###### tags: `大一微積分`
# 彥傑
$\begin{align}\end{align}$
[Latex語法](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)
## 0220
定積分
1. $\begin{align}\int_{-1}^6|x-3| dx\end{align}$
$\begin{align}=\int_{-1}^33-x dx+\int_{3}^6x-3 dx=12-4+\frac{27}{2}-9=\frac{25}{2}\end{align}$
2. $\begin{align}\int_{-1}^1\frac{\tan x}{x^4-x^2+1} dx\end{align}$
$\begin{align}偶函數=0\end{align}$
3. $\begin{align}\int_1^2\sqrt{x^2-1}dx\end{align}$
$\begin{align}令x=\sec\theta, dx=\sec\theta\tan\theta d\theta\end{align}$
$\begin{align}=\int_0^\frac{\pi}{3}\tan\theta\cdot\sec\theta\tan\theta d\theta=\int_0^\frac{\pi}{3}\sec^3\theta-\sec\theta d\theta=[\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|]_0^\frac{\pi}{3}\end{align}$
$\begin{align}=\sqrt{3}+\frac{1}{2}\ln(2+\sqrt{3})\end{align}$
4. $\begin{align}\int_0^\frac{\pi}{2}\frac{1}{1+\sin x} dx\end{align}$
$\begin{align}=\int_0^\frac{\pi}{2}\frac{1-\sin x}{\cos^2 x}dx=\int_0^\frac{\pi}{2}\frac{1}{\cos^2 x}-\frac{\sin x}{\cos^2 x}dx=[\tan x-\sec x]_0^\frac{\pi}{2}\end{align}$
$\begin{align}=1\end{align}$
5. $\begin{align}\int_0^\frac{\pi}{2}\frac{\sin^3 x}{\sin^3 x+\cos^3 x} dx\end{align}$
$\begin{align}I=\int_0^\frac{\pi}{2}\frac{\sin^3 x}{\sin^3 x+\cos^3 x} dx\end{align}$
$\begin{align}令u=\frac{x}{2}-x\end{align}$
$\begin{align}=\int_\frac{\pi}{2}^0\frac{\cos^3u}{\cos^3u+\sin^3u}\cdot(-du)=\int_0^\frac{\pi}{2}\frac{\cos^3u}{\cos^3u+\sin^3u}\cdot(du)\end{align}$
$\begin{align}2I=\int_0^\frac{\pi}{2}1dx=\frac{\pi}{2} ,I=\frac{\pi}{4}\end{align}$
6. $\begin{align}\int_{-2}^1\frac{1}{x^2}dx\end{align}$
$\begin{align}=DNE\end{align}$
7. $\begin{align}\int_0^\infty\frac{\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x}dx\end{align}$

8. $\begin{align}\int_0^1\frac{\ln(x+1)}{(x^2+1)}dx\end{align}$

## 0213

判斷無窮級數的斂散性：

1. $\begin{align}\sum_{n=1}^\infty\frac{\sqrt[n]n!}{\prod\limits_{k=1}^n(2+\sqrt{k})}\end{align}$
$=\begin{align}\sum_{n=1}^\infty\frac{\sqrt[n]n!}{(2^n+\sqrt{n!})}\end{align}$
$\frac{\sqrt[n]n!}{(2^n+\sqrt{n!})}<\frac{\sqrt[n]n!}{(\sqrt{n!})}$
$\lim\limits_{n\to\infty}\frac{\sqrt[n]n!}{(\sqrt{n!})}=0\rightarrow\begin{align}\sum_{n=1}^\infty\frac{\sqrt[n]n!}{(\sqrt{n!})}conv.\rightarrow\sum_{n=1}^\infty\frac{\sqrt[n]n!}{(2^n+\sqrt{n!})}conv.\end{align}$
2. $\begin{align}\sum_{n=1}^\infty\frac{n^3(\sqrt{2}+(-1)^n)^n}{3^n}\end{align}$

3. $\begin{align}\sum_{n=1}^\infty\frac{(n!)^2}{(2n)!}\end{align}$
$\lim\limits_{n\to\infty}\frac{\frac{(n+1)!(n+1!)}{(2n+2)!}}{\frac{(n!)^2}{(2n)!}}=\lim\limits_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}=\frac{1}{4}<1$
$\begin{align}thus,\sum_{n=1}^\infty\frac{(n!)^2}{(2n)!}conv.\end{align}$
4. $\begin{align}\sum_{n=1}^\infty\frac{n!}{n^n}\end{align}$
$\lim\limits_{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{(n+1)}}}{\frac{n!}{n^n}}=\lim\limits_{n\to\infty}(\frac{n}{n+1})^n=\frac{1}{e} ,0<\frac{1}{e}<1$
$\begin{align}thus,\sum_{n=1}^\infty\frac{n!}{n^n}\end{align} conv.$
5. $\begin{align}\sum_{n=1}^\infty\frac{n^2}{(2+\frac{1}{n})^n}\end{align}$
$\lim\limits_{n\to\infty}\frac{n^2}{(1+\frac{n+1}{n})^{n}}=0\rightarrow conv.$
6. $\begin{align}\sum_{n=1}^\infty\frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}\end{align}$
$\frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}>\frac{n^{n}}{(n+\frac{1}{n})^n}$
$\lim\limits_{n\to\infty}\sqrt[n]{\frac{n^{n}}{(n+\frac{1}{n})^n}}=1\rightarrow fail$
$\frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}<\frac{n^{n+\frac{1}{n}}}{(n)^n}$
$\lim\limits_{n\to\infty}\frac{n^{n+\frac{1}{n}}}{(n)^n}=1\rightarrow fail$
$\lim\limits_{n\to\infty}{\frac{n^{n}}{(n+\frac{1}{n})^n}}=\lim\limits_{n\to\infty}[\frac{n^2}{n^2+1}]^n=$
7. $\begin{align}\sum_{n=1}^\infty\frac{[(n+1)!]^n}{2!4!\cdots(2n)!}\end{align}$
$\lim\limits_{n\to\infty}\frac{\frac{[(n+2)!]^{(n+1)}}{2!4!\cdots(2n+2)!}}{\frac{[(n+1)!]^n}{2!4!\cdots(2n)!}}=\lim\limits_{n\to\infty}\frac{\frac{[(n+2)!]^{(n+1)}}{[(n+1)!]^n}}{(2n+2)!}=\lim\limits_{n\to\infty}\frac{(n+2)^{n+1}\cdot (n+1)!}{(2n+2)!}$
8. $\begin{align}\sum_{n=1}^\infty\frac{1!+2!+\cdots+n!}{(2n)!}\end{align}$

9. $\begin{align}\sum_{n=1}^\infty\frac{(n!)^2}{2^{n^2}}\end{align}$
$\lim\limits_{n\to\infty}\frac{\frac{[(n+1)!]^2}{2^{n+1^2}}}{\frac{(n!)^2}{2^{n^2}}}=\lim\limits_{n\to\infty}\frac{(n+1)^2}{2^{2n+1}}$
10. $\begin{align}\sum_{n=1}^\infty(\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[5]{2})\cdots(\sqrt{2}-\sqrt[2n+1\;]{2})\end{align}$
$\lim\limits_{n\to\infty}\frac{(\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[5]{2})\cdots(\sqrt{2}-\sqrt[2n+3\;]{2})}{(\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[5]{2})\cdots(\sqrt{2}-\sqrt[2n+1\;]{2})}=\lim\limits_{n\to\infty}(\sqrt{2}-\sqrt[2n+3\;]{2})>1\rightarrow div.$
11. $\begin{align}\frac{1000}{1}+\frac{1000\cdot1001}{1\cdot3}+\frac{1000\cdot1001\cdot1002}{1\cdot3\cdot5}+\cdots\end{align}$
$\begin{align}=\sum_{n=1}^\infty\frac{\frac{(999+n)!}{999!}}{(2n-1)!}\end{align}$
$\begin{align}\lim\limits_{n\to\infty}\frac{\frac{\frac{(1000+n)!}{999!}}{(2n+1)!}}{\frac{\frac{(999+n)!}{999!}}{(2n-1)!}}=\lim\limits_{n\to\infty}\frac{1000+n}{2n(2n+1)}=0\rightarrow conv.\end{align}$
12. $\begin{align}\frac{1000}{1!}+\frac{1000^2}{2!}+\frac{1000^3}{3!}+\cdots+\frac{1000^n}{n!}+\cdots\end{align}$
$\begin{align}=\sum_{n=1}^\infty\frac{1000^n}{n!}\end{align}$
$\begin{align}\lim\limits_{n\to\infty}\frac{\frac{1000^{n+1}}{(n+1)!}}{\frac{1000^n}{n!}}=\lim\limits_{n\to\infty}\frac{1000}{n+1}=0\rightarrow conv.\end{align}$
13. $\begin{align}\frac{2\cdot1!}{1^1}+\frac{2^2\cdot2!}{2^2}+\frac{2^3\cdot3!}{3^3}+\cdots+\frac{2^n\cdot n!}{n^n}+\cdots\end{align}$
$\begin{align}=\sum_{n=1}^\infty\frac{2^n \cdot n!}{n^n}\end{align}$
$\begin{align}\lim\limits_{n\to\infty}\frac{\frac{2^{n+1}\cdot(n+1)!}{(n+1)^{n+1}}}{\frac{2^n\cdot n!}{n^n}}=\lim\limits_{n\to\infty}\frac{2}{(\frac{n+1}{n})^n}=\frac{2}{e}<1 \rightarrow conv.\end{align}$
14. $\begin{align}\frac{3\cdot1!}{1^1}+\frac{3^2\cdot2!}{2^2}+\frac{3^3\cdot3!}{3^3}+\cdots+\frac{3^n\cdot n!}{n^n}+\cdots\end{align}$
$\begin{align}=\sum_{n=1}^\infty\frac{3^n \cdot n!}{n^n}\end{align}$
$\begin{align}\lim\limits_{n\to\infty}\frac{\frac{3^{n+1}\cdot(n+1)!}{(n+1)^{n+1}}}{\frac{3^n\cdot n!}{n^n}}=\lim\limits_{n\to\infty}\frac{3}{(\frac{n+1}{n})^n}=\frac{3}{e}>1 \rightarrow div.\end{align}$
## 1014
1. $\begin{align}\lim\limits_{n\to\infty}\frac{n+\sqrt{n^2+1}}{\sqrt[3]{8n^3-n^2+2}}\end{align}$
$\begin{align}=1\end{align}$
2. $\begin{align}\lim\limits_{n\to\infty}\frac{n+2}{\sqrt{n^3(n+1)}-n^2}\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{n+2}{n^3}\times(\sqrt{n^3(n+1)}+n^2)\end{align}$
$\begin{align}=\frac{2}{1}=2\end{align}$
3. $\begin{align}\lim\limits_{n\to\infty}\frac{n}{\ln n}(\sqrt[n]{n}-1)\end{align}$
==$(x^n)'=nx^{n-1}$；$(n^x)'=(e^{x\ln n})'=\ln n\cdot n^x$==
==$(x^x)'=(e^{x\ln x})'=(\ln x+1)x^x$==
==$\begin{align}(x^\frac{1}{x})'=(e^{\frac{1}{x}\ln x})'=x^\frac{1}{x}\cdot(\frac{\ln x}{x})'\end{align}$==
==$\begin{align}\lim\limits_{n\to\infty}n^\frac{1}{n}=\lim\limits_{n\to\infty}e^{\frac{1}{n}\ln n}=e^{\lim\limits_{n\to\infty}\frac{\ln n}{n}}=e^0=1\end{align}$==
$\begin{align}=\lim\limits_{n\to\infty}\frac{n^{\frac{1}{n}}-1}{\frac{\ln n}{n}}=\lim\limits_{n\to\infty}\frac{n^\frac{1}{n}\cdot(\frac{\ln n}{n})'}{(\frac{\ln n}{n})'}\end{align}$
$\begin{align}=1\end{align}$
4. $\begin{align}\lim\limits_{n\to\infty}\frac{2^2+4^2+\cdots+(2n)^2}{n^3}\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{4(\frac{n(n+1)(2n+1)}{6})}{n^3}\end{align}$
$\begin{align}=\frac{4}{3}\end{align}$

