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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #132 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{11sin(4x)^{2}}{9x^{4}}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{ 11}{9}$

$\frac{ 11}{9}$

$\infty$

Correct answer:

$\infty$

Explanation:

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Example Question #133 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-7x^{2}}{19sin(9\cdot \pi x)}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{ 7}{19}$

$\infty$

$\frac{ 7}{19}$

Correct answer:

Explanation:

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Example Question #134 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-12x}{-2sin(4\cdot \pi x)^{3}}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

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Example Question #1091 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{11x}{6cos(\frac{(13\cdot \pi x)}{2}-\frac{ \pi }{2})^{3}}\end{align*}$

Possible Answers:

$-\frac{ 11}{6}$

$-\infty$

$\infty$

$\frac{ 11}{6}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{11x}{6cos(\frac{(13\cdot \pi x)}{2}-\frac{ \pi }{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{11}{-117\cdot \pi cos(\frac{(13\cdot \pi x)}{2}-\frac{ \pi }{2})^{2}sin(\frac{(13\cdot \pi x)}{2}-\frac{ \pi }{2})}\rightarrow\frac{11}{0}=\infty\\&lim_{x\rightarrow0+}\frac{11x}{6cos(\frac{(13\cdot \pi x)}{2}-\frac{ \pi }{2})^{3}}=\infty\end{align*}$

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Example Question #136 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{16sin(11\cdot \pi x)^{3}}{10x^{3}}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{(10648\cdot \pi ^{3})}{5}$

$\infty$

$\frac{(10648\cdot \pi ^{3})}{5}$

Correct answer:

$\frac{(10648\cdot \pi ^{3})}{5}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{16sin(11\cdot \pi x)^{3}}{10x^{3}}\rightarrow\frac{0}{0}\\&\frac{528\cdot \pi cos(11\cdot \pi x)sin(11\cdot \pi x)^{2}}{30x^{2}}\rightarrow\frac{0}{0}\\&\frac{11616\cdot \pi ^{2}cos(11\cdot \pi x)^{2}sin(11\cdot \pi x) - 5808\cdot \pi ^{2}sin(11\cdot \pi x)^{3}}{60x}\rightarrow\frac{0}{0}\\&\frac{127776\cdot \pi ^{3}cos(11\cdot \pi x)^{3} - 447216\cdot \pi ^{3}cos(11\cdot \pi x)sin(11\cdot \pi x)^{2}}{60}\rightarrow\frac{127776\cdot \pi ^{3}}{60}\\&lim_{x\rightarrow0-}\frac{16sin(11\cdot \pi x)^{3}}{10x^{3}}=\frac{(10648\cdot \pi ^{3})}{5}\end{align*}$

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Example Question #137 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{-17tan(5\cdot \pi x)^{3}}{11sin(10\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$\infty$

$-\frac{ 17}{11}$

$-\infty$

$\frac{ 17}{11}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-17tan(5\cdot \pi x)^{3}}{11sin(10\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{-255\cdot \pi tan(5\cdot \pi x)^{2}\cdot (tan(5\cdot \pi x)^{2} + 1)}{220\cdot \pi cos(10\cdot \pi x)sin(10\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{- 2550\cdot \pi ^{2}tan(5\cdot \pi x)\cdot (tan(5\cdot \pi x)^{2} + 1)^{2} - 2550\cdot \pi ^{2}tan(5\cdot \pi x)^{3}\cdot (tan(5\cdot \pi x)^{2} + 1)}{2200\cdot \pi ^{2}cos(10\cdot \pi x)^{2} - 2200\cdot \pi ^{2}sin(10\cdot \pi x)^{2}}\rightarrow\frac{0}{2200\cdot \pi ^{2}}=0\\&lim_{x\rightarrow1-}\frac{-17tan(5\cdot \pi x)^{3}}{11sin(10\cdot \pi x)^{2}}=0\end{align*}$

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Example Question #1092 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{5tan(4\cdot \pi x)}{4x}\end{align*}$

Possible Answers:

$5\cdot \pi$

$-5\cdot \pi$

$\infty$

$-\infty$

Correct answer:

$5\cdot \pi$

Explanation:

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Example Question #1091 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{3sin(3\cdot \pi x)}{-6cos(\frac{(11\cdot \pi x)}{2})^{3}}\end{align*}$

Possible Answers:

$\infty$

$\frac{ 1}{2}$

$-\frac{ 1}{2}$

$-\infty$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{3sin(3\cdot \pi x)}{-6cos(\frac{(11\cdot \pi x)}{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{9\cdot \pi cos(3\cdot \pi x)}{99\cdot \pi cos(\frac{(11\cdot \pi x)}{2})^{2}sin(\frac{(11\cdot \pi x)}{2})}\rightarrow\frac{-9\cdot \pi}{0}=\infty\\&lim_{x\rightarrow1-}\frac{3sin(3\cdot \pi x)}{-6cos(\frac{(11\cdot \pi x)}{2})^{3}}=\infty\end{align*}$

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Example Question #140 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{-15x^{3}}{19cos(5\cdot \pi x -\frac{ \pi }{2})}\end{align*}$

Possible Answers:

$-\frac{(15\cdot \pi )}{19}$

$\infty$

$-\infty$

$\frac{(15\cdot \pi )}{19}$

Correct answer:

Explanation:

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Example Question #141 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{-18sin(10x)}{17x}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{180}{17}$

$-\frac{180}{17}$

Correct answer:

$-\frac{180}{17}$

Explanation:

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #132 : Limits

Example Question #133 : Limits

Example Question #134 : Limits

Example Question #1091 : Calculus 3

Example Question #136 : Limits

Example Question #137 : Limits

Example Question #1092 : Calculus 3

Example Question #1091 : Calculus 3

Example Question #140 : Limits

Example Question #141 : Limits

All Calculus 3 Resources

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