# 009A Sample Final 3, Problem 7

Compute

(a)  $\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}$ (b)  $\lim _{x\rightarrow \pi }{\frac {\sin x}{\pi -x}}$ (c)  $\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}$ Foundations:
L'Hôpital's Rule, Part 1

Let  $\lim _{x\rightarrow c}f(x)=0$ and  $\lim _{x\rightarrow c}g(x)=0,$ where  $f$ and  $g$ are differentiable functions

on an open interval  $I$ containing  $c,$ and  $g'(x)\neq 0$ on  $I$ except possibly at  $c.$ Then,   $\lim _{x\rightarrow c}{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow c}{\frac {f'(x)}{g'(x)}}.$ Solution:

(a)

Step 1:
We begin by noticing that we plug in  $x=0$ into
${\frac {x}{3-{\sqrt {9-x}}}},$ we get   ${\frac {0}{0}}.$ Step 2:
Now, we multiply the numerator and denominator by the conjugate of the denominator.
Hence, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}{\frac {(3+{\sqrt {9-x}})}{(3+{\sqrt {9-x}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x(3+{\sqrt {9-x}})}{9-(9-x)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x(3+{\sqrt {9-x}})}{x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {3+{\sqrt {9-x}}}{1}}}\\&&\\&=&\displaystyle {\frac {3+{\sqrt {9}}}{1}}\\&&\\&=&\displaystyle {\frac {6}{1}}\\&&\\&=&\displaystyle {6.}\end{array}}$ (b)

Step 1:
We proceed using L'Hôpital's Rule. So, we have

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \pi }{\frac {\sin(x)}{\pi -x}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \pi }{\frac {\cos(x)}{-1}}.}\end{array}}$ Step 2:
Now, we plug in  $x=\pi$ to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \pi }{\frac {\sin(x)}{\pi -x}}}&=&\displaystyle {\frac {\cos(\pi )}{-1}}\\&&\\&=&\displaystyle {\frac {-1}{-1}}\\&&\\&=&\displaystyle {1.}\end{array}}$ (c)

Step 1:
We begin by factoring the numerator and denominator. We have

$\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}\,=\,\lim _{x\rightarrow -2}{\frac {(x+2)(x-3)}{(x+2)(x^{2}-2x+4)}}.$ So, we can cancel  $x+2$ in the numerator and denominator. Thus, we have

$\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}\,=\,\lim _{x\rightarrow -2}{\frac {x-3}{x^{2}-2x+4}}.$ Step 2:
Now, we can just plug in  $x=-2$ to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}}&=&\displaystyle {\frac {-2-3}{(-2)^{2}-2(-2)+4}}\\&&\\&=&\displaystyle {-{\frac {5}{12}}.}\end{array}}$ (a)   $6$ (b)   $1$ (c)   $-{\frac {5}{12}}$ 