5. $\begin{align}\lim\limits_{n\to\infty}(\frac{n^2+3n+1}{n^2+n-3})^{n+1}\end{align}$
$\begin{align}(n+1)\ln(\frac{n^2+3n+1}{n^2+n-3})\end{align}$
$\begin{align}=\frac{\ln(\frac{n^2+3n+1}{n^2+n-3})}{\frac{1}{n+1}}=\frac{n^2+n-3}{n^2+3n+1}\times\frac{(2n+3)(n^2+n-3)-(n^2+3n+1)(2n+1)}{(n^2+n-3)^2}\times[-(n+1)^2]\end{align}$
$\begin{align}\rightarrow 2 \end{align}$
$\begin{align}ans:e^2\end{align}$
==$\begin{align}\lim\limits_{n\to\infty}(1+\frac{1}{f(n)})^{f(n)}=e，\lim\limits_{n\to\infty}f(n)=\infty\end{align}$==
$\begin{align}\left[(1+\frac{1}{\frac{n^2+n-3}{2n+4}})^{(\frac{n^2+n-3}{2n+4})}\right]^\frac{(2n+4)(n+1)}{n^2+n-3}\rightarrow e^2\end{align}$
6. $\begin{align}\lim\limits_{n\to\infty}n^2(a^\frac{1}{n}-a^\frac{1}{n+1})\end{align}$
$\begin{align}=\frac{a^{\frac{1}{n}}-a^{\frac{1}{n+1}}}{\frac{1}{n^2}}\end{align}$
$\begin{align}=\frac{a^\frac{1}{n}\times(-\ln a\cdot n^{-2})-a^\frac{1}{n+1}\times (-\ln a\cdot (n+1)^{-2})}{-\frac{1}{n^3}}\end{align}$
$\begin{align}t=\frac{1}{n+1}, n=\frac{1}{t}-1\end{align}$
$\begin{align}=\frac{a^{\frac{1}{t}-1}-a^{\frac{1}{t}}}{(\frac{t}{1-t})^2}\end{align}$
$\begin{align}\Rightarrow\frac{(a^{-1}-1)(-\ln a\times t^{-2}\times a^\frac{1}{t})}{2\times\frac{t}{1-t}\times\frac{1}{(1-t)^2}}\end{align}$
$\begin{align}=\frac{-\ln a(a^{-1}-1)\cdot a^\frac{1}{t}}{2\cdot(\frac{t}{1-t})^3}\end{align}$
7. $\begin{align}\lim\limits_{n\to\infty}(\frac{n+1}{n+2})^{n+3}\end{align}$

## 0930
1. $\begin{align}\lim\limits_{x\to\infty}\ln(1+2^x)\ln(1+\frac{3}{x})\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}\ln(1+\frac{3}{x})^{\ln(1+2^x)}\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}\ln(1+\frac{3}{x})^{\ln2^x+\ln(1+2^{-x})}\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}\ln(1+\frac{3}{x})^{x\ln2}\cdot(1+\frac{3}{x})^{\ln(1+2^{-x})}\end{align}$
$\begin{align}=\ln e^{3\ln2}\cdot1=3\ln 2\end{align}$
1. $\begin{align}\lim\limits_{x\to\infty}(x^3\ln\frac{x+1}{x-1}-2x^2)\end{align}$
$\begin{align}=\lim\limits_{y\to0}\frac{\ln\frac{1+y}{1-y}-2y}{y^3}\end{align}$
$\begin{align}=\lim\limits_{y\to0}\frac{\frac{1-y}{1+y}\times\frac{1\cdot(1-y)+1\cdot(y+1)}{(1-y)^2}-2}{(3y^2)}\end{align}$
$\begin{align}=\lim\limits_{y\to 0}\frac{2y^2}{(3y^2)(1-y^2)}\end{align}$
$\begin{align}=\frac{2}{3}\end{align}$
1. $\begin{align}\lim\limits_{x\to 0}(x^2+2xe^x+e^{2x})^\frac{2}{\sin x}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}e^{(\frac{2}{\sin x})\ln(x^2+2xe^x+e^{2x})}\end{align}$
$\begin{align}=e^{\lim\limits_{x\to 0}(\frac{2}{\sin x})\ln(x^2+2xe^x+e^{2x})}\end{align}$
$\begin{align}=e^{\lim\limits_{x\to0}\frac{2\cdot2\cdot(1+e^x)}{\cos x\cdot(x+e^x)}}\end{align}$
$\begin{align}=e^{\frac{4\times2}{1\times1}}=e^8\end{align}$
1. $\begin{align}\lim\limits_{x\to\infty}(x\tan^{-1}\frac{1}{x})^{x^2}\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}e^{x^2\ln(x\tan^{-1}\frac{1}{x})}\end{align}$
$\begin{align}=e^{\lim\limits_{x\to\infty}x^2\ln(x\tan^{-1}\frac{1}{x})}\end{align}$
$\begin{align}\lim\limits_{x\to\infty}x^2\ln(x\tan^{-1}\frac{1}{x})=\lim\limits_{y\to 0}\frac{1}{y^2}\ln\frac{\tan^{-1}(y)}{y}\end{align}$
$\begin{align}\lim\limits_{y\to 0}\ln\frac{\tan^{-1}(y)}{y}=\ln\lim\limits_{y\to 0}\frac{\tan^{-1}(y)}{y}=\ln\lim\limits_{y\to 0}\frac{1}{1+y^2}=0\end{align}$
$\begin{align}\lim\limits_{y\to 0}\frac{\ln\tan^{-1}(y)-\ln(y)}{y^2}=\lim\limits_{y\to 0}\frac{\frac{1}{\tan^{-1}(y)}\cdot\frac{1}{1+y^2}-\frac{1}{y}}{2y}\end{align}$
均值定理：$f(x)$ 在 $(a,b)$ 可微分, 存在一個 $c\in(a,b)$ 使得 $\begin{align}f'(c)=\frac{f(a)-(b)}{a-b}\end{align}$
$\begin{align}\lim\limits_{y\to 0}\frac{\ln\frac{\tan^{-1}(y)}{y}-\ln1}{y^2}=\lim\limits_{y\to 0}\frac{\frac{1}{c}\cdot(\frac{\tan^{-1}(y)}{y}-1)}{y^2}=\lim\limits_{y\to 0}\frac{\tan^{-1}(y)-y}{cy^3}\end{align}$
$\begin{align}\lim\limits_{y\to 0}\frac{\frac{1}{y^2+1}-1}{3cy^2}=\lim\limits_{y\to 0}\frac{1-(y^2+1)}{3cy^2(y^2+1)}=\lim\limits_{y\to 0}\frac{-1}{3c(y^2+1)}=-\frac{1}{3}\end{align}$
由均值定理, 存在 $c$，使得 $\begin{align}\frac{\tan^{-1}(y)}{y}<c<1\end{align}$
==$\begin{align}\sin(x)<x<\tan(x)<\sec(x)\Rightarrow \tan^{-1}(y)<\tan(\tan^{-1}(y))=y\end{align}$==
$\begin{align}\frac{1}{c}=f'(c)=\frac{\ln\frac{\tan^{-1}(y)}{y}-\ln1}{\frac{\tan^{-1}(y)}{y}-1}\end{align}$
## 0923
極限的定義：Epsilon-Delta Definition
Let $f(x)$ be a function defined on a open interval around $x_0$.

$$\begin{align}\lim\limits_{x\to x_0} f(x)=L\end{align}$$

if $\forall\;\epsilon>0，\exists\;\delta>0$ such that $\forall\;x$，$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$
1. Formal Definition of Epsilon-Delta Limits
$0<|x-x_0|<\delta\Rightarrow -\delta<x-x_0<\delta\Rightarrow x_0-\delta<x<x_0+\delta\Rightarrow (x_0-\delta,x_0+\delta)$
$|f(x)-L|<\epsilon\Rightarrow (L-\epsilon,L+\epsilon)$
\
==Ex1.== For the function $f(x)=3x^2+2x+1$, Alice wants Bob to show that $$\begin{align}\lim\limits_{x\to 2}f(x)=17\end{align}$$ using the $\epsilon-\delta$ definition of a limit. Alice says, "I bet you can't choose a real number $\delta$" so that for all $x$ in $(2-\delta, 2+\delta)$, we'll have that $|f(x)-17|<0.5$." Which of the following four choices is the largest $\delta$ that Bob could give so that he completes Alice's chanllenge?
(A)0.05 (B)0.03 ($\text{C}$)0.02 (D)0.01.
\
$\begin{align}-\frac{1}{2}<3x^2+2x+1-17<\frac{1}{2}\Rightarrow 6x^2+4x-31>0；6x^2+4x-33<0\end{align}$
$\begin{align}31<6x^2+4x<33\end{align}$
$\begin{align}31<6(x-2)^2+28(x-2)+32<33\end{align}$
(B)
\
==Ex2.== $\begin{align}\lim\limits_{x\to 2}\frac{2x^2-4}{x-1}=4\end{align}$
$\forall\;\epsilon>0$，$\exists\;\delta>0$
Claim：$0<|x-2|<\delta$ such that $\begin{align}|\frac{2x^2-4}{x-1}-4|<\epsilon\end{align}$
$\begin{align}|\frac{2x^2-4x}{x-1}|<\epsilon\Rightarrow|x-2|\cdot|\frac{2x}{x-1}|<\epsilon\Rightarrow|x-2|\cdot|2+\frac{2}{x-1}|<\epsilon\end{align}$
$\begin{align}\delta=\frac{1}{2}帶入\end{align}$
$\begin{align}-\frac{1}{2}< x-2<\frac{1}{2}\Rightarrow0.5< x-1<1.5\Rightarrow\frac{4}{3}<\frac{2}{x-1}<4\end{align}$
==$\begin{align}\frac{10}{3}<2+\frac{2}{x-1}<6\end{align}$==
$\begin{align}|x-2||\frac{2x}{x-1}|<6\le\frac{\epsilon}{6}\cdot6=\epsilon\end{align}$
$\begin{align}取\delta=\min{(\frac{1}{2},\frac{\epsilon}{6})}\end{align}$
==Ex3.==$\begin{align}\lim\limits_{x\to3}\frac{2x^2+1}{x-2}=19\end{align}$
$\begin{align}\forall\;\epsilon>0,\exists\;\delta>0\end{align}$
Claim:$0<|x-3|<\delta$ such that$\begin{align}|\frac{2x^2+1}{x-2}-19|<\epsilon\end{align}$
$\begin{align}|\frac{2x^2-19x+39}{x-2}|<\epsilon\Rightarrow|x-3||\frac{2x-13}{x-2}|<\epsilon\Rightarrow|x-3||2-\frac{9}{x-2}|<\epsilon\end{align}$
$\begin{align}\delta=\frac{1}{2}帶入\end{align}$
$\begin{align}0<|x-3|<\frac{1}{2}\Rightarrow\frac{5}{2}<x<\frac{7}{2}\Rightarrow\frac{2}{3}<\frac{1}{x-2}<2\Rightarrow-18<\frac{-9}{x-2}<-6\end{align}$
$\begin{align}4<|\frac{2x-13}{x-2}|<16\end{align}$
$\begin{align}|x-3||\frac{2x-13}{x-2}|<\delta\times16\le\frac{\epsilon}{16}\times16\end{align}$
$\begin{align}\delta取\min(\frac{1}{2},\frac{\epsilon}{16})\end{align}$
==Ex4.==$\begin{align}\lim\limits_{x\to3}x^2=9\end{align}$
$\begin{align}\forall\;\epsilon>0,\exists\;\delta>0\end{align}$
Claim:$0<|x-3|<\delta$ such that$\begin{align}|x^2-9|<\epsilon\end{align}$
$\begin{align}|(x+3)(x-3)|<\epsilon\end{align}$
$\begin{align}\delta=1代入\end{align}$
$\begin{align}2<x<4\Rightarrow5<x<7\end{align}$
$\begin{align}取\delta=\min(1,\frac{\epsilon}{7})\end{align}$
$\begin{align}|x+3||x-3|<\delta\times7\le\frac{\epsilon}{7}\times7\end{align}$
2. Infinite Limit for finite $x$
$\begin{align}\lim\limits_{x\to 1}\frac{1}{x-1}\end{align}$
3. Finite Limit at Infinity
$\begin{align}\lim\limits_{x\to\infty}\frac{x}{2x+1}\end{align}$
4. Limits Does Not Exist
5. Finding Delta given an Epsilon

