---
# System prepended metadata

title: 微積分複習
tags: [微積分]

---

# 微積分複習
###### tags: `微積分`
- [朱樺微積分講義](http://www.math.ntu.edu.tw/~hchu/Calculus/)
- [知識型計算引擎 Wolfram Alpha](https://www.wolframalpha.com/)
- [繪圖計算機 Desmos](https://www.desmos.com/calculator?lang=zh-TW)
- [電子書下載平台 Library Genesis](http://libgen.rs/) 可搜尋你們的微積分課本，獲得電子書版本
## Curve Sketching
$f(x) = -\dfrac{x^2-x+1}{x-1}$
(b) $\begin{aligned}f'(x) &= -\dfrac{(2x-1)(x-1)-(x^2-x+1)(1)}{(x-1)^2}\\&=-\dfrac{x(x-2)}{(x-1)^2}=0\end{aligned}$

$\implies x=0$ or $x=2$, local extremes

(c\) $\begin{aligned}f''(x) &= -\dfrac{(2x-2)(x-1)^2-(x^2-2x)(2x-2)}{(x-1)^4}\\&=\dfrac{-2}{(x-1)^3}\end{aligned}$, no inflection points
If $x>1$, then $f''(x)<0$ $\to$ concave down;
If $x<1$, then $f''(x)>0$ $\to$ concave up.

- (d) Asymptotes
	- vertical asymptote: 
		Because the zero of the denominator is $x=1$, $x=1$ is a vertical asymptote, i.e.$$\lim_{x\to 1^+}f(x)=-\infty$$ and $$\lim_{x\to 1^-}f(x)=\infty.$$
	- oblique asymptote:
		$x^2-x+1=x(x-1)+1$
		$f(x) = -x -\dfrac{1}{x-1}$
		$f(x)-(-x)=-\dfrac{1}{x-1}\to 0$ as $x\to \infty$ 
		Therefore, $y=-x$ is an oblique asymptote, i.e.$$\lim_{x\to\infty}f(x)=-x.$$
- (e) $f(-x) = \dfrac{x^2+x+1}{x+1}$, so $f$ is neither even nor odd.
	- $f$ is even $\iff f(x) = f(-x)$
	- $f$ is odd $\iff f(x) = -f(-x)$
- (f) 
	- points intersecting the $x$-axis: none
		$f(x)=0$ has no real roots.
	- points intersecting the $y$-axis: $(0,1)$ 
		$f(0)=-\dfrac{0^2-0+1}{0-1}=1$.
- (g) 
	- $f(x)>0\implies (x-1)<0\implies x<1$;
	- $f(x)<0\implies (x-1)>0\implies x>1$.





## Integration
5. $\int^0_{-\infty}e^{-|x|}\text{d}x=\int^0_{-\infty} e^{x}\text{d}x=(e^{x})|^0_{-\infty}=1-0=1$
## Techniques of Integration

## Covergence of Sequence
1. $a_n=\dfrac{1-5n^4}{n^4+8n^3}=\dfrac{n^{-4}-5}{1+8n^{-1}}$ converges to $\dfrac{0-5}{1+8\cdot 0}=-5$.
2. $a_n=\dfrac{3^n}{n^3}$ diverges because $3^n$ grows faster than $n^3$.
	Ratio test: $$\begin{align}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\left|\frac{\frac{3^{n+1}}{(n+1)^3}}{\frac{3^n}{n^3}}\right|=\lim_{n\to\infty}\left|\frac{3n^3}{(n+1)^3}\right|\\&=\lim_{n\to\infty}\left|\frac{3}{(1+n^{-1})^3}\right|=\lim_{n\to\infty}\left|\frac{3}{1+3n^{-1}+3n^{-3}+n^{-3}}\right|\\&=\left|\frac{3}{1+3\cdot 0+3\cdot 0+0}\right|=3>1\implies \sum^\infty_{n\to\infty}a_n~\text{diverges}\end{align}$$ 
3. $a_n=\dfrac{\sin^2n}{2^n}$ converges to $0$, because $\sin^2n$ is oscillating while $2^x$ grows exponentially.
4. $a_n=n-\sqrt{n^2-n}$
	Let $f(x)=x-\sqrt{x^2-x}=x\left(1-\sqrt{1-\frac{1}{x}}\right)$
	$\because \lim_{x\to\infty} 1=1$ and $\lim_{x\to\infty}\sqrt{1-\frac{1}{x}}=1$ $\qquad\therefore\lim_{x\to\infty}\left(1-\sqrt{1-\frac{1}{x}}\right)=0$
	$\lim_{x\to\infty}f(x)=0$
## Indeterminate form and L'Hôpital’s Rule
1. $u=1/x$
	$\lim_{u\to 0}\frac{5u^{-1}-2}{\sqrt{2u^{-2}+3}-\sqrt{u^{-2}-1}}=\lim_{u\to 0}\frac{5-2u}{\sqrt{2+3u^2}-\sqrt{1-u^2}}=\frac{5}{\sqrt{2}-\sqrt{1}}=5(\sqrt{2}+1)$
2. $\begin{align}\lim_{x\to 0}\frac{x^2\sin x}{1-\cos x}&=\lim_{x\to 0}\frac{2x\sin x+x^2\cos x}{\sin x}=\lim_{x\to 0}\left(2x+x^2\cot x\right)\\&=\lim_{x\to 0}(2x)+\lim_{x\to 0}(x^2\cot x)=0+\lim_{x\to 0}\frac{x^2}{\tan x}\\&=\lim_{x\to 0}\frac{2x}{\sec^2 x}=0\end{align}$