# 007A Sample Midterm 2, Problem 1 Detailed Solution

Evaluate the following limits.

(a) Find  ${\displaystyle \lim _{x\rightarrow 2}{\frac {{\sqrt {x^{2}+12}}-4}{x-2}}}$

(b) Find  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}}$

(c) Evaluate  ${\displaystyle \lim _{x\rightarrow 0}x^{2}\cos {\bigg (}{\frac {1}{x}}{\bigg )}}$

Background Information:
1.  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1}$
2. Squeeze Theorem
Let  ${\displaystyle f,g}$  and  ${\displaystyle h}$  be functions on an open interval  ${\displaystyle I}$  containing  ${\displaystyle c}$
such that for all  ${\displaystyle x}$  in  ${\displaystyle I,~f(x)\leq g(x)\leq h(x).}$
If  ${\displaystyle \lim _{x\rightarrow c}f(x)=L=\lim _{x\rightarrow c}h(x),}$  then  ${\displaystyle \lim _{x\rightarrow c}g(x)=L.}$

Solution:

(a)

Step 1:
We begin by noticing that if we plug in  ${\displaystyle x=2}$  into
${\displaystyle {\frac {{\sqrt {x^{2}+12}}-4}{x-2}},}$
we get   ${\displaystyle {\frac {0}{0}}.}$
Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2}{\frac {{\sqrt {x^{2}+12}}-4}{x-2}}}&=&\displaystyle {\lim _{x\rightarrow 2}{\bigg [}{\frac {({\sqrt {x^{2}+12}}-4)}{(x-2)}}\cdot {\frac {({\sqrt {x^{2}+12}}+4)}{({\sqrt {x^{2}+12}}+4)}}{\bigg ]}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x^{2}+12)-16}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x^{2}-4}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x-2)(x+2)}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x+2}{{\sqrt {x^{2}+12}}+4}}}\\&&\\&=&\displaystyle {\frac {4}{8}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}}$

(b)

Step 1:
First, we write
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\bigg [}{\frac {\sin(3x)}{x}}\cdot {\frac {x}{\sin(7x)}}{\bigg ]}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\bigg [}{\frac {3}{7}}\cdot {\frac {\sin(3x)}{3x}}\cdot {\frac {7x}{\sin(7x)}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {3}{7}}\lim _{x\rightarrow 0}{\bigg [}{\frac {\sin(3x)}{3x}}\cdot {\frac {7x}{\sin(7x)}}{\bigg ]}.}\end{array}}}$
Step 2:
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}}&=&\displaystyle {{\frac {3}{7}}\lim _{x\rightarrow 0}{\bigg [}{\frac {\sin(3x)}{3x}}\cdot {\frac {7x}{\sin(7x)}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {3}{7}}{\bigg (}\lim _{x\rightarrow 0}{\frac {\sin(3x)}{3x}}{\bigg )}\cdot {\bigg (}\lim _{x\rightarrow 0}{\frac {7x}{\sin(7x)}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3}{7}}\cdot (1)\cdot (1)}\\&&\\&=&\displaystyle {{\frac {3}{7}}.}\end{array}}}$

(c)

Step 1:
First, recall that
${\displaystyle -1\leq \cos(\theta )\leq 1}$
for all  ${\displaystyle \theta .}$
Then, for all  ${\displaystyle x\neq 0,}$
${\displaystyle -1\leq \cos {\bigg (}{\frac {1}{x}}{\bigg )}\leq 1.}$
Hence, for all  ${\displaystyle x\neq 0,}$
${\displaystyle -x^{2}\leq x^{2}\cos {\bigg (}{\frac {1}{x}}{\bigg )}\leq x^{2}.}$
Step 2:
Now, notice
${\displaystyle \lim _{x\rightarrow 0}x^{2}=0}$
and
${\displaystyle \lim _{x\rightarrow 0}-x^{2}=0.}$
Step 3:
Since
${\displaystyle \lim _{x\rightarrow 0}x^{2}=\lim _{x\rightarrow 0}-x^{2}=0,}$
we have
${\displaystyle \lim _{x\rightarrow 0}x^{2}\cos {\bigg (}{\frac {1}{x}}{\bigg )}=0}$
by the Squeeze Theorem.

(a)     ${\displaystyle {\frac {1}{2}}}$
(b)     ${\displaystyle {\frac {3}{7}}}$
(c)     ${\displaystyle 0}$