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

(b)

Step 1:
First, we write
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 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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