓
7. Converges $= 6$
   Ratio test: $\displaystyle\frac{(n+1)^2/2^{n+1}}{n^2/2^n}=\frac{(n+1)^2}{2n^2}\to\frac{1}{2}<1$. ✓ Converges.
   Exact value (bonus): Differentiate $\sum nx^{n-1}=\frac{1}{(1-x)^2}$ again, multiply by $x$: $\sum n^2 x^n = \frac{x(1+x)}{(1-x)^3}$. At $x=\frac{1}{2}$: $\frac{(1/2)(3/2)}{(1/2)^3}=6$. ✓

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Part 4 — Integration by Parts

Recall the formula: $\displaystyle\int u\,dv = uv - \int v\,du$

The trick is choosing $u$ and $dv$ wisely. A helpful mnemonic for choosing $u$: LIATE — Logarithm, Inverse trig, Algebraic, Trig, Exponential. The item highest on that list makes the best $u$.

1. $\displaystyle\int x e^x\,dx$   ($u=x$, $dv=e^x\,dx$)
2. $\displaystyle\int x\cos x\,dx$
3. $\displaystyle\int x^2 e^x\,dx$   (parts twice — or spot the pattern from #1)
4. $\displaystyle\int \ln x\,dx$   (surprising: $u=\ln x$, $dv=dx$)
5. $\displaystyle\int x\ln x\,dx$

▶ Answers — Part 4

1. $xe^x - e^x + C = e^x(x-1) + C$
   $u=x$, $dv=e^x\,dx$ $\Rightarrow$ $du=dx$, $v=e^x$.
   $\displaystyle\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$. ✓
2. $x\sin x + \cos x + C$
   $u=x$, $dv=\cos x\,dx$ $\Rightarrow$ $du=dx$, $v=\sin x$.
   $\displaystyle\int x\cos x\,dx = x\sin x - \int\sin x\,dx = x\sin x + \cos x + C$. ✓
3. $e^x(x^2 - 2x + 2) + C$
   First parts: $u=x^2$, $v=e^x$ $\Rightarrow$ $x^2 e^x - 2\displaystyle\int xe^x\,dx$.
   Apply result from #1 for the remaining integral: $-2(xe^x - e^x) = -2xe^x+2e^x$.
   Total: $e^x(x^2-2x+2)+C$. ✓
4. $x\ln x - x + C$
   $u=\ln x$, $dv=dx$ $\Rightarrow$ $du=\frac{1}{x}dx$, $v=x$.
   $\displaystyle\int\ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + C$. ✓
   This is the "sneaky" parts — $dv=dx$ looks too simple, but it works beautifully.
5. $\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$
   $u=\ln x$, $dv=x\,dx$ $\Rightarrow$ $du=\frac{1}{x}dx$, $v=\frac{x^2}{2}$.
   $\displaystyle\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int\frac{x^2}{2}\cdot\frac{1}{x}\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}+C$. ✓

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Part 5 — Volumes Around the $y$-Axis: Shell vs. Washer

Rotating around the $y$-axis — this is where the shell method really shines. For each problem, solve both ways and confirm they agree.

Problem 1.

The region bounded by $y = \sqrt{x}$, $x = 1$, and $y = 0$ is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 1

Shell method (vertical shells, integrate in $x$; radius $= x$, height $= \sqrt{x}$, from $x=0$ to $x=1$) $$ V = 2\pi\int_0^1 x \sqrt{x}\,dx = 2\pi\left[\frac{2}{5} x^{5/2} \right]_0^1 = \boxed{\dfrac{4\pi}{5}} $$ Washer method (integrate in $y$ from $0$ to $1$; for a given $y$, $x$ runs from $0$ to $y^2$, so outer radius $R = 1$, inner radius $r=y^2$) $$ V = \pi\int_0^1 \left( 1^2 - \bigl(y^2\bigr)^2\right)\,dy = \pi\int_0^1 (1-y^4)\,dy = \pi\left[y-\frac{y^5}{5}\right]_0^1 = \boxed{\dfrac{4\pi}{5}}\checkmark $$

🔍 Shell was easier.

Problem 1 — $y=\sqrt{x}$, $y=1$, rotated around $y$-axis

drag to rotate

blue curve = $y=\sqrt{x}$  |  green dashed = $y$-axis  |  blue walls = outer cylinder ($R=1$)  |  purple = inner paraboloid scoop ($r=y^2$), $V=\tfrac{4\pi}{5}$

Problem 2.

