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
Suppose that  $\lim _{x\rightarrow \infty }f(x)$  and  $\lim _{x\rightarrow \infty }g(x)$  are both zero or both  $\pm \infty .$
If  $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$  is finite or  $\pm \infty ,$
then  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

Step 1:

Step 2:

(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
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}}&=&\displaystyle {\frac {-2-3}{(-2)^{2}-2(-2)+4}}\\&&\\&=&\displaystyle {{\frac {-5}{12}}.}\end{array}}$

Final Answer:
(a)
(b)   $1$
(c)   ${\frac {-5}{12}}$

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