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. $\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$

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

(b)

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

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

(c)

Step 1:
First, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {(-2x^{3}-2x+3)}{(3x^{3}+3x^{2}-3)}}{\frac {({\frac {1}{x^{3}}})}{({\frac {1}{x^{3}}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}}{3+{\frac {3}{x}}-{\frac {3}{x^{3}}}}}}.\end{array}}$

Step 2:
Now, we use the properties of limits to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}}{3+{\frac {3}{x}}-{\frac {3}{x^{3}}}}}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow \infty }(-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}})}{\lim _{x\rightarrow \infty }(3+{\frac {3}{x}}-{\frac {3}{x^{3}}})}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow \infty }-2+\lim _{x\rightarrow \infty }{\frac {2}{x^{2}}}+\lim _{x\rightarrow \infty }{\frac {3}{x^{3}}}}{\lim _{x\rightarrow \infty }3+\lim _{x\rightarrow \infty }{\frac {3}{x}}-\lim _{x\rightarrow \infty }{\frac {3}{x^{3}}}}}\\&&\\&=&\displaystyle {\frac {-2+0+0}{3+0+0}}\\&&\\&=&\displaystyle {{\frac {-2}{3}}.}\end{array}}$

Final Answer:
(a)     $\lim _{x\rightarrow 3}f(x)=6$
(b)     ${\frac {2}{3}}$
(c)     ${\frac {-2}{3}}$

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