# 009A Sample Midterm 3, Problem 1 Detailed Solution

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

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

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

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

(c)

Step 1:
First, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {(-2x^{3}-2x+3)}{(3x^{3}+3x^{2}-3)}}{\frac {({\frac {1}{x^{3}}})}{({\frac {1}{x^{3}}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}}{3+{\frac {3}{x}}-{\frac {3}{x^{3}}}}}}.\end{array}}}$
Step 2:
Now, we use the properties of limits to get

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}}{3+{\frac {3}{x}}-{\frac {3}{x^{3}}}}}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow \infty }{\bigg (}-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}{\bigg )}}}{\displaystyle {\lim _{x\rightarrow \infty }{\bigg (}3+{\frac {3}{x}}-{\frac {3}{x^{3}}}{\bigg )}}}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow \infty }-2+\lim _{x\rightarrow \infty }{\frac {2}{x^{2}}}+\lim _{x\rightarrow \infty }{\frac {3}{x^{3}}}}}{\displaystyle {\lim _{x\rightarrow \infty }3+\lim _{x\rightarrow \infty }{\frac {3}{x}}-\lim _{x\rightarrow \infty }{\frac {3}{x^{3}}}}}}\\&&\\&=&\displaystyle {\frac {-2+0+0}{3+0+0}}\\&&\\&=&\displaystyle {-{\frac {2}{3}}.}\end{array}}}$

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