009A Sample Midterm 3, Problem 1 Detailed Solution

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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  $\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}.$ Background Information:
1. If  $\lim _{x\rightarrow a}g(x)\neq 0,$ we have
$\lim _{x\rightarrow a}{\frac {f(x)}{g(x)}}={\frac {\lim _{x\rightarrow a}f(x)}}{\lim _{x\rightarrow a}g(x)}}}.$ 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

${\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {\lim _{x\rightarrow 3}{\frac {f(x)}{2x}}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow 3}f(x)}}{\lim _{x\rightarrow 3}2x}}}\\&&\\&=&\displaystyle {{\frac {\lim _{x\rightarrow 3}f(x)}}{6}}.}\end{array}}$ 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
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {(-2x^{3}-2x+3)}{(3x^{3}+3x^{2}-3)}}{\frac {({\frac {1}{x^{3}}})}{({\frac {1}{x^{3}}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}}{3+{\frac {3}{x}}-{\frac {3}{x^{3}}}}}}.\end{array}}$ Step 2:
Now, we use the properties of limits to get

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2x^{3}-2x+3}{3x^{3}+3x^{2}-3}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}}{3+{\frac {3}{x}}-{\frac {3}{x^{3}}}}}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow \infty }{\bigg (}-2-{\frac {2}{x^{2}}}+{\frac {3}{x^{3}}}{\bigg )}}}{\lim _{x\rightarrow \infty }{\bigg (}3+{\frac {3}{x}}-{\frac {3}{x^{3}}}{\bigg )}}}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow \infty }-2+\lim _{x\rightarrow \infty }{\frac {2}{x^{2}}}+\lim _{x\rightarrow \infty }{\frac {3}{x^{3}}}}}{\lim _{x\rightarrow \infty }3+\lim _{x\rightarrow \infty }{\frac {3}{x}}-\lim _{x\rightarrow \infty }{\frac {3}{x^{3}}}}}}\\&&\\&=&\displaystyle {\frac {-2+0+0}{3+0+0}}\\&&\\&=&\displaystyle {-{\frac {2}{3}}.}\end{array}}$ (a)     $6$ (b)     ${\frac {2}{3}}$ (c)     $-{\frac {2}{3}}$ 