# 009A Sample Final 2, Problem 8

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Compute

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

(b)  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}}$

(c)  ${\displaystyle \lim _{x\rightarrow 1}{\frac {x^{3}-1}{x^{10}-1}}}$

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:
First, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {x^{-1}+x}{1+{\sqrt {1+x}}}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x}}+x}{1+{\sqrt {1+x}}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x}}+x}{1+{\sqrt {1+x}}}}{\frac {{\big (}{\frac {1}{\sqrt {x}}}{\big )}}{{\big (}{\frac {1}{\sqrt {x}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x^{3/2}}}+{\sqrt {x}}}{{\frac {1}{\sqrt {x}}}+{\sqrt {{\frac {1}{x}}+1}}}}.}\end{array}}}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {x^{-1}+x}{1+{\sqrt {1+x}}}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x^{3/2}}}+{\sqrt {x}}}{{\frac {1}{\sqrt {x}}}+{\sqrt {{\frac {1}{x}}+1}}}}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow \infty }{\big (}{\frac {1}{x^{3/2}}}+{\sqrt {x}}{\big )}}{\lim _{x\rightarrow \infty }{\big (}{\frac {1}{\sqrt {x}}}+{\sqrt {{\frac {1}{x}}+1}}{\big )}}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow \infty }{\frac {1}{x^{3/2}}}+\lim _{x\rightarrow \infty }{\sqrt {x}}}{\lim _{x\rightarrow \infty }{\frac {1}{\sqrt {x}}}+\lim _{x\rightarrow \infty }{\sqrt {{\frac {1}{x}}+1}}}}\\&&\\&=&\displaystyle {\frac {0+\lim _{x\rightarrow \infty }{\sqrt {x}}}{0+1}}\\&&\\&=&\displaystyle {\infty .}\end{array}}}$

(b)

Step 1:
First, we write
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}{\frac {(\cos x+1)}{(\cos x+1)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x(\cos x+1)}{\cos ^{2}x-1}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x(\cos x+1)}{-\sin ^{2}x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\cos x+1}{-\sin x}}.}\end{array}}}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin x}{\cos x-1}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos x+1}{-\sin x}}}\\&&\\&=&\displaystyle {-\infty }\end{array}}}$
and
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{-}}{\frac {\sin x}{\cos x-1}}}&=&\displaystyle {\lim _{x\rightarrow 0^{-}}{\frac {\cos x+1}{-\sin x}}}\\&&\\&=&\displaystyle {\infty .}\end{array}}}$
Therefore,
${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}={\text{DNE}}.}$

(c)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1}{\frac {x^{3}-1}{x^{10}-1}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow 1}{\frac {3x^{2}}{10x^{9}}}.}\end{array}}}$

Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1}{\frac {x^{3}-1}{x^{10}-1}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow 1}{\frac {3x^{2}}{10x^{9}}}}\\&&\\&=&\displaystyle {\frac {3(1)^{2}}{10(1)^{9}}}\\&&\\&=&\displaystyle {{\frac {3}{10}}.}\end{array}}}$

(a)    ${\displaystyle \infty }$
(b)    ${\displaystyle {\text{DNE}}}$
(c)    ${\displaystyle {\frac {3}{10}}}$