## 0908
1. 試求 $\begin{align}\lim\limits_{x\to\infty}\cos^{-1}(\sqrt{x^2+x}-x)\end{align}$
==若 $f(x)$ 為一連續函數 (continue)，則 $\begin{align}\lim\limits_{x\to a}f(g(x))=f(\lim\limits_{x\to a}g(x))\end{align}$==
$\begin{align}=\cos^{-1}(\lim\limits_{x\to\infty}(\sqrt{x^2+x}-x))\end{align}$
$\begin{align}=\cos^{-1}(\lim\limits_{x\to\infty}\frac{x}{\sqrt{x^2+x}+x})\end{align}$
$\begin{align}=\frac{\pi}{3}\end{align}$
1. 試求 $\begin{align}\lim\limits_{n\to\infty}(\sin(\pi\sqrt{n^2+1})+\sin^2(\pi\sqrt{n^2+n}))\end{align}$ 之值，其中 $n$ 為正整數。
==$\sin(\pi\sqrt{n^2+1})=\pm\sin(\pi\sqrt{n^2+1}-n\pi)=\begin{align}\pm\sin\frac{\pi}{\sqrt{n^2+1}+n}\end{align}$==
$\begin{align}=\lim\limits_{x\to\infty}(0+(\pm1)^2)=1\end{align}$

1. $\begin{align}\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})\end{align}$
==$\sin(x+y)=\sin x\cos y+\cos x\sin y$==
==$\sin(x-y)=\sin x\cos y-\cos x\sin y$==
==兩式相減 $\sin(x+y)-\sin(x-y)=2\cos x\sin y$==
==$x+y=A，x-y=B\Rightarrow x=\begin{align}\frac{A+B}{2}，y=\frac{A-B}{2}\end{align}$==
==$\begin{align}\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\end{align}$==
==積化和差、和差化積==
$\left\{\begin{align}2\sin x\cos y=\sin(x+y)+\sin(x-y)\\ 2\cos x\sin y=\sin(x+y)-\sin(x-y)\\ 2\cos x\cos y=\cos (x+y)+\cos(x-y)\\-2\sin x\sin y=\cos(x+y)-\cos(x-y)\end{align}\right.$
$\left\{\begin{align}\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}\\ \sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\\\cos x+\cos y=2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\\cos x-\cos y=-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}\end{align}\right.$
$\begin{align}=\lim\limits_{x\to\infty}2\cos\frac{\sqrt{x+1}+\sqrt{x}}{2}\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\end{align}$
\
==$\begin{align}-\frac{1}{x}\leq\lim\limits_{x\to\infty}(\frac{1}{x}\sin x)\leq\frac{1}{x}\end{align}$==
\
$\begin{align}-2\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\le2\cos\frac{\sqrt{x+1}+\sqrt{x}}{2}\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\le2\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\end{align}$
1. 已知 $m，n$ 為正整數，試求 $\begin{align}\lim\limits_{x\to\pi}\frac{\sin mx}{\sin nx}\end{align}$ 之值。
==$\begin{align}\lim\limits_{x\to 0}\frac{\sin x}{x}=1\end{align}$==
令 $\begin{align}y=x-\pi\Rightarrow\lim\limits_{y\to 0}\frac{\sin m(\pi+y)}{\sin n(\pi+y)}=\lim\limits_{y\to 0}\frac{(-1)^m\sin(my)}{(-1)^n\sin (ny)}\end{align}$
$\begin{align}=(-1)^{m-n}\lim\limits_{y\to 0}\frac{\sin (my)}{\sin (ny)}=(-1)^{m-n}\frac{m}{n}\end{align}$

1. 試求 $\begin{align}\lim\limits_{x\to-\infty}\frac{\sqrt{x^2+3x}}{\sqrt[3]{x^3-2x^2}}\end{align}$ 之值。
$\begin{align}=\lim\limits_{x\to-\infty}\frac{-\sqrt{\frac{x^2}{(-x)^2}+\frac{3x}{(-x)^2}}}{\sqrt[3]{\frac{x^3}{x^3}-\frac{2x^2}{x^3}}}\end{align}$
$\begin{align}=\frac{-1}{1}\end{align}$
1. 假設 $\begin{align}\lim\limits_{x\to\infty}(\sqrt[3]{1+x^2+x^3}-ax-b)=0\end{align}$，試求數對 $(a,b)$ 之值。
$\begin{align}=\lim\limits_{x\to\infty}(\sqrt[3]{1+x^2+x^3}-x+x-ax-b)\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}[(\frac{1+x^2+x^3-x^3}{(1+x^2+x^3)^{\frac{2}{3}}+(1+x^2+x^3)^{\frac{1}{3}}x+x^2})+(1-a)x-b]\end{align}$
$\begin{align}\Rightarrow a=1,b=\frac{1}{3}\end{align}$
1. 試求 $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n\cos\frac{x}{2^k}$ 之值。
$\begin{align}=\lim\limits_{n\to\infty}\cos\frac{x}{2}\cdot\cos\frac{x}{2^2}\cdots\cos\frac{x}{2^n}\end{align}$
==$\begin{align}\sin 2\theta=2\sin\theta\cos\theta\Rightarrow\cos\theta=\frac{\sin2\theta}{2\sin\theta}\end{align}$==
$=\begin{align}\lim\limits_{n\to\infty}\frac{\sin x}{2\sin\frac{x}{2}}\cdot\frac{\sin\frac{x}{2}}{2\sin\frac{x}{2^2}}\cdots\frac{\sin\frac{x}{2^{n-1}}}{2\sin\frac{x}{2^n}}\end{align}$
$=\begin{align}\lim\limits_{n\to\infty}\frac{\sin x}{2^n\cdot\sin\frac{x}{2^n}}=\frac{\sin x}{x}\lim\limits_{n\to\infty}\frac{\frac{x}{2^n}}{\sin\frac{x}{2^n}}=\frac{\sin x}{x}\end{align}$

1. $\begin{align}\lim\limits_{n\to\infty}(1+\frac{x}{n})^n\end{align}$
$\begin{align}=e\end{align}$
1. $\begin{align}\lim\limits_{x\to\infty}\ln(1+2^x)\ln(1+\frac{3}{x})\end{align}$
$\begin{align}\end{align}$
1. $\begin{align}\lim\limits_{x\to\infty}(x^3\ln\frac{x+1}{x-1}-2x^2)\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}(x^2)(\ln\frac{x+1}{e^2(x-1)})\end{align}$
1. $\begin{align}\lim\limits_{x\to 0}(x^2+2xe^x+e^{2x})^\frac{2}{\sin x}\end{align}$

1. $\begin{align}\lim\limits_{x\to\infty}(x\tan^{-1}\frac{1}{x})^{x^2}\end{align}$

## 0901
1. $\begin{align}\lim\limits_{n\to\infty}\sum_{k=1}^{4n}\frac{n}{n^2+k^2}\end{align}$
$\begin{align}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{\sqrt{k(k+1)}}{n^2}\end{align}$
$\begin{align}\sum_{k=1}^n\frac{k}{n^2}\le\sum_{k=1}^n\frac{\sqrt{k(k+1)}}{n^2}\le\sum_{k=1}^n\frac{k+1}{n^2}\end{align}$
$\begin{align}\frac{n(n+1)}{2n^2}\le\sum_{k=1}^n\frac{\sqrt{k(k+1)}}{n^2}\le\frac{(n+1)(n+2)}{2n^2}\end{align}$
$\begin{align}\lim\limits_{n\to\infty}\frac{n(n+1)}{2n^2}\le\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{\sqrt{k(k+1)}}{n^2}\le\lim\limits_{n\to\infty}\frac{(n+1)(n+2)}{2n^2}\end{align}$
$\begin{align}\frac{1}{2}\le\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{\sqrt{k(k+1)}}{n^2}\le\frac{1}{2}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{4n^2+k}}\end{align}$
$\begin{align}\sum_{k=1}^n\frac{1}{\sqrt{4n^2+1}}\le\sum_{k=1}^n\frac{1}{\sqrt{4n^2+k}}\le\sum_{k=1}^n\frac{1}{\sqrt{4n^2+n}}\end{align}$
$\begin{align}\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{4n^2+1}}\le\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{4n^2+k}}\le\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{4n^2+n}}\end{align}$
$\begin{align}\frac{1}{2}\le\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{4n^2+k}}\le\frac{1}{2}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{\sqrt{2n^2+kn}}{n^2}\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\sqrt{2+\frac{k}{n}}\end{align}$
$\begin{align}=\int_0^1\sqrt{2+x}dx=\frac{2}{3}(2+x)^{\frac{3}{2}}|_0^1=2\sqrt{3}-\frac{4\sqrt{2}}{3}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{2\sqrt{n^2+k}}\end{align}$
$\begin{align}\frac{n}{2\sqrt{n^2+n}}\le\sum_{k=1}^n\frac{1}{2\sqrt{n^2+k}+\cdots}\le\frac{n}{2\sqrt{n^2+1}}\end{align}$
$\begin{align}\frac{1}{2}\le\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{2\sqrt{n^2+k}}\le\frac{1}{2}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\frac{(1^2+2^2+\cdots+n^2)(1^5+2^5+\cdots+n^5)}{(1^3+2^3+\cdots+n^3)(1^4+2^4+\cdots+n^4)}\end{align}$
==黎曼和$\begin{align}\int_0^1f(x) dx=\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{n}f(\frac{k}{n})\end{align}$==
$\begin{align}=\lim\limits_{n\to\infty}\frac{\frac{1}{n}\cdot((\frac{1}{n})^2+(\frac{2}{n})^2+\cdots+(\frac{n}{n})^2)+\frac{1}{n}\cdot((\frac{1}{n})^5+(\frac{2}{n})^5+\cdots+(\frac{n}{n})^5)}{\frac{1}{n}\cdot((\frac{1}{n})^3+(\frac{2}{n})^3+\cdots+(\frac{n}{n})^3)+\frac{1}{n}\cdot((\frac{1}{n})^4+(\frac{2}{n})^4+\cdots+(\frac{n}{n})^4)}\end{align}$
$\begin{align}=\frac{\int_0^1x^2 dx+\int_0^1x^5 dx}{\int_0^1 x^3 dx+\int_0^1 x^4 dx}\end{align}$
$\begin{align}=\frac{\frac{1}{3}+\frac{1}{6}}{\frac{1}{4}+\frac{1}{5}}\end{align}$
$\begin{align}=\frac{10}{9}\end{align}$