The region bounded by $y = x^2$, $x = 2$, and $x$-axis is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 2

Shell method (vertical shells, integrate in $x$ from $0$ to $2$; radius $= x$, height $= x^2$) $$ V = 2\pi\int_0^2 x\bigl( x^2\bigr)\,dx = \frac{\pi}{2}\Bigl[{x^4}\Bigr]_0^2 = \boxed{8\pi} $$ Washer method (integrate in $y$ from $0$ to $4$; outer radius $R = 2$ (from $x=2$ boundary), inner radius $r = \sqrt{y}$ (from $y=x^2$)) $$ \begin{eqnarray*} V &=& \pi\int_0^4\Bigl(2^2 - \bigl(\sqrt{y}\bigr)^2\Bigr)\,dy = \pi\int_0^4\bigl(4 - y\bigr)\,dy \\ &=& \pi\left[4y - \frac{y^2}{2}\right]_0^4 = \pi(16-8) = \boxed{8\pi}\checkmark \end{eqnarray*} $$

🔍 Shell was more natural — the region is described cleanly in $x$. Washer needed the outer boundary $x=2$ rewritten as a constant radius, and the inner boundary inverted to $x=\sqrt{y}$.

Problem 2 — $y=x^2$, $y=4$, rotated around $y$-axis

drag to rotate

purple = $y=x^2$  |  green dashed = $y$-axis  |  3D washer solid $V=8\pi$

Problem 3.

The region bounded by $y = x(2-x)$ and $y = 0$ (the parabola meets the $x$-axis at $x=0$ and $x=2$) is rotated around the $\boldsymbol{y}$-axis.

▶ Solution — Problem 3

Shell method (vertical shells, integrate in $x$ from $0$ to $2$; radius $= x$, height $= x(2-x)$) $$V = 2\pi\int_0^2 x \cdot x(2-x)\,dx = 2\pi\int_0^2\bigl(2x^2 - x^3\bigr)\,dx$$ $$= 2\pi\left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2 = 2\pi\!\left(\frac{16}{3} - 4\right) = 2\pi\cdot\frac{4}{3} = \boxed{\dfrac{8\pi}{3}}$$ Washer method (integrate in $y$: need to invert $y=x(2-x) = 1-(x-1)^2$, a downward parabola with peak at $(1,1)$) Rewrite: $x-1 = \pm\sqrt{1-y}$, so the right branch is $x = 1+\sqrt{1-y}$ (outer radius) and left branch is $x = 1-\sqrt{1-y}$ (inner radius). $$ \begin{eqnarray*} V &=& \pi\int_0^1\Bigl[\bigl(1+\sqrt{1-y}\bigr)^2 - \bigl(1-\sqrt{1-y}\bigr)^2\Bigr]\,dy \\ &=& \pi\int_0^1 4\sqrt{1-y}\,dy \quad\text{(difference of squares collapses nicely!)}\\ &=& 4\pi\left[-\frac{2}{3}(1-y)^{3/2}\right]_0^1 = 4\pi\cdot\frac{2}{3} = \boxed{\dfrac{8\pi}{3}}\checkmark \end{eqnarray*} $$

🔍 Shell method is dramatically easier — one clean integral. The washer method required completing the square, inverting the parabola to find both branches, then a $u$-substitution. This is the best example yet of why shell method was invented!

Problem 3 — $y=x(2-x)$, rotated around $y$-axis

drag to rotate

orange = $y=x(2-x)$  |  green dashed = $y$-axis  |  3D solid $V=\tfrac{8\pi}{3}$ (like a donut-cap!)

March 10, 2026

📅 Wed, March 10, 2026  |  Topics – Limits (two methods) · Series · FTC with chain rule · Volumes of Revolution

Part 1 — Limits: Two Roads, Same Destination

Each limit below can be solved two ways. Try both — they should give the same answer!

