# 009A Sample Final 3, Problem 7

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
Suppose that  ${\displaystyle \lim _{x\rightarrow \infty }f(x)}$  and  ${\displaystyle \lim _{x\rightarrow \infty }g(x)}$  are both zero or both  ${\displaystyle \pm \infty .}$

If  ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}}$  is finite or  ${\displaystyle \pm \infty ,}$

then  ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$

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)   ${\displaystyle -{\frac {5}{12}}}$