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:
In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity. (a)   $\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}$ (b)   $\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}$ (c)   $\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}$

Solution:

(a)

Step 1:

Step 2:

(b)

Step 1:

Step 2:

(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)
(c)    ${\frac {3}{10}}$

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