• Method A – L’Hôpital’s theorem (when you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$)
• Method B – Recognise it as a derivative definition

\[f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]

1. $\displaystyle\lim_{h\to 0} \frac{(2+h)^3 - 8}{h}$   (hint: what function? what point?)
2. $\displaystyle\lim_{h\to 0} \frac{\sqrt{4+h} - 2}{h}$
3. $\displaystyle\lim_{x\to 0} \frac{\sin x}{x}$
4. $\displaystyle\lim_{x\to 0} \frac{\sqrt{4+x} - 2}{\sin x}$
5. $\displaystyle\lim_{x\to 0} \frac{\sqrt{4+\sin x} - 2}{x}$
6. $\displaystyle\lim_{x\to 0} \frac{\sqrt{4+\sin x} - 2}{\sin x}$
7. $\displaystyle\lim_{x\to 0} \frac{e^x - 1}{x}$
8. $\displaystyle\lim_{x\to 0} \frac{\ln(1+x)}{x}$
9. $\displaystyle\lim_{x\to 0} \frac{1 - \cos x}{x^2}$   (L'Hôpital twice, or Taylor; and derivative definition of $\cos$ at 0)
10. $\displaystyle\lim_{x\to 0} \frac{e^{3x} - 1}{x}$   (can you do this without L'Hôpital using #7?)
11. $\displaystyle\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$   (factor directly; and derivative definition)

▶ Answers — Part 1

1. $12$
   Method B (derivative): This is $f'(2)$ for $f(x)=x^3$, so $f'(2)=3\cdot 2^2=12$.
   Method A (L'Hôpital): $\frac{0}{0}$ form; diff top/bottom w.r.t. $h$: $\displaystyle\frac{3(2+h)^2}{1}\big|_{h=0} = 12$. ✓
2. $\dfrac{1}{4}$
   Method B: $f'(4)$ for $f(x)=\sqrt{x}$; $f'(x)=\frac{1}{2\sqrt{x}}$, so $f'(4)=\frac{1}{4}$.
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{1/(2\sqrt{4+h})}{1}\big|_{h=0}=\frac{1}{4}$. ✓
3. $1$
   Method B: $f'(0)$ for $f(x)=\sin x$; $f'(0)=\cos 0=1$.
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{\cos x}{1}\big|_{x=0}=1$. ✓
4. $\dfrac{1}{4}$
   No L'Hôpital needed — just multiply and divide by $x$:
   $\dfrac{\sqrt{4+x}-2}{\sin x} = \dfrac{\sqrt{4+x}-2}{x}\cdot\dfrac{x}{\sin x}$
   First factor $\to \dfrac{1}{4}$ by #2;   second factor $\to 1$ because $\dfrac{\sin x}{x}\to 1$ by #3.
   Product $= \dfrac{1}{4}\cdot 1 = \dfrac{1}{4}$. ✓
5. $\dfrac{1}{4}$
   No L'Hôpital — substitute and split:
   $\dfrac{\sqrt{4+\sin x}-2}{x} = \dfrac{\sqrt{4+\sin x}-2}{\sin x}\cdot\dfrac{\sin x}{x}$
   For the first factor, let $u = \sin x$; as $x\to 0$, $u\to 0$, so $\dfrac{\sqrt{4+u}-2}{u}\to\dfrac{1}{4}$ by #2.
   Second factor $\to 1$ by #3.   Product $= \dfrac{1}{4}\cdot 1 = \dfrac{1}{4}$. ✓
6. $\dfrac{1}{4}$
   No L'Hôpital — direct substitution, reuses only #2:
   Let $u = \sin x$; as $x\to 0$, $u\to 0$, so
   $\dfrac{\sqrt{4+\sin x}-2}{\sin x} = \dfrac{\sqrt{4+u}-2}{u}\to\dfrac{1}{4}$ directly by #2.
   No need for #3 this time! The $\sin x$ cancels itself top and bottom. ✓
7. $1$
   Method B: $f'(0)$ for $f(x)=e^x$; $f'(0)=e^0=1$.
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{e^x}{1}\big|_{x=0}=1$. ✓
8. $1$
   Method B: $f'(0)$ for $f(x)=\ln(1+x)$; $f'(x)=\frac{1}{1+x}$, so $f'(0)=1$.
   Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{1/(1+x)}{1}\big|_{x=0}=1$. ✓
9. $\dfrac{1}{2}$
   Method B: $-f''(0)/2!$ for $\cos x$... actually: rewrite as $\displaystyle\frac{1-\cos x}{x^2}$. Note $1-\cos x = 2\sin^2(x/2)$, so $\displaystyle\frac{2\sin^2(x/2)}{x^2}=\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2\to\frac{1}{2}$.
   Method A: $\frac{0}{0}$ twice: $\displaystyle\frac{\sin x}{2x}\big|_{x=0}$ → still $\frac{0}{0}$ → $\displaystyle\frac{\cos x}{2}\big|_{x=0}=\frac{1}{2}$. ✓
10. $3$
    Method B (slick): Let $u=3x$; as $x\to 0$, $u\to 0$, so $\displaystyle\frac{e^{3x}-1}{x}=3\cdot\frac{e^u-1}{u}\to 3\cdot 1=3$. Uses result from #7!
    Method A: $\frac{0}{0}$; diff: $\displaystyle\frac{3e^{3x}}{1}\big|_{x=0}=3$. ✓
11. $12$
    Method B: $f'(2)$ for $f(x)=x^3$; $f'(2)=3\cdot 4=12$. (Same as #1!)
    Direct factor: $x^3-8=(x-2)(x^2+2x+4)$; cancel $(x-2)$; plug in $x=2$: $4+4+4=12$. ✓

