009A Sample Midterm 1, Problem 1

Find the following limits:

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

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

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

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

Solution:

(a)

Step 1:
Since  $\lim _{x\rightarrow 2}x=2\neq 0,$
we have
${\begin{array}{rcl}\displaystyle {5}&=&\displaystyle {\lim _{x\rightarrow 2}{\bigg [}{\frac {4-g(x)}{x}}{\bigg ]}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow 2}(4-g(x))}}{\displaystyle {\lim _{x\rightarrow 2}x}}}\\&&\\&=&\displaystyle {{\frac {\displaystyle {\lim _{x\rightarrow 2}(4-g(x))}}{2}}.}\end{array}}$

Step 2:
If we multiply both sides of the last equation by  $2,$  we get
$10=\lim _{x\rightarrow 2}(4-g(x)).$
Now, using linearity properties of limits, we have
${\begin{array}{rcl}\displaystyle {10}&=&\displaystyle {\lim _{x\rightarrow 2}4-\lim _{x\rightarrow 2}g(x)}\\&&\\&=&\displaystyle {4-\lim _{x\rightarrow 2}g(x).}\\\end{array}}$

Step 3:
Solving for  $\lim _{x\rightarrow 2}g(x)$  in the last equation,
we get
$\lim _{x\rightarrow 2}g(x)=-6.$

(b)

Step 1:
First, we write
$\lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}=\lim _{x\rightarrow 0}{\frac {4}{5}}{\bigg (}{\frac {\sin(4x)}{4x}}{\bigg )}.$

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

(c)

Step 1:
When we plug in values close to  $-3$  into   ${\frac {x}{x^{2}-9}},$
we get a small denominator, which results in a large number.
Thus,
$\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}$
is either equal to  $\infty$  or  $-\infty .$

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

Final Answer:
(a)     $-6$
(b)     ${\frac {4}{5}}$
(c)     $\infty$

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