# Limit
###### tags: `Calculus`
## Definition
$$
f\;is\;define\;at\;open\;interval\;contains\;a\;except\;a\;itself.\\if\;for\;every\;number\;\epsilon>0\\there\;is\;a\;\delta>0\;s.t.\;if\;0<\left|x-a\right|<\delta\;then\;\left|f(x)-L\right|<\epsilon\\we\;can\;write\;\lim_{x\rightarrow a}f(x)=L
$$
## Continuity
- ### Definition
$$
\;f\;is\;continuous\;at\;x_0\;in\;its\;domain\;if\;for\;every\;\varepsilon\;>\;0\\there\;is\;a\;\delta\;>\;0\;s.t.\;whenever\;x\;is\;in\;the\;domain\;of\;f\\and\;\left|x\;-\;x_0\right|\;<\;\delta,we\;have\;\left|f(x)\;-\;f(x_0)\right|<\varepsilon\\Again,\;we\;say\;f\;is\;continuous\;if\;it\;is\;continuous\;at\;every\;point\;in\;its\;domain.
$$
## Inf Limit
- ### Definition
$$
f\;is\;define\;on\;open\;interval\;that\;contains\;a\;expect\;a\;itself\\Then\;\lim_{x\rightarrow a}f(x)=\infty\;\\means\;for\;every\;positive\;M\;there\;is\;a\;\delta\;s.t.\\if\;0<\left|x-a\right|<\delta\;then\;f(x)>M\\
$$
- ### Example
$$
Prove \lim_{x\rightarrow0}\frac1{x^2}=\infty\\
$$
$$
Let\;M\;be\;a\;given\;number.\;We\;want\;to\;find\;\delta\;s.t.\\if\;0<\left|x\right|<\delta\;\;\;then\;\;\;\frac1{x^2}>M\\Choose\;\delta=\frac1{\sqrt M}\;and\;0<\left|x\right|<\delta\;=\frac1{\sqrt M},\;then\;\frac1{x^2}>M\\This\;show\;that\;\frac1{x^2}\rightarrow\infty\;as\;x\rightarrow0\\
$$
## Previou Exam
- ### Problem 1
$$
Let\;f(x)\;be\;a\;continuous\;function.It\;is\;given\;that\;\lim_{h\rightarrow0}\;\frac{f(h)}h=2020.\\(a)\;Compute\;f(0).\;Then,\;prove\;that\;f\;is\;differentiable\;at\;x=0\;and\;compute\;f′(0).\\(b)\;Suppose\;in\;addition\;that\;f\;is\;twice\;differentiable\;and\;that\;f″(x)\geq2\;for\;all\;x>0.\\U\sin g\;Mean\;Value\;Theorem,\;or\;otherwise,\;prove\;that\;f(x)\geq2020x+x2\;for\;all\;x\geq0.
$$
### Solution
#### (a)
$$
Since\;f(x)\;is\;continuous,\\f(0)=\lim_{h\rightarrow0}f(h)=\lim_{h\rightarrow0}\frac{f(h)}h\cdot h=2020\cdot0=0\\f'(0)=\lim_{h\rightarrow0}\frac{f(h)-f(0)}{h-0}=2020
$$
#### (b)
$$
\forall x>0\;\;\;,\frac{f'(x)-f'(0)}{x-0}=f''(c)\;for\;some\;c\;\in(0,x)\\f'(x)=x\cdot f''(c)+f'(0)\geq2x+2020\\Let\;g(x)=f(x)-2020x-x^2\;\;,\forall x>0\\g(0)=0\\\frac{g(x)-g(0)}{x-0}=g'(d)\;for\;some\;d\in(0,x)\\g(x)=x\cdot g'(d)+g(0)=x\cdot(f'(x)-2020-2x)\geq0\\\Rightarrow f(x)\geq x^2+2020x
$$
##

- ### Problem 2
$$
\left|f(x)-f(y)\right|\leq\left|x-y\right|^2\;,\;\forall x,y\in\mathbb{R}\\(a).show\;f\;is\;differentiable\;everywhere\\(b).f=?\\
$$
### Solution
#### (a)
$$
f'(x)=\lim_{x\rightarrow y}\frac{f(x)-f(y)}{x-y}\\\Rightarrow\left|f'(x)\right|=\lim_{x\rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right|\leq\left|x-y\right|\;,\;\forall x,y\in\mathbb{R}\;\;(By\;definition\;of\;f)
$$
#### (b)
$$
as\;x\rightarrow y\;,\;\left|f'(x)\right|\rightarrow0\Rightarrow f'(x)\equiv0,\;\forall x\;(By\;squeeze\;Thm)\\\Rightarrow f(x)\;is\;cons\tan t(f(x)\equiv C,c\in\mathbb{R})
$$