1. $\begin{align}\lim\limits_{n\to\infty}\frac{n(1^2+2^2+\cdots+n^2)}{1^3+2^3+\cdots+n^3}\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}n\cdot\frac{n(n+1)(2n+1)}{6}\cdot[\frac{2}{n(n+1)}]^2=\frac{4}{3}\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{\frac{1}{n}\cdot((\frac{1}{n})^2+(\frac{2}{n})^2+\cdots+(\frac{n}{n})^2)}{\frac{1}{n}\cdot((\frac{1}{n})^3+(\frac{2}{n})^3+\cdots+(\frac{n}{n})^3)}\end{align}$
=$\begin{align}\frac{\int_0^1x^2dx}{\int_0^1x^3dx}=\frac{4}{3}\end{align}$
1. $\begin{align}a_n=\frac{\sqrt{1\times 2}+\sqrt{2\times 3}+\cdots+\sqrt{n(n+1)}}{n^2}\end{align}$
$\begin{align}\frac{n(n+1)}{2n^2}<a_n<\frac{n(n+3)}{2n^2}\end{align}$
$\begin{align}\frac{1}{2}\leq\lim\limits_{n\to\infty}a_n\leq\frac{1}{2}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}(1+\frac{1}{2}+\cdots+\frac{1}{2^n})(1+\frac{1}{3}+\cdots+\frac{1}{3^n})\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}3(1-(\frac{1}{2})^n)(1-(\frac{1}{3})^n)=3\end{align}$
1. 已知 $\begin{align}\lim\limits_{n\to\infty}(\frac{3n^2+2n+5}{2n^2+n-1}+6a_n)=4\end{align}$，求 $\begin{align}\lim\limits_{n\to\infty}a_n\end{align}$
$\begin{align}a_n=\frac{5}{12}+...\end{align}$
$\begin{align}\lim\limits_{n\to\infty}a_n=\frac{5}{12}\end{align}$
1. 設 $n$ 為正整數，坐標平面上有一等腰三角形，它的三頂點分別是 $(0,2)、(\frac{1}{n},0)、(-\frac{1}{n},0)$。假設此三角形的外接圓直徑長等於 $D_n$，則 $\begin{align}\lim\limits_{n\to\infty}D_n\end{align}$
$\begin{align}O=(0,2-r_n)\end{align}$
$\begin{align}(r_n)^2=(\frac{1}{n})^2+(2-r_n)^2\end{align}$
$\begin{align}2D_n=\frac{1}{n^2}+4\end{align}$
$\begin{align}\lim\limits_{n\to\infty}D_n=2\end{align}$
1. $\begin{align}a=\frac{\pi}{3}，b=\sqrt{2}，c=0.99，x_n=\frac{2a^n}{3+4a^n}+\frac{5b^n}{6+7b^n}+\frac{8c^n}{9+10c^n}\end{align}$，求$\begin{align}\lim\limits_{n\to\infty}x_n\end{align}$
$\begin{align}\lim\limits_{n\to\infty}x_n=\frac{1}{2}+\frac{5}{7}+0=\frac{17}{14}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\frac{1+4+7+\cdots+(3n-2)}{n^2}\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{\frac{3n(n+1)}{2}-2n}{n^2}=\frac{3}{2}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}(\frac{n^2-1}{n+1}-\frac{n^2+2}{n+2})\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{2n^2-n^2+\cdots}{(n+1)(n+2)}=1\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}(\sqrt{n^2+2n+3}-\sqrt{n^2-5n+1})\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\frac{7n-2}{\sqrt{n^2+2n+3}+\sqrt{n^2-5n+1}}=\frac{7}{2}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\sqrt{n}(\sqrt{n+3}-\sqrt{n-5})\end{align}$
$\begin{align}=\lim\limits_{n\to\infty}\sqrt{n}\cdot\frac{8}{\sqrt{n-3}+\sqrt{n-5}}=4\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\frac{7\cdot 4^n+6\cdot 2^n}{5\cdot 4^n+2\cdot 3^n}\end{align}$
$\begin{align}=\frac{7}{5}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\frac{3^n+2^{3n-2}}{5^{n-1}}\end{align}$
$\begin{align}=\infty\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\frac{3^{2n-1}-2^{4n+1}}{4^{2n+2}+5^{n+3}}\end{align}$
$\begin{align}=\frac{-2}{16}=-\frac{1}{8}\end{align}$
1. $\begin{align}\lim\limits_{n\to\infty}\sqrt[n]{2^n+3^n+2\cdot5^n}\end{align}$
$\begin{align}=5\cdot1=5\end{align}$
## 0825
1. 已知 $\begin{align}x\in(0,\frac{\pi}{2})\end{align}$，試比較 $\begin{align}\frac{\tan x}{x}\end{align}$ 和 $\begin{align}\frac{x}{\sin x}\end{align}$ 兩者的大小關係。
若 $\begin{align}\frac{\tan x}{x}>\frac{x}{\sin x}\Leftrightarrow \frac{\sin^2 x}{\cos x}>x^2\Leftrightarrow \frac{\sin x}{\sqrt{\cos x}}-x>0\end{align}$
$\begin{align}f(x)=\frac{\sin x}{\sqrt{\cos x}}-x\end{align}$
(1) $f'(x)>0$ 單調遞增 (2) $f(0)>0$
$\begin{align}f'(x)=\frac{\cos x\sqrt{\cos x}-\sin x\cdot\frac{-\sin x}{2\sqrt{\cos x}}}{\cos x}-1\end{align}$
$\begin{align}f'(x)=\frac{\cos^2 x+1-2\cos x\sqrt{\cos x}}{2\cos x\sqrt{\cos x}}\end{align}$
$\begin{align}f'(x)=\frac{(\cos x\sqrt{\cos x}-1)^2+\cos^2 x(1-\cos x)}{2\cos x\sqrt{\cos x}}>0\end{align}$
$\begin{align}f(x)>f(0)=0\end{align}$
由MVT，存在 $0<c_x<x$，使得 $\begin{align}\frac{\sin x-\sin 0}{x-0}=\cos c_x<1\end{align}$
$\begin{align}\frac{\sin x}{x}<1\end{align}$
由MVT，存在 $0<d_x<x$，使得 $\begin{align}\frac{\tan x -\tan 0}{x-0}=\sec^2 d_x>1\end{align}$
$\begin{align}\frac{\tan x}{x}>1\end{align}$

1. $200^{199}$、$199^{200}$
$200^\frac{1}{200}<199^\frac{1}{199}$
$200^{199}<199^{200}$

1. 比較 $e^\pi$ 和 $\pi^e$ 的大小關係
若 $\begin{align}e^\pi>\pi^e\Leftrightarrow e^\frac{1}{e}>\pi^\frac{1}{\pi}\end{align}$
$\begin{align}(2^x)'=(e^{x\ln 2})'=(\ln 2)\cdot 2^x\end{align}$
$\begin{align}f(x)=x^\frac{1}{x}=e^{\frac{1}{x}\ln x}\end{align}$
$\begin{align}f'(x)=x^\frac{1}{x}\cdot(\frac{1-\ln x}{x^2})=0\end{align}$
$\begin{align}\ln x=1\Rightarrow x=e\end{align}$

1. $\begin{align}\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})\end{align}$
由 MVT，存在 $\sqrt{x}<c<\sqrt{x+1}$，使得
$\begin{align}\cos c=\frac{\sin\sqrt{x+1}-\sin\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}\end{align}$
$\begin{align}\sin\sqrt{x+1}-\sin\sqrt{x}=\cos c(\sqrt{x+1}-\sqrt{x})\end{align}$
$\begin{align}0\leq|\sin\sqrt{x+1}-\sin\sqrt{x}|\leq|\sqrt{x+1}-\sqrt{x}|\end{align}$
$\begin{align}\lim\limits_{x\to\infty}\sqrt{x+1}-\sqrt{x}=\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0\end{align}$
因此原式 $=0$

1. $\begin{align}\lim\limits_{x\to 0^+}\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}\end{align}$
$\begin{align}f(x)=\tan x\end{align}$
由 MVT，存在 $\sin x<c<\tan x$，使得
$\begin{align}\sec^2 c=\frac{\tan(\tan x)-\tan (\sin x)}{\tan x-\sin x}\end{align}$
$\begin{align}\tan(\tan x)-\tan(\sin x)=\sec^2 c(\tan x-\sin x)>1\cdot(\tan x-\sin x)\end{align}$
$\begin{align}\lim\limits_{x\to 0^+}\tan x-\sin x=0\end{align}$
## 0817
1. $\begin{align}\lim\limits_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{(\tan(\tan x)-\tan x)+(\tan x -x)-(\sin(\sin x)-\sin x)-(\sin x-x)}{(\tan x-x)-(\sin x-x)}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{\frac{\tan^3 x}{x^3}\frac{(\tan(\tan x)-\tan x)}{\tan ^3 x}+\frac{(\tan x -x)}{x^3}-\frac{\sin^3 x}{x^3}\frac{(\sin(\sin x)-\sin x)}{\sin^3 x}-\frac{(\sin x-x)}{x^3}}{\frac{(\tan x-x)}{x^3}-\frac{(\sin x-x)}{x^3}}\end{align}$
$\begin{align}=\frac{1\cdot\frac{1}{3}+\frac{1}{3}-1\cdot(-\frac{1}{6})+(\frac{1}{6})}{\frac{1}{3}+\frac{1}{6}}=2\end{align}$
$\begin{align}\lim\limits_{x\to 0}\frac{\tan x-x}{x^3}=\frac{1}{3}\end{align}$
$\begin{align}\tan x=0+x+0+\frac{1}{3}x^3+0+\frac{2}{15}x^5+O(x)\end{align}$
在 $x=0$ 處的泰勒展開式我們稱為
Maclaurin series（麥克勞林級數）
1. $\begin{align}\lim\limits_{x\to 0}\frac{x^2 e^x}{\cos x-1}\end{align}$
$\begin{align}=-2\end{align}$(cos泰勒後是2次 )
1. $\begin{align}\lim\limits_{x\to 0}\frac{\frac{x^2}{2}-1+\cos x}{x^4}\end{align}$
$\begin{align}=\frac{1}{24}\end{align}$
$\begin{align}f(x)=1+0+\frac{-1}{2!}x^2+0+\frac{1}{4!}x^4+O(x)\end{align}$
1. $\begin{align}\lim\limits_{x\to 0}\frac{\sin x-x}{x^3}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{\frac{-1}{3!}x^3+\frac{1}{5!}x^5+O(x)}{x^3}\end{align}$
$\begin{align}=-\frac{1}{6}\end{align}$
$\begin{align}f(x)=0+1x+0+\frac{-1}{3!}x^3+0+\frac{1}{5!}x^5+O(x)\end{align}$
1. $\begin{align}\lim\limits_{x\to 0}\frac{e^x-1-x}{x^2}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{(1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+O(x))-1-x}{x^2}=\frac{1}{2}\end{align}$
$\begin{align}e^x=1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3\end{align}$
1. $\begin{align}\lim\limits_{x\to 0}\frac{x-\sin x}{x^2\sin x}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{x-(x-\frac{1}{3!}x^3+\frac{1}{5!}x^5+O(x))}{x^2(x-\frac{1}{3!}x^3+\frac{1}{5!}x^5+O(x))}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{\frac{1}{6}-\frac{1}{5!}x^2+O(x)}{1-\frac{1}{6}x^2+O(x)}=\frac{1}{6}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{1-\cos x}{2x\sin x+x^2\cos x}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{\sin x}{(2\sin x+2x\cos x)+(2x\cos x+x^2(-\sin x))}\end{align}$
$\begin{align}=\lim\limits_{x\to 0}\frac{\cos x}{2\cos x+2\cos x+2x(-\sin x)+2\cos x+2x(-\sin x)+2x(-\sin x)+x^2(-\cos x)}\end{align}$
$\begin{align}=\frac{1}{6}\end{align}$
Taylor series $f(x)$ 在 $x=a$ 處的泰勒展開式：
$\begin{align}f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n+O(x)\end{align}$
$\begin{align}\sin x=0+1x+0+\frac{-1}{3!}x^3+0+\frac{1}{5!}x^5=\sum_{k=1}^n\frac{(-1)^{k+1}}{(2k-1)!}x^{2k-1}+O(x)\end{align}$
1. $\begin{align}\lim\limits_{x\to\infty}(\sqrt[3]{x^3+3x}-\sqrt{x^2-2x})\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}(\sqrt[3]{x^3+3x}-x)+(x-\sqrt{x^2-2x})\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}\frac{3x}{(x^3+3x)^{\frac{2}{3}}+\sqrt{x^3+3x}\cdot x+x^2}+\frac{2x}{x+\sqrt{x^2-2x}}\end{align}$
$\begin{align}=0+1\end{align}$
:::info
另法：因式分解的公式
$(a^6-b^6)=(a-b)(a^5+a^4b+a^3b^2+\cdots+ab^4+b^5)$
$(a-b)^3=a^3-3a^2b+3ab^2-b^3$
$a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)^3+3ab(a-b)$
$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)^3-3ab(a+b)$
$(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$
$\begin{align}=\lim\limits_{x\to\infty}\frac{(x^3+3x)^2-(x^2-2x)^3}{[(x^3+3x)^\frac{5}{3}+(x^3+3x)^\frac{4}{3}(x^2-2x)^\frac{1}{2}+\cdots+(x^2-2x)^\frac{5}{2}]}\end{align}$
$\begin{align}=\lim\limits_{x\to\infty}\frac{x^6+6x^4+9x^2-x^6+6x^5-12x^4+8x^3}{[(x^3+3x)^\frac{5}{3}+(x^3+3x)^\frac{4}{3}(x^2-2x)^\frac{1}{2}+\cdots+(x^2-2x)^\frac{5}{2}]}\end{align}$
$\begin{align}=\frac{6}{6}\end{align}$
:::

