# Thms of Limits
###### tags: `Calculus`
## Definition
$$
A\;function\;f\;has\;a\;maximum\;in\;D\\\exists P\in D\;s.t.\;f(p)\geq f(x)\;,\;\forall x\in D\\then\;We\;call\;P\;the\;maximum\;point
$$
## Thm
### Extreme Value Thm
$$
if\;f\;is\;a\;continuous\;function\;a\;
''closed\;finite''\;interval\;[a,b]\\Then\; f\;has\;a\;maximum \;and\;minimum\;value
$$
### Fermat Thm
$$
if\;f\;has\;a\;maximum/minimum\;at\;a\;inerior\;point\;p\\and\;f'(p)\;exists\;then\;f'(p)=0
$$

#### <Proof>
$$
assume\;that\;P\;is\;a\;maximum\;point\\f'(p)=\left\{\begin{array}{l}\lim_{h\rightarrow0^+}\frac{f(p+h)-f(p)}h\leq0\\\lim_{h\rightarrow0^-}\frac{f(p+h)-f(p)}h\geq0\end{array}\Rightarrow f'(p)=0(By\;squeeze\;Thm)\right.
$$

### Racetrack Principle
$$
if\;f'(x)>g'(x)\;\forall x>0\;,and\;if\;f(0)=g(0)\;,then\;{f(x)>g(x)\;}\forall x>0
$$
#### <Proof>
$$
let\;h(x)=g(x)-f(x)\Rightarrow h'(x)=g'(x)-f'(x)>0\\and\;h(0)=g(0)-f(0)=0\\we\;can\;use\;the\;M.V.T\;on\;the\;interval\;\lbrack0,x\rbrack\;\Rightarrow\;h'(x_0)=\frac{h(x)-h(0)}{x-0}=\frac{f(x)-g(x)}x>0.\\By\;assumption,x>0\;multiply\;both\;sides\;by\;x\;gives\;f(x)-g(x)>0.\\This\;implies\;f(x)>g(x)
$$

### Local Maximum and Minimum

#### Definition
$$
f\;has\;a\;local\;maximum\;at\;p\\if\;\exists\delta>0\;s.t.f(p)\geq f(x),\;\forall x\in(p-\delta,p+\delta)
$$

### Rolle's Thm
$$
Suppose\;that\;f\;is\;\left\{\begin{array}{l}continuous\;on\;\lbrack a,b\rbrack\\differentiable\;on\;(a,b)\\f(a)=f(b)\end{array}\right.\\Then\;\exists c\in(a,b)\;s.t.\;f'(c)=0(critical\;point)
$$
#### <Proof>
#### case 1:
$$
The\;maximum\;or\;minimum\;of\;f\;is\;attained\;at\;an\;interior\;point\Rightarrow Fermal's\;Thm
$$
#### case 2:
$$
Both\;the\;maximum\;and\;minimum\;of\;f\;and\;attained\;at\;boundary\;point\\\Rightarrow max\lbrack a,b\rbrack\;f\;=\;min\lbrack a,b\rbrack f\;=f(a)=f(b)\\f\;is\;a\;cons\tan t\;on\;\lbrack a,b\rbrack
$$
### Mean Value Thm(IMPORTANT!!!!)
$$
Suppose\;that\;f\;is\;\left\{\begin{array}{l}continuous\;on\;\lbrack a,b\rbrack\\differentiable\;on\;(a,b)\end{array}\right.\\Then\;\exists c\in(a,b)\;s.t.\;f'(c)=\frac{f(b)-f(a)}{b-a}\Leftrightarrow f(b)-f(a)=f'(c)(b-a)
$$
#### <Proof>
$$
Let\;g(x)=f(x)-(f(a)+\frac{f(b)-f(a)}{b-a}(x-a))\\g\;is\;\left\{\begin{array}{l}continuous\;on\;\lbrack a,b\rbrack\\differentiable\;on\;(a,b)\\g(a)=g(b)=0\end{array}\right.\\\\By\;Rolle's\;Thm\\\exists c\in(a,b)\;s.t.\;g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0\Rightarrow f'(c)=\frac{f(b)-f(a)}{b-a}
$$
### Thm
$$
f'(x)=0\;\forall x\in(a,b)\Leftrightarrow f\;is\;cons\tan t\;on\;(a,b)\\
$$
#### <Proof>
$$
(\Rightarrow)Given\;a<x_1<x_2<b\;\\f(x_2)-f(x_1)\overset{M.V.T}{=\;}f'(x_3)(x_2-x_1)=0\\x_1<x_3<x_2\\
$$
### Cauchy's M.V.T(Hard!!!!)
$$
(x,y)=(f(t),g(t))\\\exists c\in(a,b)\;s.t.\;(f(b)-f(a),g(b)-g(a))//(f'(c),g'(c))\Leftrightarrow\\\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}
$$
#### <Proof>
$$
Let\;h(x)=(f(b)-f(a))(g(x)-g(a))-(f(x)-f(a))(g(b)-g(a))\\h(a)=h(b)=0\\\overset{Rolle's\;Thm}\Rightarrow h'(c)=0\;for\;some\;c\in(a,b)
$$
### Intermediate Value Thm
$$
if\;f\;is\;continuous\;at\;\lbrack a,b\rbrack\;,\;f(a)<f(b)\\\forall u\vert\;f(a)<u<f(b)\;\exists\;c\in(a,b)\;s.t.\;f(c)=u
$$