---

Part 2 — Limits at Infinity

These all involve $x\to\infty$ or $n\to\infty$. One of them secretly reuses a result from Part 1 — can you spot which one?

1. $\displaystyle\lim_{x\to\infty} x\sin\dfrac{1}{x}$   (hint: try the substitution $u = \frac{1}{x}$)
2. $\displaystyle\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{\!n}$   (this one you should just know by heart!)
3. $\displaystyle\lim_{n\to\infty} \left(1+\frac{3}{n}\right)^{\!2n}$   (variation on #2 — what changes?)
4. $\displaystyle\lim_{x\to\infty} \dfrac{x^{100}}{e^x}$   (L'Hôpital 100 times? or is there a slicker way?)
5. $\displaystyle\lim_{x\to\infty} \left(\sqrt{x+1} - \sqrt{x}\right)$   (try multiplying by the conjugate)

▶ Answers — Part 2

1. $1$
   Let $u=\dfrac{1}{x}$; as $x\to\infty$, $u\to 0^+$, so $x\sin\dfrac{1}{x} = \dfrac{\sin u}{u} \to 1$ by Part 1 #3! ✓
   That's the hidden connection — the same $\displaystyle\frac{\sin\theta}{\theta}\to 1$ just wearing a disguise.
2. $e \approx 2.718\ldots$
   This is the definition of $e$. Memorise it — it appears everywhere.
   More generally: $\displaystyle\lim_{n\to\infty}\!\left(1+\frac{a}{n}\right)^n = \lim_{n\to\infty}\!\left(\left(1+\frac{a}{n}\right)^{\frac{n}{a}}\right)^{a} = e^a$.
3. $e^6$
   Rewrite the exponent: $\left(1+\dfrac{3}{n}\right)^{2n} = \left(\left(1+\dfrac{3}{n}\right)^{\frac{n}{3}}\right)^6$.
   Inner limit $\to e$ by the formula from #2 with $a=1$; $(e)^6 = e^6$.
4. $0$
   Exponential always beats any polynomial at $\infty$ — no matter how large the power.
   Formally: L'Hôpital applied 100 times gives $\dfrac{100!}{e^x}\to 0$.
   Slicker: $e^x > \dfrac{x^{101}}{101!}$ for all large $x$ (from the Taylor series), so $\dfrac{x^{100}}{e^x} < \dfrac{101!}{x} \to 0$.
5. $0$
   Multiply by the conjugate $\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$:
   $\sqrt{x+1}-\sqrt{x} = \dfrac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \dfrac{1}{\sqrt{x+1}+\sqrt{x}} \to \dfrac{1}{\infty} = 0$. ✓

Part 3 — Series

Converge or diverge? Give exact value when possible.

1. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$   (hint: partial fractions!)
2. $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n}$
3. $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$
4. $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$
5. $\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n!}$
6. $\displaystyle\sum_{n=1}^{\infty} \frac{n^2+1}{n^3+1}$
7. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$

