007A Sample Midterm 3, Problem 1 Detailed Solution

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) Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow 2}{\frac {2-x}{x^{2}-4}}.}

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

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Background Information:
1. If  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow a}g(x)\neq 0,}   we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1}

Solution:

(a)

Step 1:
First, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow 3}{\frac {f(x)}{2x}}=1.}

Step 2:
Since  $\lim _{x\rightarrow 3}2x=6\neq 0,$  we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 6,}   we 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 \lim_{x\rightarrow 3} f(x)=6.}

(b)

Step 1:
First, we write
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 2} \frac{2-x}{x^2-4}} & = & \displaystyle{\lim_{x\rightarrow 2} \frac{2-x}{(x-2)(x+2)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow 2} \frac{-(x-2)}{(x-2)(x+2)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow 2} \frac{-1}{x+2}.} \end{array}}

Step 2:
Now, 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 2} \frac{2-x}{x^2-4}} & = & \displaystyle{\lim_{x\rightarrow 2} \frac{-1}{x+2}}\\ &&\\ & = & \displaystyle{-\frac{1}{4}.} \end{array}}

(c)

Step 1:
First, we write
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 0} \frac{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\lim_{x\rightarrow 0} \bigg[\frac{\sin(4x)}{\cos(4x)}\cdot \frac{1}{\sin(6x)}\bigg]}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow 0} \bigg[\frac{4}{6} \cdot \frac{\sin(4x)}{4x}\cdot \frac{6x}{\sin(6x)}\cdot\frac{1}{\cos(4x)}\bigg]}\\ &&\\ & = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \bigg[\frac{\sin(4x)}{4x}\cdot \frac{6x}{\sin(6x)}\cdot \frac{1}{\cos(4x)}\bigg].} \end{array}}

Step 2:
Now, 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 0} \frac{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \bigg[\frac{\sin(4x)}{4x}\cdot\frac{6x}{\sin(6x)}\cdot\frac{1}{\cos(4x)}\bigg]}\\ &&\\ & = & \displaystyle{\frac{4}{6}\bigg(\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}\bigg)\cdot\bigg(\lim_{x\rightarrow 0} \frac{6x}{\sin(6x)}\bigg)\cdot\bigg(\lim_{x\rightarrow 0} \frac{1}{\cos(4x)}\bigg)}\\ &&\\ & = & \displaystyle{\frac{4}{6} \cdot (1)\cdot (1)\cdot(1)}\\ &&\\ & = & \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 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{1}{4}}
(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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