## 0804
1. $\begin{align}\int\frac{\sqrt{1-\sqrt[3]{x}}}{\sqrt{x}}dx\end{align}$
令 $\begin{align}x=\sin^6\theta，\int\frac{\cos\theta}{\sin^3\theta}\cdot 6\sin^5\theta\cos\theta d\theta=\int6\sin^2\theta\cos^2\theta d\theta\end{align}$
$\begin{align}=\frac{3}{2}\int\sin^22\theta d\theta=\frac{3}{2}\int\frac{1-\cos4\theta}{2}d\theta=\frac{3}{4}\theta-\frac{3}{16}\sin4\theta+c\end{align}$
$\begin{align}=\frac{3}{4}\sin^{-1}\sqrt[6]{x}-\frac{3}{16}\sin(4\sin^{-1}\sqrt[6]{x})+c\end{align}$

## 0728
1. $\int\frac{xdx}{(3-2x-x^2)^\frac{3}{2}}$
$=\int\frac{xdx}{[-(x+1)^2+4]^{\frac{3}{2}}}$
#先配方後 可把分母簡單化
令 $x+1=2\sin \theta，dx=2\cos\theta d\theta$
$=\int\frac{2\sin \theta-1}{8\cos^3 \theta}\cdot 2\cos \theta d\theta$
$=\int\frac{2\sin\theta-1}{4\cos^2 \theta}d\theta$
$=\int\frac{\sin \theta}{2\cos^2 \theta}-\frac{1}{4\cos^2 \theta}d\theta$
$=\frac{1}{2}\sec \theta-\frac{1}{4}\tan\theta+c$
$=\frac{1}{\sqrt{3-2x-x^2}}-\frac{x+1}{4\sqrt{3-2x-x^2}}+c$
:::warning
$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$
:::

2. $\int x\sin(2x) dx$
$=x\cdot(-\frac{1}{2}\cos 2x)-\int(-\frac{1}{2}\cos 2x)dx$
$=x\cdot(-\frac{1}{2}\cos 2x)+\frac{1}{4}\sin2x+c$
$=\frac{1}{4}(\sin 2x-2x\cos 2x)+c$
:::danger
$\int udv=uv-\int vdu$
$\int x\sin xdx=\frac{1}{2}x^2\cdot\sin x-\int\frac{1}{2}x^2\cdot(\cos x) dx$
$\int x\sin xdx=x(-\cos x)-\int(-\cos x) dx$
:::
$\int x\cos 2xdx=x(\frac{1}{2}\sin 2x)-\int\frac{1}{2}\sin 2xdx=x\frac{1}{2}\sin 2x+\frac{1}{4}\cos 2x+C$

3. $\int_0^\infty x\sin (2x)e^{-x} dx$

| $v$ | $du$ |
|:----:|:----:|
| $e^{-x}$ | $\sin(2x)$ |
| $-e^{-x}$|$-\frac{1}{2}\cos(2x)$
|$e^{-x}$|$-\frac{1}{4}\sin(2x)$|

$\begin{align}\int e^{-x}\sin(2x) dx=-\frac{1}{2}e^{-x}\cos(2x)-\frac{1}{4}e^{-x}\sin(2x)-\frac{1}{4}\int e^{-x}\sin(2x)dx\end{align}$
$\begin{align}\Rightarrow \int_0^\infty e^{-x}\sin(2x)dx=-\frac{2}{5}e^{-x}\cos(2x)-\frac{1}{5}e^{-x}\sin(2x)=\frac{2}{5}\end{align}$

$\begin{align}\int e^{-x}\cos 2x dx=\frac{1}{2}e^{-x}\sin 2x-\frac{1}{4}e^{-x}\cos 2x-\frac{1}{4}\int e^{-x}\cos 2x dx\end{align}$
$\begin{align}\Rightarrow\int_0^\infty e^{-x}\cos 2x dx=\frac{2}{5}e^{-x}\sin 2x-\frac{1}{5}e^{-x}\cos 2x=\frac{1}{5}\end{align}$

$\begin{align}原式=x\cdot(-\frac{2}{5}e^{-x}\cos(2x)-\frac{1}{5}e^{-x}\sin(2x))-\int_0^\infty(-\frac{2}{5}e^{-x}\cos(2x)-\frac{1}{5}e^{-x}\sin(2x))dx\end{align}$
$\begin{align}=0+\frac{2}{5}\int_0^\infty e^{-x}\cos(2x) dx+\frac{1}{5}\int_0^\infty e^{-x}\sin(2x) dx\end{align}$
$\begin{align}=\frac{2}{5}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{2}{5}=\frac{4}{25}\end{align}$

4. $\int\frac{2x^2}{2x^4+6x^3+9x^2+6x+2} dx$
分母：$\begin{align}=(2x^2+2x+1)(x^2+2x+2)\end{align}$
:::info
除法原理：被除式$f(x)=$除式$g(x)\times$商式$Q(x)+$餘式$R(x)$，$\deg R(x)<\deg g(x)$，或 $R(x)=0$
$\begin{align}\frac{1}{(x+2)(x+3)}=\frac{1}{x+2}-\frac{1}{x+3}\end{align}$
$\begin{align}1=1\cdot(x+3)-1\cdot (x+2)\end{align}$
$\begin{align}(x+2)=(x+3)\cdot 1+1\end{align}$
:::
$\int\frac{ax+b}{x^2+2x+2}+\frac{cx+d}{2x^2+2x+1}dx$
$2x^2=(ax+b)(2x^2+2x+1)+(cx+d)(x^2+2x+2)$
$\left\{\begin{align}2a+c=0\\2a+2b+2c+d=2\\a+2b+2c+2d=0\\b+2d=0\end{align}\right.\Rightarrow$ $\begin{align}a=\frac{4}{5}，b=\frac{12}{5}，c=-\frac{8}{5}，d=-\frac{6}{5}\end{align}$
$\begin{align}\frac{4}{5}\int\frac{x+3}{x^2+2x+2}dx-\frac{2}{5}\int\frac{4x+3}{2x^2+2x+1}dx\end{align}$
$=\begin{align}\frac{2}{5}\int\frac{2x+2+4}{x^2+2x+2}dx-\frac{2}{5}\int\frac{4x+2-1}{2x^2+2x+1}dx\end{align}$
$\begin{align}=\frac{2}{5}\ln(x^2+2x+2)-\frac{2}{5}\ln(2x^2+2x+1)+\frac{8}{5}\int\frac{1}{(x+1)^2+1}dx+\frac{2}{5}\int\frac{1}{2(x+\frac{1}{2})^2+\frac{1}{2}}dx\end{align}$
$\begin{align}\int\frac{1}{(x+1)^2+1}dx=\int\frac{1}{\sec^2 \theta}\cdot\sec^2 \theta d\theta=\tan^{-1}(x+1)+c\end{align}$
$\begin{align}\int\frac{1}{2(x+\frac{1}{2})^2+\frac{1}{2}}dx=\frac{1}{2}\int\frac{1}{\frac{1}{4}\cdot\sec^2 \theta}\cdot\frac{1}{2}\sec^2 \theta d\theta=\tan^{-1}(2x+1)+c\end{align}$

## 0721
1. $\int_0^\frac{\pi}{2}\frac{\sin x}{\sin x+\cos x}dx$
$令\tan\frac{x}{2}=t$,$dt=\frac{1}{2}\sec^2 \frac{x}{2} dx$
$=\int_0^1\frac{\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}dt$
$=\int_0^1\frac{4t}{(t^2+1)(-t^2+2t+1)}dt$
$=\frac{1}{2}\ln(t^2+1)+\tan^{-1}t-\frac{1}{2}\ln|t^2-2t-1||_0^1$
$=\frac{\pi}{4}$
$\frac{4t}{(t^2+1)(t^2-2t-1)}=\frac{at+b}{t^2+1}+\frac{ct+d}{t^2-2t-1}$
$4t=(at+b)(t^2-2t-1)+(ct+d)(t^2+1)$
$a+c=0，-2a+b+d=0，-a-2b+c=4，-b+d=0$
$a=b=d=-1，c=1$
$\frac{4t}{(t^2+1)(t^2-2t-1)}=\frac{-t-1}{t^2+1}+\frac{t-1}{t^2-2t-1}=-\frac{1}{2}\frac{2t}{t^2+1}-\frac{1}{t^2+1}+\frac{1}{2}\frac{2t-2}{t^2-2t-1}$
令 $\theta=\frac{\pi}{2}-x$
原式 $I=\int_0^\frac{\pi}{2}\frac{\cos \theta}{\cos\theta+\sin\theta} d\theta=\int_0^\frac{\pi}{2}\frac{\cos x}{\sin x+\cos x} dx$
$I=\int_a^bf(x)dx+\int_a^b g(x)dx=\int_a^b(f(x)+g(x))dx$
$2I=\int_0^\frac{\pi}{2} 1 dx=x|_0^\frac{\pi}{2}=\frac{\pi}{2}\Rightarrow I=\frac{\pi}{4}$