▶ Answers — Part 3

1. Converges $= 1$
   Partial fractions: $\displaystyle\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. Telescoping sum: all middle terms cancel, leaving $\displaystyle 1 - \lim_{N\to\infty}\frac{1}{N+1} = 1$.
2. Converges $= 2$
   Ratio test: $\displaystyle\frac{(n+1)/2^{n+1}}{n/2^n}=\frac{n+1}{2n}\to\frac{1}{2}<1$. Value: differentiate the geometric series $\displaystyle\sum x^n = \frac{1}{1-x}$ to get $\displaystyle\sum nx^{n-1}=\frac{1}{(1-x)^2}$; multiply by $x$: $\displaystyle\sum nx^n = \frac{x}{(1-x)^2}$; at $\displaystyle x=\frac{1}{2}$: $\displaystyle\frac{1/2}{1/4}=2$.
3. Converges (value not elementary)
   Leibniz alternating series test: $\displaystyle\frac{1}{\sqrt{n}}$ is decreasing and $\to 0$. ✓ (Value $\approx -0.6049$, related to Dirichlet eta function.)
4. Diverges ($\infty$)
   Integral test: $\displaystyle\int_2^\infty \frac{dx}{x\ln x} = \ln(\ln x)\Big|_2^\infty = \infty$. (Or: $p$-series-like with $p=1$ — just barely diverges!)
5. Converges $= e^3 - 1 \approx 19.09$
   Ratio test: $\displaystyle\frac{3^{n+1}/(n+1)!}{3^n/n!}=\frac{3}{n+1}\to 0 < 1$. Value: $\displaystyle\sum_{n=0}^{\infty}\frac{3^n}{n!}=e^3$, minus the $n=0$ term ($=1$): answer $= e^3-1$.
6. Diverges ($\infty$)
   Limit comparison with $\displaystyle\frac{1}{n}$: $\displaystyle\frac{(n^2+1)/(n^3+1)}{1/n}=\frac{n(n^2+1)}{n^3+1}\to 1>0$. Since $\sum\dfrac{1}{n}$ diverges, so does this.
7. Converges
   $p$-series with $p=\dfrac{3}{2}>1$. (Value $= \zeta(3/2)\approx 2.612$, no simple closed form.)

Part 4 — FTC with a Twist

These use FTC Part 1 + chain rule — the upper limit is a function of $x$, not just $x$ itself!

\[\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x)\]

1. $\displaystyle\frac{d}{dx}\int_0^{x^2} e^{-t^2}\, dt$
2. $\displaystyle\frac{d}{dx}\int_x^{x^2} \sin(t^2)\, dt$   (both limits move! split it into two integrals first)

▶ Answers — Part 4

1. $2x\,e^{-x^4}$
   Chain rule: $f(g(x))\cdot g'(x)$ where $f(t)=e^{-t^2}$ and $g(x)=x^2$.
   $= e^{-(x^2)^2}\cdot 2x = 2x\,e^{-x^4}$.
2. $2x\sin(x^4) - \sin(x^2)$
   Split: $\displaystyle\int_x^{x^2} = \int_0^{x^2} - \int_0^{x}$.
   Differentiate each: $\displaystyle\frac{d}{dx}\int_0^{x^2}\sin(t^2)\,dt = \sin(x^4)\cdot 2x$;  $\displaystyle\frac{d}{dx}\int_0^{x}\sin(t^2)\,dt = \sin(x^2)\cdot 1$.
   Answer: $\displaystyle 2x\sin(x^4) - \sin(x^2)$.

Part 5 — Volumes of Revolution: Two Methods Every Time

For each problem, compute the volume both ways — disc/washer method and shell method. Verify they agree, then reflect: which method felt easier, and why?

Problem 1.

The region bounded by $y = x^2$, $x = 0$, $x = 2$, and $y = 0$ is rotated around the $x$-axis.

▶ Solution — Problem 1

Disc method (integrate along $x$, natural here) $$V = \pi\int_0^2 y^2\,dx = \pi\int_0^2 \left(x^2\right)^2\,dx = \pi\int_0^2 x^4\,dx = \pi\left[\frac{x^5}{5}\right]_0^2 = \boxed{\dfrac{32\pi}{5}}$$ Shell method (horizontal shells, integrate along $y$ from $0$ to $4$) Each shell at height $y$ has radius $= y$, height $= 2 - \sqrt{y}$ (rightmost $x$ is $2$, leftmost is $\sqrt{y}$). $$V = 2\pi\int_0^4 y\bigl(2-\sqrt{y}\bigr)\,dy = 2\pi\int_0^4\bigl(2y - y^{3/2}\bigr)\,dy$$ $$= 2\pi\left[y^2 - \frac{2}{5}y^{5/2}\right]_0^4 = 2\pi\!\left(16 - \frac{2}{5}\cdot 32\right) = 2\pi\cdot\frac{16}{5} = \boxed{\dfrac{32\pi}{5}}\checkmark$$

🔍 Disc was easier here — one clean integral vs a fractional power. Shell required inverting $y=x^2$ to $x=\sqrt{y}$ and thinking in $y$.

