# 007A Sample Midterm 3, Problem 1 Detailed Solution

Find the following limits:

(a) If  ${\displaystyle \lim _{x\rightarrow 3}{\bigg (}{\frac {f(x)}{2x}}+1{\bigg )}=2,}$  find  ${\displaystyle \lim _{x\rightarrow 3}f(x).}$

(b) Evaluate  ${\displaystyle \lim _{x\rightarrow 2}{\frac {2-x}{x^{2}-4}}.}$

(c) Find  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\tan(4x)}{\sin(6x)}}.}$

Background Information:
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:
First, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {2}&=&\displaystyle {\lim _{x\rightarrow 3}{\bigg (}{\frac {f(x)}{2x}}+1{\bigg )}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 3}{\frac {f(x)}{2x}}+\lim _{x\rightarrow 3}1}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 3}{\frac {f(x)}{2x}}+1.}\end{array}}}$
Therefore,
${\displaystyle \lim _{x\rightarrow 3}{\frac {f(x)}{2x}}=1.}$
Step 2:
Since  ${\displaystyle \lim _{x\rightarrow 3}2x=6\neq 0,}$  we have

${\displaystyle {\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {\lim _{x\rightarrow 3}{\frac {f(x)}{2x}}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow 3}f(x)}}{\displaystyle {\lim _{x\rightarrow 3}2x}}}\\&&\\&=&\displaystyle {{\frac {\displaystyle {\lim _{x\rightarrow 3}f(x)}}{6}}.}\end{array}}}$

Multiplying both sides by  ${\displaystyle 6,}$  we get
${\displaystyle \lim _{x\rightarrow 3}f(x)=6.}$

(b)

Step 1:
First, we write
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2}{\frac {2-x}{x^{2}-4}}}&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {2-x}{(x-2)(x+2)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {-(x-2)}{(x-2)(x+2)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {-1}{x+2}}.}\end{array}}}$
Step 2:
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2}{\frac {2-x}{x^{2}-4}}}&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {-1}{x+2}}}\\&&\\&=&\displaystyle {-{\frac {1}{4}}.}\end{array}}}$

(c)

Step 1:
First, we write
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\tan(4x)}{\sin(6x)}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\bigg [}{\frac {\sin(4x)}{\cos(4x)}}\cdot {\frac {1}{\sin(6x)}}{\bigg ]}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\bigg [}{\frac {4}{6}}\cdot {\frac {\sin(4x)}{4x}}\cdot {\frac {6x}{\sin(6x)}}\cdot {\frac {1}{\cos(4x)}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {4}{6}}\lim _{x\rightarrow 0}{\bigg [}{\frac {\sin(4x)}{4x}}\cdot {\frac {6x}{\sin(6x)}}\cdot {\frac {1}{\cos(4x)}}{\bigg ]}.}\end{array}}}$
Step 2:
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\tan(4x)}{\sin(6x)}}}&=&\displaystyle {{\frac {4}{6}}\lim _{x\rightarrow 0}{\bigg [}{\frac {\sin(4x)}{4x}}\cdot {\frac {6x}{\sin(6x)}}\cdot {\frac {1}{\cos(4x)}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {4}{6}}{\bigg (}\lim _{x\rightarrow 0}{\frac {\sin(4x)}{4x}}{\bigg )}\cdot {\bigg (}\lim _{x\rightarrow 0}{\frac {6x}{\sin(6x)}}{\bigg )}\cdot {\bigg (}\lim _{x\rightarrow 0}{\frac {1}{\cos(4x)}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {4}{6}}\cdot (1)\cdot (1)\cdot (1)}\\&&\\&=&\displaystyle {{\frac {2}{3}}.}\end{array}}}$

(a)     ${\displaystyle 6}$
(b)     ${\displaystyle -{\frac {1}{4}}}$
(c)     ${\displaystyle {\frac {2}{3}}}$