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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 #152 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{6ln((x + 1)^{3})}{14x^{3}}\end{align*}$

Possible Answers:

$\frac{(3\cdot \pi )}{7}$

$\infty$

$-\frac{(3\cdot \pi )}{7}$

$-\infty$

Correct answer:

$\infty$

Explanation:

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Example Question #153 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{19tan(16x)^{2}}{sin(4\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$\frac{304}{\pi ^{2}}$

$-\frac{304}{\pi ^{2}}$

$-\infty$

$\infty$

Correct answer:

$\frac{304}{\pi ^{2}}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{19tan(16x)^{2}}{sin(4\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{38tan(16x)\cdot (16tan(16x)^{2} + 16)}{8\cdot \pi cos(4\cdot \pi x)sin(4\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{1216tan(16x)^{2}\cdot (16tan(16x)^{2} + 16) + 38\cdot (16tan(16x)^{2} + 16)^{2}}{32\cdot \pi ^{2}cos(4\cdot \pi x)^{2} - 32\cdot \pi ^{2}sin(4\cdot \pi x)^{2}}\rightarrow\frac{9728}{32\cdot \pi ^{2}}\\&lim_{x\rightarrow0-}\frac{19tan(16x)^{2}}{sin(4\cdot \pi x)^{2}}=\frac{304}{\pi ^{2}}\end{align*}$

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Example Question #154 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{-4ln((x + 1)^{2})^{3}}{7cos(\pi x -\frac{ \pi }{2})}\end{align*}$

Possible Answers:

$\infty$

$-\frac{(4\cdot \pi )}{7}$

$-\infty$

$\frac{(4\cdot \pi )}{7}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-4ln((x + 1)^{2})^{3}}{7cos(\pi x -\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{-\frac{(12ln((x + 1)^{2})^{2}\cdot (2x + 2))}{(x + 1)^{2}}}{-7\cdot \pi sin(\pi x -\frac{ \pi }{2})}\rightarrow\frac{0}{7\cdot \pi}=0\\&lim_{x\rightarrow0+}\frac{-4ln((x + 1)^{2})^{3}}{7cos(\pi x -\frac{ \pi }{2})}=0\end{align*}$

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Example Question #155 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{9sin(15\cdot \pi x)^{2}}{13\cdot (x - 1)^{2}}\end{align*}$

Possible Answers:

$-\frac{(2025\cdot \pi ^{2})}{13}$

$-\infty$

$\frac{(2025\cdot \pi ^{2})}{13}$

$\infty$

Correct answer:

$\frac{(2025\cdot \pi ^{2})}{13}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{9sin(15\cdot \pi x)^{2}}{13\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{270\cdot \pi cos(15\cdot \pi x)sin(15\cdot \pi x)}{26x - 26}\rightarrow\frac{0}{0}\\&\frac{4050\cdot \pi ^{2}cos(15\cdot \pi x)^{2} - 4050\cdot \pi ^{2}sin(15\cdot \pi x)^{2}}{26}\rightarrow\frac{4050\cdot \pi ^{2}}{26}\\&lim_{x\rightarrow1+}\frac{9sin(15\cdot \pi x)^{2}}{13\cdot (x - 1)^{2}}=\frac{(2025\cdot \pi ^{2})}{13}\end{align*}$

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Example Question #156 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{-10ln((x + 1)^{3})^{2}}{14x^{3}}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{(5\cdot \pi )}{7}$

$-\frac{(5\cdot \pi )}{7}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-10ln((x + 1)^{3})^{2}}{14x^{3}}\rightarrow\frac{0}{0}\\&\frac{-\frac{(60ln((x + 1)^{3}))}{(x + 1)}}{42x^{2}}\rightarrow\frac{0}{0}\\&\frac{\frac{(60ln((x + 1)^{3}))}{(x + 1)^{2}}-\frac{ 180}{(x + 1)^{2}}}{84x}\rightarrow\frac{-180}{0}=-\infty\\&lim_{x\rightarrow0+}\frac{-10ln((x + 1)^{3})^{2}}{14x^{3}}=-\infty\end{align*}$

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Example Question #157 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{6\cdot (x - 1)^{2}}{7ln(x^{2})^{3}}\end{align*}$

Possible Answers:

