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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1100 : Calculus 3

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

Possible Answers:

$-\infty$

$\infty$

$\frac{180}{17}$

$-\frac{180}{17}$

Correct answer:

$-\frac{180}{17}$

Explanation:

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Example Question #141 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{-11sin(17\cdot \pi x)}{10\cdot (x - 1)^{2}}\end{align*}$

Possible Answers:

$-\frac{11}{10}$

$\infty$

$\frac{11}{10}$

$-\infty$

Correct answer:

$-\infty$

Explanation:

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Example Question #1102 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-11sin(3\cdot \pi x)^{2}}{-20cos(\frac{(19\cdot \pi x)}{2})^{3}}\end{align*}$

Possible Answers:

$-\frac{11}{20}$

$\infty$

$\frac{11}{20}$

$-\infty$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-11sin(3\cdot \pi x)^{2}}{-20cos(\frac{(19\cdot \pi x)}{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{-66\cdot \pi cos(3\cdot \pi x)sin(3\cdot \pi x)}{570\cdot \pi cos(\frac{(19\cdot \pi x)}{2})^{2}sin(\frac{(19\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{198\cdot \pi ^{2}sin(3\cdot \pi x)^{2} - 198\cdot \pi ^{2}cos(3\cdot \pi x)^{2}}{5415\cdot \pi ^{2}cos(\frac{(19\cdot \pi x)}{2})^{3} - 10830\cdot \pi ^{2}cos(\frac{(19\cdot \pi x)}{2})sin(\frac{(19\cdot \pi x)}{2})^{2}}\rightarrow\frac{-198\cdot \pi ^{2}}{0}=\infty\\&lim_{x\rightarrow1+}\frac{-11sin(3\cdot \pi x)^{2}}{-20cos(\frac{(19\cdot \pi x)}{2})^{3}}=\infty\end{align*}$

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Example Question #141 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{11sin(3\cdot \pi x)}{-9cos(\frac{(5\cdot \pi x)}{2})^{3}}\end{align*}$

Possible Answers:

$\frac{(11\cdot \pi )}{9}$

$\infty$

$-\frac{(11\cdot \pi )}{9}$

$-\infty$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{11sin(3\cdot \pi x)}{-9cos(\frac{(5\cdot \pi x)}{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{33\cdot \pi cos(3\cdot \pi x)}{\frac{(135\cdot \pi cos(\frac{(5\cdot \pi x)}{2})^{2}sin(\frac{(5\cdot \pi x)}{2}))}{2}}\rightarrow\frac{-33\cdot \pi}{0}=-\infty\\&lim_{x\rightarrow1+}\frac{11sin(3\cdot \pi x)}{-9cos(\frac{(5\cdot \pi x)}{2})^{3}}=-\infty\end{align*}$

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Example Question #142 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{-19tan(20x)^{2}}{2x^{3}}\end{align*}$

Possible Answers:

$\infty$

$\frac{(19\cdot \pi )}{2}$

$-\frac{(19\cdot \pi )}{2}$

$-\infty$

Correct answer:

$-\infty$

Explanation:

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Example Question #143 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{14 - 14x}{4sin(8\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{7}{2}$

$\infty$

$\frac{7}{2}$

Correct answer:

$\infty$

Explanation:

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Example Question #144 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-13tan(12x)^{2}}{8sin(13\cdot \pi x)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$-\frac{13\cdot\pi}{8}$

$\frac{13\cdot\pi}{8}$

Correct answer:

Explanation:

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Example Question #1101 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{7sin(12\cdot \pi x)^{3}}{-12cos(\frac{(7\cdot \pi x)}{2})^{3}}\end{align*}$

Possible Answers:

$\frac{1152}{49}$

$-\frac{1152}{49}$

$\infty$

$-\infty$

Correct answer:

$-\frac{1152}{49}$

Explanation:

