007A Sample Midterm 3, Problem 1 Detailed Solution

Find the following limits:

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

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

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

---

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

Step 2:
Since  $\lim _{x\rightarrow 3}2x=6\neq 0,$  we have
${\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {\lim _{x\rightarrow 3}{\frac {f(x)}{2x}}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow 3}f(x)}}{\displaystyle {\lim _{x\rightarrow 3}2x}}}\\&&\\&=&\displaystyle {{\frac {\displaystyle {\lim _{x\rightarrow 3}f(x)}}{6}}.}\end{array}}$
Multiplying both sides by  $6,$  we get
$\lim _{x\rightarrow 3}f(x)=6.$

(b)

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

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

(c)

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

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

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

Return to Sample Exam