009A Sample Midterm 3, Problem 1

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

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

Foundations:
1. Linearity rules of limits
2. lim sin(x)/x

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

(b)

Step 1:

Step 2:

(c)

Step 1:

Step 2:

Final Answer:
(a)     $\lim _{x\rightarrow 3}f(x)=6$
(b)
(c)

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