1. $\int\sin^3 xdx$
cos x的微分是-sin x
$=\int(1-\cos^2 x)\sin x dx$
$=-\cos x+\frac{1}{3}\cos^3 x+c$

1. $\int_0^\pi\sqrt{\sin^3 x-\sin ^5 x}dx$
$=\int_0^\pi\sqrt{\sin^3 x(1-\sin^2 x)}dx$
$=\int_0^\pi\sqrt{\sin^3 x(\cos^2 x)}dx$
$=\int_0^\pi\sin^{\frac{3}{2}}x |\cos x| dx$
分段來去絕對值
$=\int_0^\frac{\pi}{2}\sin^{\frac{3}{2}} x\cos x dx-\int_\frac{\pi}{2}^\pi\sin^{\frac{3}{2}}x\cos x dx$
$=\frac{2}{5}\sin^{\frac{5}{2}}x|_0^\frac{\pi}{2}-\frac{2}{5}\sin^{\frac{5}{2}}x|_\frac{\pi}{2}^\pi$
$=\frac{2}{5}+\frac{2}{5}$
$=\frac{4}{5}$
:::warning
$(u^n)^\prime=n\cdot u^{n-1}\cdot u^\prime$
$\int u^n\cdot u^\prime dx=\frac{1}{n+1}u^{n+1}+c$
$\int\sin^n x\cos xdx=\frac{1}{n+1}\sin^{n+1} x+c$
:::

1. $\int\frac{1}{a+b\sin x} dx$，$a>0,b>0$
令 $t=\tan\frac{x}{2}，\sin x=\frac{2t}{1+t^2}$
$\frac{1}{2}\sec^2 \frac{x}{2}dx=dt$
$=\int\frac{1}{a+\frac{2bt}{1+t^2}}\cdot\frac{2}{1+t^2}dt$
$=\int\frac{2}{at^2+2bt+a}dt$
$=\frac{2}{a}\int\frac{1}{(t+\frac{b}{a})^2+(1-\frac{b^2}{a^2})}dt$
配方後湊tan^2+1
令 $t+\frac{b}{a}=\sqrt{1-\frac{b^2}{a^2}}\tan\theta$
$=\frac{2}{a}\int\frac{1}{(1-\frac{b^2}{a^2})\sec^2 \theta}dt$
$=\frac{2}{a}\cdot\frac{1}{\sqrt{1-\frac{b^2}{a^2}}}\theta+c$

2. $\int_\frac{\pi}{4}^\frac{\pi}{3}\frac{\tan x}{\ln\cos x} dx$
$\int\tan xdx=\int\frac{\sin x}{\cos x} dx=-\ln|\cos x|+c$
$=-\ln|\ln \cos x| |_\frac{\pi}{4}^\frac{\pi}{3}$
$=-\ln\ln2+\ln\ln\sqrt{2}$
3. $\int\sec^3x dx$

$u'=\sec^2 x,v=\sec x$

$=\sec x\tan x-\int\sec x\tan^2 xdx$
$\int\sec^3 xdx=\frac{1}{2}(\sec x\tan x-\ln|\sec x+\tan x|)+c$

1. $\int e^x\sin x dx$
$=-e^x\cos x+e^x\sin x-\int e^x\sin x dx$
$\Rightarrow 2\int e^x\sin x dx=e^x(\sin x-\cos x)$
$\Rightarrow \int e^x\sin x dx=\frac{1}{2}e^x(\sin x-\cos x)+c$

2. $\int(\sin^{-1}x)^2 dx$
令 $\sin^{-1}x=\theta$
$\sin\theta=x , dx=\cos\theta d\theta$
$=\int\theta^2\cos\theta d\theta$
$=\theta^2\cdot\sin\theta+2\theta\cos\theta-2\int\cos\theta d\theta+c$
$=x(\sin^{-1}x)^2+2\sin^{-1}x\cdot\sqrt{1-x^2}-2x+c$

3. $\int \sin^{-1}x dx$
令 $\sin^{-1}x=\theta$
$\sin\theta=x , dx=\cos\theta d\theta$
$=\int\theta\cdot\cos\theta d\theta$
$(uv)'=u'v+uv'\Rightarrow uv'=(uv)'-u'v$
$\Rightarrow \int udv=uv-\int vdu$
$=\theta\cdot\sin\theta-\int\sin\theta d\theta$
$=x\sin^{-1}x+\sqrt{1-x^2}+c$

4. $\int\frac{x^7}{x^4+2} dx$
$=\int x^3+\frac{-2x^3}{x^4+2}dx$
$=\frac{1}{4}x^4-\frac{1}{2}\ln(x^4+2)+c$
## 0714 極值作業
1. $\lim\limits_{n\to\infty}\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}$
$\frac{n}{\sqrt{n^2+n}}<\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}<\frac{n}{\sqrt{n^2+1}}$
$=1$
$\frac{1}{4\times 5}=\frac{1}{4}-\frac{1}{5}$
1. $\lim\limits_{x\to0}\frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}$
$=\lim\limits_{x\to0}\frac{1+x\sin x-\cos x}{(x\tan x)(\sqrt{1+x\sin x}+\sqrt{\cos x})}$
$=\lim\limits_{x\to0}\frac{2\sin^2 \frac{x}{2}+x\sin x}{(x\tan x)(\sqrt{1+x\sin x}+\sqrt{\cos x})}$
$=\frac{\frac{1}{2}+1}{1\cdot2}$
$=\frac{3}{4}$
::: info
$\lim\limits_{x\to0}\frac{x \tan x}{\sin^2 x}=\lim\limits_{x\to0}\frac{\frac{\tan x}{x}}{\frac{\sin^2 x}{x^2}}=1$
:::

3. $\lim\limits_{x\to0}\frac{1+\sin x-\cos x}{1+\sin px-\cos px}$，$p$ 為常數
$=\lim\limits_{x\to0}\frac{\sin x+2\sin^2 \frac{x}{2}}{\sin px+2\sin^2 \frac{px}{2}}$
$=\lim\limits_{x\to0}\frac{\frac{\sin x}{x}+\frac{2\sin^2 \frac{x}{2}}{x}}{\frac{\sin px}{x}+\frac{2\sin^2 \frac{px}{2}}{x}}$
$=\frac{1}{p}$
:::info
$\lim\limits_{x\to0}\frac{\sin^2 x}{x}=0$
:::

4. $\lim\limits_{x\to0}\frac{\sqrt{2-2\cos x}}{x}$
$=\lim\limits_{x\to0}\frac{\sqrt{4\sin^2 \frac{x}{2}}}{x}$
$=\lim\limits_{x\to0}\frac{|2\sin\frac{x}{2}|}{x}$
$\lim\limits_{x\to0^+}\frac{2\sin\frac{x}{2}}{x}=1$
$\lim\limits_{x\to0^-}\frac{-2\sin\frac{x}{2}}{x}=-1$
:::info
$\begin{align}\lim\limits_{x\to0^+}\frac{|x|}{x}=\lim\limits_{x\to0^+}\frac{x}{x}=1，\lim\limits_{x\to0^-}\frac{|x|}{x}=\lim\limits_{x\to0^-}\frac{-x}{x}=-1\end{align}$
:::

5. $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{\sin x+1}}{x^3}$
$=\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3(\sqrt{\tan x+1}+\sqrt{\sin x+1})}$
$=\lim\limits_{x\to0}\frac{\sin x}{x}(\frac{1}{x^2}\cdot\frac{1-\cos x}{\cos x})\frac{1}{\sqrt{\tan x+1}+\sqrt{\sin x+1}}$
$=1\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$

:::info
$\lim\limits_{x\to0}\frac{\sin x}{x}=1$，$\lim\limits_{x\to0}\frac{1-\cos x}{x}=0$，$\lim\limits_{x\to0}\frac{\tan x}{x}=1$
半角公式 $\sin^2 x=\frac{1-\cos 2x}{2}$，$\cos^2 x=\frac{1+\cos 2x}{2}$
$\begin{align}\lim\limits_{x\to0}\frac{1-\cos x}{x^2}=\lim\limits_{x\to0}\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{2}\lim\limits_{x\to0}\frac{\sin^2\frac{x}{2}}{(\frac{x}{2})^2}=\frac{1}{2}\end{align}$
:::

6. $\lim\limits_{x\to a}\frac{(2x-a)^m-a^m}{x^n-a^n}$，$m，n$ 為自然數
==因式分解的公式 $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+b^{n-1})$==
$=\lim\limits_{x\to a}\frac{[(2x-a)-a][(2x-a)^{m-1}+(2x-1)^{m-2}a+\cdots +a^{m-1}]}{(x-a)(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1})}$
$=\frac{2(ma^{m-1})}{na^{n-1}}=\frac{2m}{n}a^{m-n}$
$=\lim\limits_{x\to a}\frac{2m(2x-a)^{m-1}}{nx^{n-1}}$

7. $\lim\limits_{x\to0}\frac{(1+2x)^5-(1+4x)^3}{x}$
$=\lim\limits_{x\to 0}\frac{32x^5+16x^4-56x^3-12x^2-2x}{x}$
$=-2$ ==可以用羅必達試試==
$=\lim\limits_{x\to 0}\frac{10(1+2x)^4-12(1+4x)^2}{1}$
$=-2$
8. $\lim\limits_{x\to0}\frac{2x}{\sqrt{x+5}-\sqrt{5}}$
$=\lim\limits_{x\to 0}\frac{2x(\sqrt{x+5}+\sqrt{5})}{x}$
$=4\sqrt{5}$ ==對嗎？==
9. $\lim\limits_{x\to2}\frac{\sqrt{5x-1}-\sqrt{2x+5}}{x^2-4}$
$=\lim\limits_{x\to 2}\frac{3(x-2)}{(x^2-4)(\sqrt{5x-1}+\sqrt{2x+5})}$
$=\frac{1}{8}$

1. $\lim\limits_{x\to2}\frac{\sqrt[3]{3x+2}-2}{x-2}$
$=\lim\limits_{x\to 2}\frac{3(x-2)}{(x-2)(\sqrt[3]{(3x+2)^2}+2(\sqrt[3]{3x+2})+4)}$
$=\frac{1}{4}$

1. $\lim\limits_{n\to\infty}n\left(\sqrt[3]{\frac{n-1}{n+2}}-1\right)$
$=\lim\limits_{n\to \infty}\frac{n(\frac{-3}{n+2})}{\sqrt[3]{(\frac{n-1}{n+2})^2}+\sqrt[3]{\frac{n-1}{n+2}}+1}$
1. $\lim\limits_{n\to\infty}n\left(1-\sqrt{\frac{2n-1}{2n}}\right)$
$=\lim\limits_{n\to \infty}n(\frac{1}{2n(1-\sqrt{\frac{2n-1}{2n}})})$

1. $\lim\limits_{n\to\infty}\frac{\sqrt{n^4+3n^3-6}-(n-1)(n+1)}{n}$
$=\lim\limits_{n\to \infty}\frac{n^4+3n^3-6-n^4+2n^2-1}{n(\sqrt{n^4+3n^2-6}+n^2-1)}$
$=\frac{5}{2}$

