009A Sample Final 3, Problem 7 Detailed Solution

Compute

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

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

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

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Background Information:
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
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}\cdot {\frac {(3+{\sqrt {9-x}})}{(3+{\sqrt {9-x}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x(3+{\sqrt {9-x}})}{9-(9-x)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x(3+{\sqrt {9-x}})}{x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {3+{\sqrt {9-x}}}{1}}}\\&&\\&=&\displaystyle {\frac {3+{\sqrt {9}}}{1}}\\&&\\&=&\displaystyle {\frac {6}{1}}\\&&\\&=&\displaystyle {6.}\end{array}}$

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

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