009A Sample Final 2, Problem 1

Compute

(a)  ${\displaystyle \lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}}$

(b)  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin ^{2}x}{3x}}}$

(c)  ${\displaystyle \lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}+2}}{2x-1}}}$

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 if we plug in  ${\displaystyle x=4}$  into
${\displaystyle {\frac {{\sqrt {x+5}}-3}{x-4}},}$
we get   ${\displaystyle {\frac {0}{0}}.}$
Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}}&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}{\frac {({\sqrt {x+5}}+3)}{({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {(x+5)-9}{(x-4)({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {x-4}{(x-4)({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {1}{{\sqrt {x+5}}+3}}}\\&&\\&=&\displaystyle {\frac {1}{{\sqrt {9}}+3}}\\&&\\&=&\displaystyle {{\frac {1}{6}}.}\end{array}}}$

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin ^{2}(x)}{3x}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow 0}{\frac {2\sin(x)\cos(x)}{3}}.}\end{array}}}$

Step 2:
Now, we plug in  ${\displaystyle x=0}$  to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin ^{2}(x)}{3x}}}&=&\displaystyle {\frac {2\sin(0)\cos(0)}{3}}\\&&\\&=&\displaystyle {\frac {2(0)(1)}{3}}\\&&\\&=&\displaystyle {0.}\end{array}}}$

(c)

Step 1:
First, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}+2}}{2x-1}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}(1+{\frac {2}{x^{2}}})}}{2x-1}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {|x|{\sqrt {1+{\frac {2}{x^{2}}}}}}{2x-1}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-x{\sqrt {1+{\frac {2}{x^{2}}}}}}{x(2-{\frac {1}{x}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-1{\sqrt {1+{\frac {2}{x^{2}}}}}}{(2-{\frac {1}{x}})}}.}\end{array}}}$
Step 2:
Now,
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}+2}}{2x-1}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-1{\sqrt {1+{\frac {2}{x^{2}}}}}}{(2-{\frac {1}{x}})}}}\\&&\\&=&\displaystyle {\frac {-{\sqrt {1+0}}}{(2-0)}}\\&&\\&=&\displaystyle {-{\frac {1}{2}}.}\end{array}}}$

(a)   ${\displaystyle {\frac {1}{6}}}$
(b)   ${\displaystyle 0}$
(c)   ${\displaystyle -{\frac {1}{2}}}$