1. $\lim\limits_{n\to\infty}\left[\sqrt{n^2+4n+5}-(n-1)\right]$
$=\lim\limits_{n\to \infty}\frac{6n+4}{\sqrt{n^2+4n+5}+(n-1)}$
$=3$
1. $\lim\limits_{n\to\infty}\sqrt{n}(\sqrt{n+2}-\sqrt{n+1})$
$=\lim\limits_{n\to \infty}\sqrt{n}\cdot\frac{1}{\sqrt{n+2}+\sqrt{n+1}}$
$=\frac{1}{2}$
1. $\lim\limits_{n\to\infty}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{(2n-1)(2n+1)}\right]$
$=\lim\limits_{n\to \infty}\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+...+(\frac{1}{2n-1}-\frac{1}{2n+1})]$
$=\frac{1}{2}$
1. $\lim\limits_{x\to 0}\frac{\tan x-\sin x}{\sin^3 x}$
$=\lim\limits_{x\to 0}\frac{\sec^2x-\cos x}{3\sin^2x\cdot\cos x}$
$=$
1. $\lim\limits_{x\to 0}\frac{\tan(\frac{x}{2})}{x}=$？
$=\lim\limits_{x\to 0}\frac{\frac{1}{2}\sec^2x}{1}$
$=\frac{1}{2}$
1. $\lim\limits_{x\to\infty} x(\sqrt{x^2+1}-x)=$？
$=\infty$
1. $\lim\limits_{x\to \infty}(\sqrt{4x^2+3x+1}-\sqrt{4x^2-3x-2})=$？
$=\lim\limits_{x\to \infty}\frac{6x+3}{\sqrt{4x^2+3x+1}-\sqrt{4x^2-3x-2}}$
1. $\lim\limits_{x\to 1}\frac{x+x^2+\cdots+x^n-n}{x-1}$
$=\lim\limits_{x\to 1}\frac{(x-1)[x^{n-1}+(n-1)\cdot x^{n-2}+...+n]}{x-1}$
$= \frac{n(n+1)}{2}$
$=\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+(x^3-1)+\cdots+(x^n-1)}{x-1}$
$=1+2+3+\cdots+n$
1. $\lim\limits_{x\to 1}\frac{x^k-1}{x-1}=$？
$=\lim\limits_{x\to1}(x^{k-1}+x^{k-2}+...+1)$
$= k$

1. $\lim\limits_{x\to 0}\frac{\sqrt[3]{x+1}-1}{x}=$？
$=\lim\limits_{x\to 0}\frac{x}{x((x+1)^\frac{2}{3}+(x+1)^\frac{1}{3}+1)}$
$=\frac{1}{3}$
1. $\lim\limits_{x\to\infty}\frac{(2x-3)^{20}(3x+2)^{30}}{(5x+1)^{50}}=$？
$=\frac{2^{20}3^{30}}{5^{50}}$
1. $\lim\limits_{x\to 1}(\frac{1}{1-x}-\frac{3}{1-x^3})=$？
$=\lim\limits_{x\to1}\frac{(1-x)^2(-x-2)}{(1-x^2)(x^2+x+1)}$
$=-1$
1. $\lim\limits_{x\to \infty}(\frac{x^3}{2x^2-1}-\frac{x^2}{2x-1})=$？
$=\lim\limits_{x\to\infty}\frac{x^3-x^2}{4x^3-2x^2-2x+1}$
$=\frac{1}{4}$
1. $\lim\limits_{x\to 1}\frac{\sqrt{3-x}-\sqrt{1+x}}{x^2-1}=$？
$=\lim\limits_{x\to1}\frac{2-2x}{(x^2-1)(\sqrt{3-x}+\sqrt{1+x})}$
$=-\frac{\sqrt{2}}{4}$
1. $\lim\limits_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}=$？
$=\lim\limits_{x\to4}\frac{(2)(x-4)(\sqrt{x-2}+\sqrt{2})}{(x-4)(\sqrt{2x+1}+3)}$
$=\frac{2\sqrt{2}}{3}$
1. $\lim\limits_{n\to\infty}(\sqrt{n^2+2n-1}-\sqrt[3]{n^3+2n^2-1})$

$\lim\limits_{n\to\infty}(\sqrt{n^2+2n-1}$<font color="#f00">$-n+n$</font>$-\sqrt[3]{n^3+2n^2-1})$
$=\lim\limits_{n\to\infty}\frac{2n-1}{\sqrt{n^2+2n-1}+n}+\frac{-2n^2+1}{n^2+n\sqrt[3]{n^3+2n^2-1}+\sqrt[3]{(n^3+2n^2-1)^2}}$
$=1-\frac{2}{3}=\frac{1}{3}$
::: danger
這樣根號消不掉，三次根號要三次方才可以消掉，
所以請用立方差公式 $(a^3-b^3)=(a-b)(a^2+ab+b^2)$
:::

## 0713
1. $\int\frac{1}{x\ln x\ln(\ln x)}dx$
$令u=\ln\ln x$
$du=\frac{1}{x\ln x}dx$
$=\ln| \ln\ln x|+c$
1. $\int\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}dx$
$令u=\sqrt{x+1}+1$
$du=\frac{1}{2\sqrt{x+1}} dx=\frac{1}{2u-2}dx$
$=\int\frac{u-2}{u}\cdot(2u-2)du$
$=u^2-6u+4\ln|u|+c$
$=x-4\sqrt{x+1}+4\ln(\sqrt{x+1}+1)+c$
1. $\int\frac{x^5}{\sqrt{1-x^2}}dx$
$令x=\sin \theta$
$dx=\cos \theta d\theta$
$=\int\frac{\sin^5 \theta}{\cos \theta}\cdot\cos \theta d\theta$
$=\int\sin^5 \theta d\theta$
$=\int(-(1-\cos^2\theta)^2)(-\sin\theta d\theta)$
$=\int(-1+2\cos^2\theta-\cos^4\theta)(-\sin\theta d\theta)$
$=-\cos\theta+\frac{2}{3}\cos^3\theta-\frac{1}{5}\cos^5\theta+c$
1. $\int\frac{x^2+2}{(x+1)^3} dx$
$令u=x+1$
$du=dx$
$=\int\frac{u^2-2u+3}{u^3}du$
$=\ln|u|+\frac{2}{u}-\frac{3}{2u^2}+c$
1. $\int\frac{x}{4+x^4} dx$
$\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}$
$令 x^2=2\tan\theta$
$2xdx=2\sec^2\theta d\theta$
$xdx=\sec^2\theta d\theta$
$=\int\frac{1}{4\sec^2\theta}\cdot\sec^2\theta d\theta$
$=\frac{1}{4}\tan^{-1}\frac{x^2}{2}+c$

1. $\int\frac{3x+1}{\sqrt{x^2+2x-5}}dx$
$=\int\frac{3x+1}{\sqrt{(x+1)^2-6}}dx$
$令x+1=\sqrt{6}\sec\theta$
$dx=\sqrt{6}\sec\theta\tan\theta d\theta$
$=\int\frac{3\sqrt{6}\sec \theta-2}{\sqrt{6}\tan \theta}\cdot\sqrt{6}\sec \theta\tan \theta d\theta$
$=3\sqrt{6}\tan \theta-2\ln|\sec \theta+\tan \theta|+c$
1. $\int\frac{1}{x^2\sqrt{x^2-1}}dx$
$令x=\sec \theta$
$dx=\sec \theta\tan \theta d\theta$
$=\sin \theta+C$
$=\frac{\sqrt{x^2-1}}{x}+c$

1. $\int\sqrt{1+\sqrt{x}}dx$
$令u=1+\sqrt{x}$
$du=\frac{1}{2\sqrt{x}}dx，dx=(2u-2)du$
$=\int\sqrt{u}\cdot(2u-2)du$
$=\frac{4}{5}u^{\frac{5}{2}}-\frac{4}{3}u^{\frac{3}{2}}+c$
1. $\int \frac{1}{1+\cos x}dx$
可能方向：
$1.半角公式 1+\cos x=2\cos^2\frac{x}{2}$
$=\int \frac{1}{2}\sec^2\frac{x}{2}dx$
$=\tan\frac{x}{2}+c$
$2.乘共軛因子$
$=\int\frac{1-\cos x}{\sin^2 x} dx$
$=\int\csc^2 x-\csc x\cot x dx$
$=-\cot x+\csc x+c$
$3.萬能公式$
$=\int\frac{1}{1+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}dx$
$=\int \frac{1+\tan^2\frac{x}{2}}{2}dx$
1. $\int\frac{1}{1+\sin x}dx$
$=\int \frac{1-\sin x}{\cos^2x}dx$
$=\int\sec^2x-\tan x\sec xdx$
$=\tan x-\sec x+c$
## 0707
1. $\int \frac{1}{(x^2+1)^2} dx$
$x=\tan\theta$
$dx=\sec^2\theta d\theta$
$=\int \cos^2\theta d\theta$
$=\int\frac{1+\cos2\theta}{2}d\theta$
$=\frac{1}{2}\theta+\frac{1}{4}\sin2\theta+c$
$=\frac{1}{2}\tan^{-1}x+\frac{x}{2x^2+2}+c$
1. $\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx$
$x^{\frac{1}{6}}=u$
$du=\frac{1}{6}x^{-\frac{5}{6}}dx$
$=\int\frac{1}{u^3+u^2}\cdot6x^{\frac{5}{6}}du$
$=\int\frac{1}{u^3+u^2}\cdot6u^5du$
$=\int\frac{6u^3}{u+1}du$
$=\int 6(u^2-u+1)-\frac{6}{u+1}du$
$=2u^3-3u^2+6u-6\ln|u+1|+c$
1. $\int \frac{1}{\sqrt{a^2+x^2}}dx$
$x=a\tan\theta$
$dx=a\sec^2\theta d\theta$
$=\int\frac{1}{a\sec\theta}\cdot a\sec^2\theta d\theta$
$=\ln|\sec\theta+\tan\theta|+c$
$=\ln|\frac{\sqrt{x^2+a^2}+x}{a}|+c$
1. $\int \sqrt{a^2-x^2}dx$
$x=a\sin \theta$
$dx=a\cos \theta d\theta$
$=\int (a\cos \theta\cdot)^2d\theta$
$=a^2\int\frac{1+\cos2\theta}{2}d\theta$
$=\frac{1}{2}a^2(\theta+\frac{1}{2}\sin2\theta)+c$
1. $\int \frac{1-\ln x}{(x-\ln x)^2}dx$
$=\int \frac{x-\ln x+(1-x)}{(x-\ln x)^2} dx$
$=\int\frac{1}{x-\ln x}+\frac{1-x}{(x-\ln x)^2} dx$
$令u=x-\ln x$
$du=1-\frac{1}{x}dx$
$dx=\frac{x}{x-1}du\Rightarrow (x-1)dx=x du$
$原式=\int\frac{1}{u}-\frac{x}{u^2} du$
$=\ln|u|+\frac{x}{u}-\int\frac{1}{u}+c$
$=\frac{x}{x-\ln x}+c$

---
$\frac{x}{x-\ln x}=\frac{1}{1-\frac{\ln x}{x}}$
觀察 $\frac{d}{dx}(1-\frac{\ln x}{x})=-\frac{1-\ln x}{x^2}$
原式 $\int\frac{\frac{1-\ln x}{x^2}}{(1-\frac{\ln x}{x})^2}dx=\int\frac{-u'}{u^2} du$

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1. $\int \frac{x+1}{x^2-3x+1}dx$
$=\frac{1}{2}\int\frac{(2x-3)+5}{x^2-3x+1}dx$
$=\frac{1}{2}\ln|x^2-3x+1|+\frac{\sqrt{5}}{2}\ln|\frac{2x-3-\sqrt{5}}{2x-3+\sqrt{5}}|+c$

