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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+2}   in the numerator and denominator. Thus, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-2}   to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Final Answer:
(a)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -6}
(b)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1}
(c)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{5}{12}}

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