# 007A Sample Midterm 1, Problem 1 Detailed Solution

Find the following limits:

(a) Find  ${\displaystyle \lim _{x\rightarrow 2}g(x),}$  provided that  ${\displaystyle \lim _{x\rightarrow 2}{\bigg [}{\frac {4-g(x)}{x}}{\bigg ]}=5.}$

(b) Find  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}}$

(c) Evaluate  ${\displaystyle \lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}}$

Foundations:
1. If  ${\displaystyle \lim _{x\rightarrow a}g(x)\neq 0,}$  we have
${\displaystyle \lim _{x\rightarrow a}{\frac {f(x)}{g(x)}}={\frac {\displaystyle {\lim _{x\rightarrow a}f(x)}}{\displaystyle {\lim _{x\rightarrow a}g(x)}}}.}$
2. Recall
${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1}$

Solution:

(a)

Step 1:
Since  ${\displaystyle \lim _{x\rightarrow 2}x=2\neq 0,}$
we have
${\displaystyle {\begin{array}{rcl}\displaystyle {5}&=&\displaystyle {\lim _{x\rightarrow 2}{\bigg [}{\frac {4-g(x)}{x}}{\bigg ]}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow 2}(4-g(x))}}{\displaystyle {\lim _{x\rightarrow 2}x}}}\\&&\\&=&\displaystyle {{\frac {\displaystyle {\lim _{x\rightarrow 2}(4-g(x))}}{2}}.}\end{array}}}$
Step 2:
If we multiply both sides of the last equation by  ${\displaystyle 2,}$  we get
${\displaystyle 10=\lim _{x\rightarrow 2}(4-g(x)).}$
Now, using linearity properties of limits, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {10}&=&\displaystyle {\lim _{x\rightarrow 2}4-\lim _{x\rightarrow 2}g(x)}\\&&\\&=&\displaystyle {4-\lim _{x\rightarrow 2}g(x).}\\\end{array}}}$
Step 3:
Solving for  ${\displaystyle \lim _{x\rightarrow 2}g(x)}$  in the last equation,
we get

${\displaystyle \lim _{x\rightarrow 2}g(x)=-6.}$

(b)

Step 1:
First, we write
${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}=\lim _{x\rightarrow 0}{\bigg (}{\frac {4}{5}}\cdot {\frac {\sin(4x)}{4x}}{\bigg )}.}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}}&=&\displaystyle {{\frac {4}{5}}\lim _{x\rightarrow 0}{\frac {\sin(4x)}{4x}}}\\&&\\&=&\displaystyle {{\frac {4}{5}}(1)}\\&&\\&=&\displaystyle {{\frac {4}{5}}.}\end{array}}}$

(c)

Step 1:
When we plug in values close to  ${\displaystyle -3}$  into   ${\displaystyle {\frac {x}{x^{2}-9}},}$
we get a small denominator, which results in a large number.
Thus,
${\displaystyle \lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}}$
is either equal to  ${\displaystyle \infty }$  or  ${\displaystyle -\infty .}$
Step 2:
To figure out which one, we factor the denominator to get
${\displaystyle \lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}=\lim _{x\rightarrow -3^{+}}{\frac {x}{(x-3)(x+3)}}.}$
We are taking a right hand limit. So, we are looking at values of  ${\displaystyle x}$
a little bigger than  ${\displaystyle -3.}$  (You can imagine values like  ${\displaystyle x=-2.9.}$ )
For these values, the numerator will be negative.
Also, for these values,  ${\displaystyle x-3}$  will be negative and  ${\displaystyle x+3}$  will be positive.
Therefore, the denominator will be negative.
Since both the numerator and denominator will be negative (have the same sign),
${\displaystyle \lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}=\infty .}$

(a)     ${\displaystyle -6}$
(b)     ${\displaystyle {\frac {4}{5}}}$
(c)     ${\displaystyle \infty }$