$\frac{6}{7}$

$-\infty$

$\infty$

$-\frac{6}{7}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{6\cdot (x - 1)^{2}}{7ln(x^{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{12x - 12}{\frac{(42ln(x^{2})^{2})}{x}}\rightarrow\frac{0}{0}\\&\frac{12}{\frac{(168ln(x^{2}))}{x^{2}}-\frac{ (42ln(x^{2})^{2})}{x^{2}}}\rightarrow\frac{12}{0}=-\infty\\&lim_{x\rightarrow1-}\frac{6\cdot (x - 1)^{2}}{7ln(x^{2})^{3}}=-\infty\end{align*}$

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Example Question #158 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{9ln(x^{8})^{2}}{9sin(3\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$\infty$

$-\frac{64}{(9\cdot \pi ^{2})}$

$\frac{64}{(9\cdot \pi ^{2})}$

$-\infty$

Correct answer:

$\frac{64}{(9\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{9ln(x^{8})^{2}}{9sin(3\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{\frac{(144ln(x^{8}))}{x}}{54\cdot \pi cos(3\cdot \pi x)sin(3\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{\frac{1152}{x^{2}}-\frac{ (144ln(x^{8}))}{x^{2}}}{162\cdot \pi ^{2}cos(3\cdot \pi x)^{2} - 162\cdot \pi ^{2}sin(3\cdot \pi x)^{2}}\rightarrow\frac{1152}{162\cdot \pi ^{2}}\\&lim_{x\rightarrow1-}\frac{9ln(x^{8})^{2}}{9sin(3\cdot \pi x)^{2}}=\frac{64}{(9\cdot \pi ^{2})}\end{align*}$

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Example Question #159 : Partial Derivatives

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

Possible Answers:

$-\infty$

$\frac{4}{(9\cdot \pi ^{2})}$

$-\frac{4}{(9\cdot \pi ^{2})}$

$\infty$

Correct answer:

$-\frac{4}{(9\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-11x^{2}}{11cos(\frac{(3\cdot \pi x)}{2}-\frac{ \pi }{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{-22x}{-33\cdot \pi cos(\frac{(3\cdot \pi x)}{2}-\frac{ \pi }{2})sin(\frac{(3\cdot \pi x)}{2}-\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{-22}{\frac{(99\cdot \pi ^{2}sin(\frac{(3\cdot \pi x)}{2}-\frac{ \pi }{2})^{2})}{2}-\frac{ (99\cdot \pi ^{2}cos(\frac{(3\cdot \pi x)}{2}-\frac{ \pi }{2})^{2})}{2}}\rightarrow\frac{-22}{\frac{(99\cdot \pi ^{2})}{2}}\\&lim_{x\rightarrow0+}\frac{-11x^{2}}{11cos(\frac{(3\cdot \pi x)}{2}-\frac{ \pi }{2})^{2}}=-\frac{4}{(9\cdot \pi ^{2})}\end{align*}$

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Example Question #160 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{-6tan(15x)^{2}}{x^{2}}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

1350

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-6tan(15x)^{2}}{x^{2}}\rightarrow\frac{0}{0}\\&\frac{-12tan(15x)\cdot (15tan(15x)^{2} + 15)}{2x}\rightarrow\frac{0}{0}\\&\frac{- 360tan(15x)^{2}\cdot (15tan(15x)^{2} + 15) - 12\cdot (15tan(15x)^{2} + 15)^{2}}{2}\rightarrow\frac{-2700}{2}\\&lim_{x\rightarrow0+}\frac{-6tan(15x)^{2}}{x^{2}}=-1350\end{align*}$

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Example Question #161 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-9tan(18x)}{ln((x + 1)^{4})^{3}}\end{align*}$

Possible Answers:

$-9\cdot \pi$

$9\cdot \pi$

$-\infty$

$\infty$

Correct answer:

$-\infty$

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 #152 : Partial Derivatives

Example Question #153 : Partial Derivatives

Example Question #154 : Partial Derivatives

Example Question #155 : Partial Derivatives

Example Question #156 : Partial Derivatives

Example Question #157 : Partial Derivatives

Example Question #158 : Partial Derivatives

Example Question #159 : Partial Derivatives

Example Question #160 : Partial Derivatives

Example Question #161 : Partial Derivatives

All Calculus 3 Resources

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