$\begin{align*}\\&\frac{7sin(12\cdot \pi x)^{3}}{-12cos(\frac{(7\cdot \pi x)}{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{252\cdot \pi cos(12\cdot \pi x)sin(12\cdot \pi x)^{2}}{126\cdot \pi cos(\frac{(7\cdot \pi x)}{2})^{2}sin(\frac{(7\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{6048\cdot \pi ^{2}cos(12\cdot \pi x)^{2}sin(12\cdot \pi x) - 3024\cdot \pi ^{2}sin(12\cdot \pi x)^{3}}{441\cdot \pi ^{2}cos(\frac{(7\cdot \pi x)}{2})^{3} - 882\cdot \pi ^{2}cos(\frac{(7\cdot \pi x)}{2})sin(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{72576\cdot \pi ^{3}cos(12\cdot \pi x)^{3} - 254016\cdot \pi ^{3}cos(12\cdot \pi x)sin(12\cdot \pi x)^{2}}{3087\cdot \pi ^{3}sin(\frac{(7\cdot \pi x)}{2})^{3} -\frac{ (21609\cdot \pi ^{3}cos(\frac{(7\cdot \pi x)}{2})^{2}sin(\frac{(7\cdot \pi x)}{2}))}{2}}\rightarrow\frac{72576\cdot \pi ^{3}}{-3087\cdot \pi ^{3}}\\&lim_{x\rightarrow1-}\frac{7sin(12\cdot \pi x)^{3}}{-12cos(\frac{(7\cdot \pi x)}{2})^{3}}=-\frac{1152}{49}\end{align*}$

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Example Question #145 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{-4\cdot (x - 1)^{2}}{9sin(2\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$\frac{1}{(9\cdot \pi ^{2})}$

$-\frac{1}{(9\cdot \pi ^{2})}$

$-\infty$

$\infty$

Correct answer:

$-\frac{1}{(9\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-4\cdot (x - 1)^{2}}{9sin(2\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{8 - 8x}{36\cdot \pi cos(2\cdot \pi x)sin(2\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{-8}{72\cdot \pi ^{2}cos(2\cdot \pi x)^{2} - 72\cdot \pi ^{2}sin(2\cdot \pi x)^{2}}\rightarrow\frac{-8}{72\cdot \pi ^{2}}\\&lim_{x\rightarrow1-}\frac{-4\cdot (x - 1)^{2}}{9sin(2\cdot \pi x)^{2}}=-\frac{1}{(9\cdot \pi ^{2})}\end{align*}$

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Example Question #1109 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-14tan(3\cdot \pi x)^{3}}{9\cdot (x - 1)^{3}}\end{align*}$

Possible Answers:

$-\infty$

$-42\cdot \pi ^{3}$

$\infty$

$42\cdot \pi ^{3}$

Correct answer:

$-42\cdot \pi ^{3}$

Explanation:

$\begin{align*}\\&\frac{-14tan(3\cdot \pi x)^{3}}{9\cdot (x - 1)^{3}}\rightarrow\frac{0}{0}\\&\frac{-126\cdot \pi tan(3\cdot \pi x)^{2}\cdot (tan(3\cdot \pi x)^{2} + 1)}{27\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{- 756\cdot \pi ^{2}tan(3\cdot \pi x)\cdot (tan(3\cdot \pi x)^{2} + 1)^{2} - 756\cdot \pi ^{2}tan(3\cdot \pi x)^{3}\cdot (tan(3\cdot \pi x)^{2} + 1)}{54x - 54}\rightarrow\frac{0}{0}\\&\frac{- 2268\cdot \pi ^{3}\cdot (tan(3\cdot \pi x)^{2} + 1)^{3} - 4536\cdot \pi ^{3}tan(3\cdot \pi x)^{4}\cdot (tan(3\cdot \pi x)^{2} + 1) - 15876\cdot \pi ^{3}tan(3\cdot \pi x)^{2}\cdot (tan(3\cdot \pi x)^{2} + 1)^{2}}{54}\rightarrow\frac{-2268\cdot \pi ^{3}}{54}\\&lim_{x\rightarrow1+}\frac{-14tan(3\cdot \pi x)^{3}}{9\cdot (x - 1)^{3}}=-42\cdot \pi ^{3}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1100 : Calculus 3

Example Question #141 : Limits

Example Question #1102 : Calculus 3

Example Question #141 : Limits

Example Question #142 : Limits

Example Question #143 : Limits

Example Question #144 : Limits

Example Question #1101 : Calculus 3

Example Question #145 : Limits

Example Question #1109 : Calculus 3

All Calculus 3 Resources

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