## 0629
作業：整理分式多項式的積分，針對分母是2次，分子是$n$次。
1. $\int\frac{1}{x^2-3x+1} dx$
$=\int\frac{1}{(x-\frac{3}{2})^2-\frac{5}{4}}dx$
$=\int\frac{1}{(x-\frac{3}{2}+\frac{\sqrt{5}}{2})(x-\frac{3}{2}-\frac{\sqrt{5}}{2}))}dx$
$=\int\frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}-\frac{1}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right)dx$
$=\frac{1}{\sqrt{5}}\ln\left|\frac{2x-3-\sqrt{5}}{2x-3+\sqrt{5}}\right|+c$
3. $\int\frac{2x-3}{x^2-3x+1} dx$
$=\ln|x^2-3x+1|+c$
3. $\int\frac{1+\ln x}{2+(x\ln x)^2}dx$
$=\frac{\sqrt{2}}{2}\arctan \frac{x\ln x}{\sqrt{2}}+c$
$\int\frac{1}{1+x^2} dx=\tan^{-1}x$
$\int\frac{1}{a^2+x^2} dx=\frac{1}{a}\tan^{-1}\frac{x}{a}$
3. $\int\frac{\tan^{-1}\sqrt{x}}{\sqrt{x}(1+x)}dx$
$u=\tan^{-1}\sqrt{x}$，$\tan u=\sqrt{x}$，$\sec^2 u du=\frac{1}{2\sqrt{x}} dx$，$dx=2\tan u\sec^2 u du$
$=\int\frac{u}{\tan u\sec^2 u}\cdot 2\tan u\sec^2 udu$
$=\int 2u du$
$=u^2+c=(\tan^{-1}\sqrt{x})^2+c$
3. $\int\frac{1}{\sin2x\cos x}dx$
$=\int\frac{\sin^2x+\cos^2x}{2\sin x\cos^2 x} dx$
$=\int\frac{\sin x}{2\cos^2x}+\frac{1}{2\sin x} dx$
$=\frac{1}{2}\sec x-\frac{1}{2}\ln|\csc x+\cot x|+c$
3. $\int\frac{\sin x}{1+\sin x} dx$
$=\int\frac{1+\sin x-1}{1+\sin x} dx=\int 1-\frac{1}{1+\sin x} dx$
令 $u=\tan\frac{x}{2}$，$du=\frac{1}{2}\sec^2\frac{x}{2} dx=\frac{1}{2}(1+u^2) dx\Rightarrow dx=\frac{2}{1+u^2} du$
$=\int \left(1-\frac{1}{1+\frac{2u}{1+u^2}}\right)\cdot\frac{2}{1+u^2} du$
$=\int\frac{2}{1+u^2}-\frac{2}{(1+u)^2}du$
$=\frac{2}{1+u}+2\tan^{-1}u+c=\frac{2}{1+\tan\frac{x}{2}}+x+c$
3. $\int\cos 3x\cos 2x dx$
$=\frac{1}{10}\sin 5x+\frac{1}{2}sin x+c$
回憶公式：
$\cos(3x+2x)=\cos3x\cos2x-\sin3x\sin2x$
$\cos(3x-2x)=\cos3x\cos2x+\sin3x\sin2x$
兩式相加
$\frac{1}{2}(\cos5x+\cos x)=\cos3x\cos2x$

1. $\int\frac{\ln\tan x}{\sin x\cos x} dx$
要先考慮 $u'$ 是誰
$u'=\frac{\sec^2x}{\tan x}=\frac{1}{\sin x \cdot\cos x}$
$原式=\int u du=\frac{1}{2}\cdot(\ln\tan x)^2+c$

## 0609

作業問題
1. $\int\sec^3 x\tan^5 xdx$
令 $\sec x=u$，$\sec x\tan x dx=du$
$=\int u^2(u^2-1)^2 du$
$=\frac{1}{7}u^7-\frac{2}{5}u^5+\frac{1}{3}u^3+C$
$=\frac{1}{7}\sec^7x-\frac{2}{5}\sec^5x+\frac{1}{3}\sec^3x+C$

1. $\int\sin^4x dx$
$=\int(\frac{1-\cos2x}{2})^2dx$
$=\frac{1}{4}\int1-2\cos 2x+\frac{1+\cos 4x}{2}dx$
$=\frac{1}{4}x-\frac{1}{2}\cdot\frac{1}{2}\sin 2x+\frac{1}{8}x+\frac{1}{8}\cdot\frac{1}{4}\sin 4x+C$

1. $\int\sin^5 x dx$ 整理, $\int\sin^{2k+1}xdx$、$\int\sin^{2k}xdx$、$\int\sin^{2k+1}x\cos^{2n+1}xdx$
$\int\sin^3x\cos^5xdx=-\int(1-\cos^2x)\cos^5xd\cos x$
$=\int-(1-u^2)\cdot u^5du$

1. $\int\sin^3 x dx$ 由公式 $\sin 3x=3\sin x-4sin^3 x$
$=-\frac{1}{4}\int(\sin 3x-3\sin x)dx$
$=\frac{1}{12}\cos 3x-\frac{3}{4}\cos x+c$
[另法] 令 $u=\cos x$，$du=-\sin x dx$
$=\int-(1-u^2)du$
$=\frac{1}{3}u^3-u+C$
$=\frac{1}{3}\cos^3x-\cos x+C$

1. $\int\frac{1}{\sqrt{a^2-x^2}} dx$
$=\int\frac{1}{a\cos\theta}\cdot a\cos\theta d\theta$
$=\theta+c$
$=\sin^{-1}\frac{x}{a}+c$

1. $\int\frac{1}{a^2-x^2}dx$
$=\int-\frac{1}{2a}\cdot(\frac{1}{x-a}-\frac{1}{x+a}) dx$
$=-\frac{1}{2a}\ln|\frac{x-a}{x+a}|+c$

1. $\int\frac{1}{a^2+x^2}dx$
$=\int\frac{1}{a^2\sec^2\theta}\cdot a\sec^2\theta d\theta$
$=\frac{1}{a}\tan^{-1}\frac{x}{a}+c$

3. $\int\frac{1}{a^2\cos^2 x+b^2\sin^2 x}dx$
$=\int\frac{\sec^2 x}{a^2+b^2\tan^2 x} dx$
$=\int\frac{1}{a^2+b^2 u^2} du$
令 $bu=a\tan\theta\Rightarrow b du=a\sec^2\theta d\theta$
$=\int\frac{1}{a^2\sec^2\theta}\cdot \frac{a}{b}\sec^2\theta d\theta$
$=\frac{1}{ab}\tan^{-1}\frac{b\tan x}{a}+C$

1. $\int\frac{1}{x(1+2\ln x)}dx$
$=\frac{1}{2}\ln|1+2\ln x|+c$

1. $\int\frac{1}{x\ln ^2 x} dx$
$=\int\frac{1}{u^2}du$
$=-\frac{1}{u}+c$
$=-\frac{1}{\ln x}+c$

17. $\int\frac{1}{x\sqrt{1-(\ln x)^2}} dx$
$=\int\frac{1}{\sqrt{1-u^2}} du$
$=\int\frac{1}{\cos\theta}\cdot\cos\theta d\theta$
$=\sin^{-1}(\ln x)+C$

16. $\int\frac{1}{1+e^x} dx$
$=\int\frac{1}{1+u}\cdot\frac{1}{u}du$
$=\int\frac{1}{u}-\frac{1}{u+1}du$
$=\ln|u|-\ln|u+1|+c$
$=\ln e^x+\ln (e^x+1)+c$
$=x+\ln(e^x+1)+c$

15. $\int \frac{e^{2x}}{1+e^x} dx$
$=\int \frac{u}{1+u} du$
$=u-\ln|1+u|+c$
$=e^x-\ln(1+e^x)+c$
14. $\int \tan ^2x dx$
$=\int\sec^2x-1dx$
$=\tan x-x+c$

13. $\int \tan^4 xdx$
$=\int(\sec^2x-1)^2 dx$
$=\int\sec^4 x-2\sec^2 x+1 dx$
$=\int\sec^2 x\tan^2x-\sec^2x+1 dx$
$=\frac{1}{3}\tan^3 x-\tan x+x+C$

12. $\int\frac{\sin x}{\sqrt{5+\cos x}} dx$
$=-2\sqrt {5+\cos x}+C$

11. $\int\cos(2x) dx$
$=\frac{1}{2}\sin 2x+C$

10. $\int\frac{1}{x(1+x^6)} dx$
$=\int\frac{1}{x^7(x^{-6}+1)} dx$
$=-\frac{1}{6}\ln |x^{-6}+1|+C$

9. $\int\frac{e^{3\sqrt{x}}}{\sqrt{x}}dx$
$=\frac{2}{3}e^{3\sqrt{x}}+C$

8. $\int\frac{1}{x^2}\sin\frac{1}{x} dx$
$=-\frac{1}{2}\cdot\sin^2\frac{1}{x}+C$

7. $\int x\sqrt{1-x^2}dx$
$=-\frac{1}{2}\cdot\frac{2}{3}\cdot (1-x^2)^{\frac{3}{2}}+C$
$=-\frac{1}{3}(1-x^2)^{\frac{3}{2}}+C$

6. $\int x\sqrt{1-x^2}dx$
$=-\frac{1}{2}\cdot\frac{2}{3}\cdot (1-x^2)^{\frac{3}{2}}+C$
$=-\frac{1}{3}(1-x^2)^{\frac{3}{2}}+C$

5. $\int xe^{x^2} dx$
$=\frac{1}{2}e^{x^2}+C$

4. $\int e^{2x+3} dx$
$=\frac{1}{2}e^{2x+3}+C$

3. $\int\frac{1}{5x+3} dx$
$=\frac{1}{5}\ln |5x+3| +C$

2. $\int\frac{1}{\sqrt{1-(x+1)^2}}dx$
令 $x+1=\sin\theta，dx=\cos\theta d\theta$
$=\int\frac{1}{\cos\theta}\cdot\cos\theta d\theta$
$=\theta+C=\sin^{-1}(x+1)$

1. $\int\frac{1}{x\sqrt{4x^2-1}} dx$
令 $x=\frac{1}{2}\sec\theta$，$dx=\frac{1}{2}\sec\theta\tan\theta d\theta$
$=\int\frac{1}{\frac{1}{2}\sec\theta\tan\theta}\cdot\frac{1}{2}\sec\theta\tan\theta d\theta$
$=\theta +C=\sec^{-1}2x+C$

證明<1>
令$\tan \frac{x}{2}=t$
$\frac{1}{2} \sec^2 \frac{x}{2}dx=dt$
$dx=\frac{2}{1+t^2}dt$
$\int\sec x dx=\int \frac{1+t^2}{1-t^2}\frac{2}{1+t^2}dt=\int \frac{2}{1-t^2}dt$
$=2 \int \frac{1}{2(1+t)}+ \frac{1}{2(1-t)}dt=\ln\vert(1+t)\vert-\ln\vert(1-t)\vert+c=\ln\vert\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\vert+c$
$=\ln \vert \frac {\tan^2 \frac {x}{2}+1+2 \tan \frac{x}{2}}{1- \tan^2 \frac{x}{2}} \vert+C$
$=\ln\vert\sec x+\tan x\vert+C$

證明<2>
$\int\sec x dx=\int\sec x(\frac{\sec x+\tan x}{\sec x+\tan x})dx$
$=\int\frac{\sec^2 x+\sec x\tan x}{\sec x+\tan x}dx$
令$u=\sec x+\tan x$
$u'=\sec x\tan x+\sec^2 x$
$\int\sec xdx=\int\frac{\sec^2 x+\sec x\tan x}{\sec x+\tan x}dx$
$=\int\frac{u'}{u}dx$
$=\ln\vert u\vert+C$
$=\ln\vert\sec x+\tan x\vert+C$

證明<3>