Problem 1 — $y=x^2$, $x\in[0,2]$, rotated around $x$-axis

drag to rotate

blue region = area rotated  |  red dashed = axis of rotation ($x$-axis)  |  3D solid = disc stack $V=\frac{32\pi}{5}$

Problem 2.

The region bounded by $y = \sqrt{x}$ and $y = x$ (they meet at $(0,0)$ and $(1,1)$) is rotated around the $x$-axis.

▶ Solution — Problem 2

Washer method (integrate along $x$; outer radius $R=\sqrt{x}$, inner radius $r=x$) $$ \begin{eqnarray*} V &=& \pi\int_0^1\Bigl((\sqrt{x})^2 - x^2\Bigr)\,dx = \pi\int_0^1(x - x^2)\,dx \\ &=& \pi\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \boxed{\dfrac{\pi}{6}} \end{eqnarray*} $$ Shell method (horizontal shells, integrate along $y$; for a given $y$ - $x$ runs from $y^2$ to $y$) Shell radius $= y$, shell height $= y - y^2$. $$ \begin{eqnarray*} V &=& 2\pi\int_0^1 y\bigl(y - y^2\bigr)\,dy = 2\pi\int_0^1(y^2 - y^3)\,dy \\ &=& 2\pi\left[\frac{y^3}{3}-\frac{y^4}{4}\right]_0^1 = 2\pi\cdot\frac{1}{12} = \boxed{\dfrac{\pi}{6}}\checkmark \end{eqnarray*} $$

🔍 Both methods are equally clean here! Washer thinks in $x$ (natural for rotation around $x$-axis); shell thinks in $y$ and requires rewriting $y=\sqrt{x}\Rightarrow x=y^2$ and $y=x\Rightarrow x=y$.

Problem 2 — $y=\sqrt{x}$ and $y=x$, rotated around $x$-axis

drag to rotate

green = $y=\sqrt{x}$ (outer)  |  orange = $y=x$ (inner)  |  red dashed = axis  |  3D washer = hollow solid $V=\frac{\pi}{6}$

Problem 3.

The region bounded by $y = x^2$ and $y = 2x$ (they meet at $(0,0)$ and $(2,4)$) is rotated around the $y$-axis.

▶ Solution — Problem 3

Shell method (vertical shells, integrate along $x$; shell radius $= x$, height $= 2x - x^2$) $$V = 2\pi\int_0^2 x\bigl(2x - x^2\bigr)\,dx = 2\pi\int_0^2(2x^2 - x^3)\,dx$$ $$= 2\pi\left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2 = 2\pi\!\left(\frac{16}{3} - 4\right) = 2\pi\cdot\frac{4}{3} = \boxed{\dfrac{8\pi}{3}}$$ Washer method (integrate along $y$; for a given $y$ - outer radius from $y=x^2\Rightarrow x=\sqrt{y}$, inner radius from $y=2x\Rightarrow x=y/2$) $$ \begin{eqnarray*} V &=& \pi\int_0^4\Bigl((\sqrt{y})^2 - \left(\dfrac{y}{2}\right)^2\Bigr)\,dy = \pi\int_0^4\!\left(y - \frac{y^2}{4}\right)\!dy \\ &=& \pi\left[\frac{y^2}{2}-\frac{y^3}{12}\right]_0^4 = \pi\!\left(8 - \frac{64}{12}\right) = \pi\cdot\frac{8}{3} = \boxed{\dfrac{8\pi}{3}}\checkmark \end{eqnarray*} $$

🔍 Shell was much easier here — rotating around the $y$-axis with functions given in $x$ is exactly what shell method is designed for. Washer forced us to invert both curves and integrate in $y$.

Problem 3 — $y=x^2$ and $y=2x$, rotated around $y$-axis

drag to rotate

orange = $y=2x$ (inner boundary)  |  purple = $y=x^2$ (outer boundary)  |  green dashed = $y$-axis (axis of rotation)

March 9, 2026

📅 Mon, March 9, 2026  |  Topics – Limits at 0 and $\infty$ · Series convergence · FTC Part 1

Part 1 — Limits

Find the value of each limit (write $\infty$, $-\infty$, or DNE when appropriate).

