009A Sample Final 2, Problem 8

Compute

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

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

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

Foundations:
L'Hôpital's Rule
Suppose that  $\lim _{x\rightarrow \infty }f(x)$  and  $\lim _{x\rightarrow \infty }g(x)$  are both zero or both  $\pm \infty .$
If  $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$  is finite or  $\pm \infty ,$
then  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

Step 1:

Step 2:

(b)

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

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

(c)

Step 1:
We proceed using L'Hôpital's Rule. So, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1}{\frac {x^{3}-1}{x^{10}-1}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow 1}{\frac {3x^{2}}{10x^{9}}}.}\end{array}}$

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

Final Answer:
(a)
(b)    ${\text{DNE}}$
(c)    ${\frac {3}{10}}$

Return to Sample Exam