009A Sample Midterm 3, Problem 1

Find the following limits:

(a) If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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. 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. $\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$

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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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{\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)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3} f(x)=6}
(b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{3}}
(c)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{2}{3}}

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