1. $\displaystyle\lim_{x\to\infty} \dfrac{x^2 - 3x}{-3x^2 + 4x + 1}$
2. $\displaystyle\lim_{x\to\infty} \dfrac{x^2 - 3x}{-3x^2 + 4x}$
3. $\displaystyle\lim_{x\to 0} \dfrac{x^2 - 3x}{-3x^2 + 4x}$
4. $\displaystyle\lim_{x\to 0^+} x\ln x$
5. $\displaystyle\lim_{x\to 0^+} x^2 \ln x$
6. $\displaystyle\lim_{x\to 0^+} x^{10} \ln x$
7. $\displaystyle\lim_{x\to 0^+} x^{0.01} \ln x$
8. $\displaystyle\lim_{x\to 0^+} x^{-1} \ln x$    (careful — which dominates?)
9. $\displaystyle\lim_{x\to 0^+} x^{-0.01} \ln x$    (subtle!)
10. $\displaystyle\lim_{x\to\infty} x\ln x$
11. $\displaystyle\lim_{x\to\infty} x^2 \ln x$
12. $\displaystyle\lim_{x\to\infty} x^{10} \ln x$
13. $\displaystyle\lim_{x\to\infty} x^{0.01} \ln x$
14. $\displaystyle\lim_{x\to\infty} x^{-1} \ln x$
15. $\displaystyle\lim_{x\to\infty} x^{-0.01} \ln x$    (compare growth rates carefully)
16. $\displaystyle\lim_{x\to\infty} \dfrac{\ln(100 + e^{3x}) + 10^{10^{10}}}{5x}$
17. $\displaystyle\lim_{x\to\infty} \dfrac{\ln(100 + e^{3x}) - \ln 101}{5x}$
18. $\displaystyle\lim_{x\to 0} \dfrac{\ln(100 + e^{3x}) - \ln 101}{5x}$
19. $\displaystyle\lim_{x\to 0} \dfrac{\ln(100 + e^{3x})}{5x}$    (does this even exist?)
20. $\displaystyle\lim_{n\to\infty} (-1)^n$

▶ Answers — Part 1

1. $-\dfrac{1}{3}$ — divide top and bottom by $x^2$; only leading coefficients survive
2. $-\dfrac{1}{3}$ — same; the $+1$ in the denominator is irrelevant at $\infty$
3. $-\dfrac{3}{4}$ — factor out $x$: $\frac{x(x-3)}{x(-3x+4)}$, cancel $x$, plug in $x=0$: $\frac{0-3}{0+4} = -\frac{3}{4}$
4. $0$ — rewrite as $\frac{\ln x}{x^{-1}}$; L'Hôpital: $\frac{1/x}{-x^{-2}} = -x \to 0$
5. $0$ — $x^2$ kills $\ln x$ even faster; for any $p>0$: $\lim_{x\to 0^+} x^p\ln x = 0$
6. $0$ — same rule, $p=10$
7. $0$ — same rule, $p=0.01>0$; polynomial always beats log
8. $-\infty$ — $\frac{\ln x}{x} \xrightarrow{x\to 0^+} \frac{-\infty}{0^+} = -\infty$ (denominator $\to 0^+$ makes it blow up)
9. $-\infty$ — $p=-0.01<0$ so $x^{-0.01}\to +\infty$ while $\ln x\to -\infty$; product $\to -\infty$
10. $\infty$ — both $x\to\infty$ and $\ln x\to\infty$; no competition
11. $\infty$
12. $\infty$
13. $\infty$ — even a tiny positive power $\times \ln x \to \infty$
14. $0$ — $\frac{\ln x}{x}\to 0$; polynomial growth crushes logarithm at $\infty$
15. $0$ — $\frac{\ln x}{x^{0.01}}\to 0$; any polynomial power beats $\ln x$ at $\infty$
16. $\dfrac{3}{5}$ — the constant $10^{10^{10}}$ is irrelevant at $\infty$; $\ln(100+e^{3x})\approx 3x$ for large $x$, so limit $=\frac{3x}{5x}=\frac{3}{5}$
17. $\dfrac{3}{5}$ — $-\ln 101$ is a fixed constant, irrelevant at $\infty$; same reasoning
18. $\dfrac{3}{505}$ — at $x=0$: $\frac{0}{0}$ form; L'Hôpital: $\dfrac{3e^{3x}/(100+e^{3x})}{5}\Big|_{x=0} = \dfrac{3/101}{5} = \dfrac{3}{505}$
19. DNE — numerator $\to\ln 101\neq 0$ while denominator $\to 0$; from the right $\to+\infty$, from the left $\to-\infty$
20. DNE — $(-1)^n$ oscillates between $+1$ and $-1$ forever; no single value

Part 2 — Series

– Converge or Diverge?

