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. Linearity rules of limits
2. Limit sin(x)/x
3. Left and right hand limits

Solution:

(a)

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

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:

Step 2:

Final Answer:
(a)     $\lim _{x\rightarrow 2}g(x)=-6$
(b)
(c)

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