# 009A Sample Final 3, Problem 7

Compute

(a)  ${\displaystyle \lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}}$

(b)  ${\displaystyle \lim _{x\rightarrow \pi }{\frac {\sin x}{\pi -x}}}$

(c)  ${\displaystyle \lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}}$

Foundations:
L'Hôpital's Rule, Part 1

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

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

Solution:

(a)

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

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \pi }{\frac {\sin(x)}{\pi -x}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \pi }{\frac {\cos(x)}{-1}}.}\end{array}}}$

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

(c)

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

${\displaystyle \lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}\,=\,\lim _{x\rightarrow -2}{\frac {(x+2)(x-3)}{(x+2)(x^{2}-2x+4)}}.}$

So, we can cancel  ${\displaystyle x+2}$  in the numerator and denominator. Thus, we have

${\displaystyle \lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}\,=\,\lim _{x\rightarrow -2}{\frac {x-3}{x^{2}-2x+4}}.}$

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

(a)   ${\displaystyle 6}$
(b)   ${\displaystyle 1}$
(c)   ${\displaystyle -{\frac {5}{12}}}$