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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 #121 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow-1}\frac{(x + 1)}{(10x + x^{2} + 9)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{ 1}{8}$

Correct answer:

$\frac{ 1}{8}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-1\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(x + 1)}{(10x + x^{2} + 9)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x + 1)}{((x + 1)\cdot (x + 9))}\\&\frac{1}{(x + 9)}\\&lim_{x\rightarrow-1}\frac{(x + 1)}{(10x + x^{2} + 9)}=\frac{ 1}{8}\end{align*}$

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Example Question #122 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow-6}\frac{(10x + x^{2} + 24)}{(x + 6)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-6\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(10x + x^{2} + 24)}{(x + 6)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x + 4)\cdot (x + 6))}{(x + 6)}\\&x + 4\\&lim_{x\rightarrow-6}\frac{(10x + x^{2} + 24)}{(x + 6)}= -2\end{align*}$

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Example Question #123 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow-9}\frac{(x + 9)}{(3x + x^{2} - 54)}\end{align*}$

Possible Answers:

$-\frac{1}{15}$

$\frac{ 1}{15}$

$\infty$

$-\infty$

Correct answer:

$-\frac{1}{15}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-9\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(x + 9)}{(3x + x^{2} - 54)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x + 9)}{((x - 6)\cdot (x + 9))}\\&\frac{1}{(x - 6)}\\&lim_{x\rightarrow-9}\frac{(x + 9)}{(3x + x^{2} - 54)}= -\frac{1}{15}\end{align*}$

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Example Question #124 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{4x}{20cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi}{2})^{4}}\end{align*}$

Possible Answers:

$-\frac{ 16}{11}$

$\infty$

$-\infty$

$\frac{ 16}{11}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{From examination of this problem, it is apparent that both}\\&\text{the numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{4x}{20cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi}{2})^{4}}\rightarrow\frac{0}{0}\\&\frac{4}{-760\cdot \pi cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi}{2})^{3}sin(\frac{(19\cdot \pi x)}{2}-\frac{ \pi}{2})}\rightarrow\frac{4}{0}=\infty\\&lim_{x\rightarrow0+}\frac{4x}{20cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi}{2})^{4}}=\infty\end{align*}$

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Example Question #125 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{2x}{17cos(4\cdot \pi x -\frac{ \pi }{2})^{4}}\end{align*}$

Possible Answers:

$-\frac{ 80}{19}$

$-\infty$

$\frac{ 80}{19}$

$\infty$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{From examination of this problem, it is apparent that both}\\&\text{the numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{2x}{17cos(4\cdot \pi x -\frac{ \pi }{2})^{4}}\rightarrow\frac{0}{0}\\&\frac{2}{-272\cdot \pi cos(4\cdot \pi x -\frac{ \pi }{2})^{3}sin(4\cdot \pi x -\frac{ \pi }{2})}\rightarrow\frac{2}{0}=\infty\\&lim_{x\rightarrow0+}\frac{2x}{17cos(4\cdot \pi x -\frac{ \pi }{2})^{4}}=\infty\end{align*}$

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Example Question #124 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-9sin(14\cdot \pi x)^{3}}{13cos(\frac{(21\cdot \pi x)}{2})^{2}}\end{align*}$

Possible Answers:

$\frac{ 9}{13}$

$\infty$

$-\frac{ 9}{13}$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{Looking at this problem, we see that}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-9sin(14\cdot \pi x)^{3}}{13cos(\frac{(21\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{-378\cdot \pi cos(14\cdot \pi x)sin(14\cdot \pi x)^{2}}{-273\cdot \pi cos(\frac{(21\cdot \pi x)}{2})sin(\frac{(21\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{5292\cdot \pi ^{2}sin(14\cdot \pi x)^{3} - 10584\cdot \pi ^{2}cos(14\cdot \pi x)^{2}sin(14\cdot \pi x)}{\frac{(5733\cdot \pi ^{2}sin(\frac{(21\cdot \pi x)}{2})^{2})}{2}-\frac{ (5733\cdot \pi ^{2}cos(\frac{(21\cdot \pi x)}{2})^{2})}{2}}\rightarrow\frac{0}{\frac{(5733\cdot \pi ^{2})}{2}}=0\\&lim_{x\rightarrow1+}\frac{-9sin(14\cdot \pi x)^{3}}{13cos(\frac{(21\cdot \pi x)}{2})^{2}}=0\end{align*}$

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Example Question #127 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{20x - 20}{cos(\frac{(5\cdot \pi x)}{2})^{4}}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{20x - 20}{cos(\frac{(5\cdot \pi x)}{2})^{4}}\rightarrow\frac{0}{0}\\&\frac{20}{-10\cdot \pi cos(\frac{(5\cdot \pi x)}{2})^{3}sin(\frac{(5\cdot \pi x)}{2})}\rightarrow\frac{20}{0}=\infty\\&lim_{x\rightarrow1+}\frac{20x - 20}{cos(\frac{(5\cdot \pi x)}{2})^{4}}=\infty\end{align*}$

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Example Question #128 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{14sin(19x)^{4}}{15cos(10\cdot \pi x -\frac{ \pi }{2})}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{ 14}{15}$

$\infty$

$\frac{ 14}{15}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{14sin(19x)^{4}}{15cos(10\cdot \pi x -\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{1064cos(19x)sin(19x)^{3}}{-150\cdot \pi sin(10\cdot \pi x -\frac{ \pi }{2})}\rightarrow\frac{0}{150\cdot \pi}=0\\&lim_{x\rightarrow0-}\frac{14sin(19x)^{4}}{15cos(10\cdot \pi x -\frac{ \pi }{2})}=0\end{align*}$

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Example Question #129 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{10x^{2}}{cos(2\cdot \pi x -\frac{ \pi }{2})^{3}}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{10x^{2}}{cos(2\cdot \pi x -\frac{ \pi }{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{20x}{-6\cdot \pi cos(2\cdot \pi x -\frac{ \pi }{2})^{2}sin(2\cdot \pi x -\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{20}{24\cdot \pi ^{2}cos(2\cdot \pi x -\frac{ \pi }{2})sin(2\cdot \pi x -\frac{ \pi }{2})^{2} - 12\cdot \pi ^{2}cos(2\cdot \pi x -\frac{ \pi }{2})^{3}}\rightarrow\frac{20}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{10x^{2}}{cos(2\cdot \pi x -\frac{ \pi }{2})^{3}}=-\infty\end{align*}$

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Example Question #131 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{16\cdot (x - 1)^{4}}{-6cos(\frac{(3\cdot \pi x)}{2})}\end{align*}$

Possible Answers:

$-\frac{ 8}{3}$

$-\infty$

$\infty$

$\frac{ 8}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{16\cdot (x - 1)^{4}}{-6cos(\frac{(3\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{64\cdot (x - 1)^{3}}{9\cdot \pi sin(\frac{(3\cdot \pi x)}{2})}\rightarrow\frac{0}{-9\cdot \pi}=0\\&lim_{x\rightarrow1+}\frac{16\cdot (x - 1)^{4}}{-6cos(\frac{(3\cdot \pi x)}{2})}=0\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #121 : Partial Derivatives

Example Question #122 : Partial Derivatives

Example Question #123 : Partial Derivatives

Example Question #124 : Limits

Example Question #125 : Limits

Example Question #124 : Partial Derivatives

Example Question #127 : Limits

Example Question #128 : Limits

Example Question #129 : Limits

Example Question #131 : Limits

All Calculus 3 Resources

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