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# System prepended metadata

title: Problems
tags: [Calculus]

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# Problems
###### tags: `Calculus`
$$
find\;\;\lim_{x\rightarrow0}\frac{\tan\;x}{\sqrt{1-\cos\;3x}}
$$
$$
\lim_{x\rightarrow0}\frac{\tan\;x}{\sqrt{1-\cos\;3x}}=\lim_{x\rightarrow0}(\frac{\tan\;x}x)\cdot x\cdot\frac{\sqrt{{(3x)}^2}}{\sqrt{1-\cos\;3x}}\cdot\frac1{\sqrt{{(3x)}^2}}\\=\lim_{x\rightarrow0}(\frac{\tan\;x}x)\cdot\sqrt{\frac{{(3x)}^2}{1-\cos\;3x}}\cdot\frac x{\left|3x\right|}=1\cdot\sqrt2\cdot\frac13sgn(x)=DNE.
$$
<font color="#f00"> $$Note:\;remember\;\sqrt{x^2}\;is\;\left|x\right|\;instead\;of\;x $$</font>
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![](https://i.imgur.com/nFkID4x.png)
$$
implicit\;differentiation\Rightarrow y'=-\frac x{5y}\\suppose\;the\;point\;of\;tangency\;is\;(a,b)\\tan gent\;line:\;(y-b)=-\frac a{5b}(x-a)\\Plug\;in\;(-6,0)\;,We\;can\;have\;two\;equation\left\{\begin{array}{l}a^2+5b^2=6\\-b=-\frac a{5b}(6-a)\end{array}\Rightarrow a=1\;,\;b=-1\right.\\tangent\;line:\;y-1=\frac15(x+1)\\y-1=\frac15(x+1)\vert_{x=4}\Rightarrow y=2
$$
<font color="#f00"> $$Note:Don't\;Use\;too\;much\;variables$$</font>
## 
$$
(a)\;Find\;the\;linearization\;of\;f(x)\;=\sin x\;at\;\frac\pi6\\(b)\;Use\;M.V.T\;to\;prove\;\sin x<\frac12+\frac{\sqrt3}2(x-\frac\pi6)\;for\;x\in(\frac\pi6,\frac\pi2\rbrack
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$$
(a)\;f(x)\approx f(\frac\pi6)+f'(\frac\pi6)\cdot(x-\frac\pi6)=\frac12+\frac{\sqrt3}2(x-\frac\pi6)
$$
$$
Sol:\\
(b)\;f(x)=\sin x\;is\;diiferentiable\;on\;\mathbb{R}\\so\;it's\;diiferentiable\;on\;(\frac\pi6,\frac\pi2)\;and\;continuous\;on\;\lbrack\frac\pi6,\frac\pi2\rbrack\\for\;x\in(\frac\pi6,\frac\pi2)\;\exists c\in(\frac\pi6,x)\;s.t.\;f'(c)=\frac{f(x)-f(\frac\pi6)}{x-\frac\pi6}\\\sin ce\;f'(c)=\cos\;c\;is\;strictly\;decrea\sin g\;on\;(\frac\pi6,\frac\pi2)\\\cos\;c\;<\frac{\sqrt3}2\;for\;c\in(\frac\pi6,\frac\pi2)\;\\f'(c)=\frac{f(x)-f(\frac\pi6)}{x-\frac\pi6}<\frac{\sqrt3}2\;for\;x\;\in(\frac\pi6,\frac\pi2\rbrack\Rightarrow\sin x<\frac{\sqrt3}2(x-\frac\pi6)+\frac12
$$
## 
$$
find\;the\;n^{th}\;derivative\;of\;the\;fucntion\;f(x)=\frac{x^n}{1-x}\\Sol:\\f(x)=\frac{x^n}{1-x}=\frac{x^n-1}{1-x}+\frac1{1-x}=-(x^{n-1}+x^{n-2}+\cdots+1)+\frac1{1-x}\\f^{(n)}(x)=0+{(\frac1{1-x})}^{(n)}{(\frac1{1-x})}'=\frac1{{(1-x)}^2}\\{(\frac1{1-x})}^{(n)}=\frac{n!}{{(1-x)}^{n+1}}=f^{(n)}(x)
$$
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$$
Suppose\;that\;3\;points\;on\;the\;parabola\;y=x^2\;\;have\;the\;property\;that\;their\;normal\\lines\;intersect\;at\;a\;common\;point.\;Show\;that\;their\;sum\;of\;x-coordinates\;is\;0\\Sol:\\let\;(x_1,x_1^2),(x_2,x_2^2),(x_3,x_3^2)\;be\;such\;3\;points\\normal\;line:\;y-x_1^2=\frac{-1}{2x_1}(x-x_1)....\\\Rightarrow x=-2x_1x_2(x_1+x_2)=-2x_2x_3(x_2+x_3)=-2x_1x_3(x_1+x_3)\\\Rightarrow x_1+x_2+x_3=0\\still\;need\;to\;consider\;case\;about\;one\;of\;x_i=0
$$