For each series – does it converge or diverge? If it converges, give the exact value if you can. If it diverges, write $\infty$, $-\infty$, or “oscillates.”

1. $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}$
2. $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}$
3. $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^4}$
4. $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}$
5. $\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{2n+1}$
6. $\displaystyle\sum_{n=1}^{\infty} (-1)^n$
7. $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n!}$
8. $\displaystyle\sum_{n=1}^{\infty} 2^{n-1}$
9. $\displaystyle\sum_{n=1}^{\infty} (0.5)^{n-1}$
10. $\displaystyle\sum_{n=1}^{\infty} (0.25)^{n-1}$
11. $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^{n-1}$
12. $\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{e^n}$
13. $\displaystyle\sum_{n=1}^{\infty} \dfrac{e^n}{n!}$
14. $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^2}{e^n}$
15. $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^{1000}}{e^n}$

▶ Answers — Part 2

1. Diverges ($\infty$) — the harmonic series; famously diverges (grows like $\ln n$)
2. Converges $= \dfrac{\pi^2}{6} \approx 1.6449$ — Basel problem (Euler, 1734); $p$-series with $p=2>1$
3. Converges $= \dfrac{\pi^4}{90} \approx 1.0823$ — $p$-series with $p=4>1$
4. Converges $= \ln 2 \approx 0.6931$ — alternating harmonic series (Leibniz test)
5. Converges $= \dfrac{\pi}{4} \approx 0.7854$ — Gregory–Leibniz formula for $\pi$! (from $\arctan 1 = \pi/4$)
6. Diverges (oscillates) — partial sums alternate $-1, 0, -1, 0, \ldots$; no limit
7. Converges $= e-1 \approx 1.7183$ — Taylor series for $e^x$ at $x=1$, minus the $n=0$ term
8. Diverges ($\infty$) — geometric with ratio $r=2>1$
9. Converges $= 2$ — geometric: $\dfrac{1}{1-0.5} = 2$
10. Converges $= \dfrac{4}{3}$ — geometric: $\dfrac{1}{1-0.25} = \dfrac{4}{3}$
11. Converges $= \dfrac{3}{2}$ — geometric: $\dfrac{1}{1-\frac{1}{3}} = \dfrac{3}{2}$
12. Diverges ($\infty$) — ratio test: $\dfrac{(n+1)!}{e^{n+1}}\cdot\dfrac{e^n}{n!} = \dfrac{n+1}{e}\to\infty$
13. Converges $= e^e - 1 \approx 14.77$ — ratio test: $\dfrac{e}{n+1}\to 0<1$; relates to Taylor expansion of $e^e$
14. Converges — ratio test: $\dfrac{(n+1)^2}{n^2 \cdot e}\to\dfrac{1}{e}<1$
15. Converges — ratio test: limit $\to\dfrac{1}{e}<1$; exponential always beats any polynomial power

Part 3 — FTC, Part 1

Find each derivative. Hint – you do NOT need to evaluate the integral!

1. $\displaystyle\frac{d}{dx} \int_{0}^{x} \sqrt{\sin^4 t + 234}\; dt$
2. $\displaystyle\frac{d}{dx} \int_{-1000}^{x} \sqrt{\sin^4 t + 234}\; dt$
3. $\displaystyle\frac{d}{dx} \int_{10^{10}}^{x} \sqrt{\sin^4 t + 234}\; dt$

👀 Look carefully at all three answers. What do you notice about the lower limit?

▶ Answers — Part 3

All three give the same answer: $$\frac{d}{dx}\int_a^x \sqrt{\sin^4 t + 234}\;dt \;=\; \sqrt{\sin^4 x + 234}$$

The lower limit $a$ — whether $0$, $-1000$, or $10^{10}$ — only contributes a fixed constant to the antiderivative, and constants vanish under differentiation. FTC Part 1: $\;\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)\;$ for any constant $a$.

Also: inside the integral the dummy variable is $t$, not $x$. Writing $\sqrt{\sin^4 x+234}$ inside the integral while differentiating with respect to $x$ would be a sign confusion — that's exactly why the dummy variable matters!