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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #421 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{5tan(10\cdot \pi x)}{7ln(x^{6})}\end{align*}$

Possible Answers:

$\infty$

$-\frac{(25\cdot \pi )}{21}$

$-\infty$

$\frac{(25\cdot \pi )}{21}$

Correct answer:

$\frac{(25\cdot \pi )}{21}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{5tan(10\cdot \pi x)}{7ln(x^{6})}\rightarrow\frac{0}{0}\\&\frac{50\cdot \pi \cdot (tan(10\cdot \pi x)^{2} + 1)}{\frac{42}{x}}\rightarrow\frac{50\cdot \pi}{42}\\&lim_{x\rightarrow1-}\frac{5tan(10\cdot \pi x)}{7ln(x^{6})}=\frac{(25\cdot \pi )}{21}\end{align*}$

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Example Question #422 : Calculus Ii

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

Possible Answers:

$-\infty$

$\frac{20}{(81\cdot \pi ^{2})}$

$\infty$

$-\frac{20}{(81\cdot \pi ^{2})}$

Correct answer:

$\frac{20}{(81\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{5ln(x^{2})^{2}}{sin(9\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{\frac{(20ln(x^{2}))}{x}}{18\cdot \pi cos(9\cdot \pi x)sin(9\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{\frac{40}{x^{2}}-\frac{ (20ln(x^{2}))}{x^{2}}}{162\cdot \pi ^{2}cos(9\cdot \pi x)^{2} - 162\cdot \pi ^{2}sin(9\cdot \pi x)^{2}}\rightarrow\frac{40}{162\cdot \pi ^{2}}\\&lim_{x\rightarrow1+}\frac{5ln(x^{2})^{2}}{sin(9\cdot \pi x)^{2}}=\frac{20}{(81\cdot \pi ^{2})}\end{align*}$

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Example Question #423 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-7ln(x^{8})^{2}}{23\cdot (x - 1)^{2}}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{448}{23}$

$\infty$

$\frac{448}{23}$

Correct answer:

$-\frac{448}{23}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-7ln(x^{8})^{2}}{23\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{-\frac{(112ln(x^{8}))}{x}}{46x - 46}\rightarrow\frac{0}{0}\\&\frac{\frac{(112ln(x^{8}))}{x^{2}}-\frac{ 896}{x^{2}}}{46}\rightarrow\frac{-896}{46}\\&lim_{x\rightarrow1+}\frac{-7ln(x^{8})^{2}}{23\cdot (x - 1)^{2}}=-\frac{448}{23}\end{align*}$

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Example Question #424 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-3sin(20\cdot \pi x)^{3}}{41ln((x + 1)^{8})}\end{align*}$

Possible Answers:

$-\frac{(375\cdot \pi ^{3})}{551368}$

$\infty$

$-\infty$

$\frac{(375\cdot \pi ^{3})}{551368}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-3sin(20\cdot \pi x)^{3}}{41ln((x + 1)^{8})}\rightarrow\frac{0}{0}\\&\frac{-180\cdot \pi cos(20\cdot \pi x)sin(20\cdot \pi x)^{2}}{\frac{328}{(x + 1)}}\rightarrow\frac{0}{328}=0\\&lim_{x\rightarrow0-}\frac{-3sin(20\cdot \pi x)^{3}}{41ln((x + 1)^{8})}=0\end{align*}$

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Example Question #425 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{46sin(15\cdot \pi x)}{-13cos(\frac{(15\cdot \pi x)}{2})^{3}}\end{align*}$

Possible Answers:

$\infty$

$-\frac{(690\cdot \pi )}{(-\frac{(43875\cdot \pi ^{3})}{8})^{(\frac{1}{3})}}$

$-\infty$

$\frac{(690\cdot \pi )}{(-\frac{(43875\cdot \pi ^{3})}{8})^{(\frac{1}{3})}}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{46sin(15\cdot \pi x)}{-13cos(\frac{(15\cdot \pi x)}{2})^{3}}\rightarrow\frac{0}{0}\\&\frac{690\cdot \pi cos(15\cdot \pi x)}{\frac{(585\cdot \pi cos(\frac{(15\cdot \pi x)}{2})^{2}sin(\frac{(15\cdot \pi x)}{2}))}{2}}\rightarrow\frac{-690\cdot \pi}{0}=\infty\\&lim_{x\rightarrow1+}\frac{46sin(15\cdot \pi x)}{-13cos(\frac{(15\cdot \pi x)}{2})^{3}}=\infty\end{align*}$

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Example Question #426 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-27x^{3}}{32cos(10\cdot \pi x -\frac{ \pi }{2})^{2}}\end{align*}$

Possible Answers:

$-\frac{27}{(3200\cdot \pi ^{2})^{(\frac{3}{2})}}$

$\frac{27}{(3200\cdot \pi ^{2})^{(\frac{3}{2})}}$

$\infty$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-27x^{3}}{32cos(10\cdot \pi x -\frac{ \pi }{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{-81x^{2}}{-640\cdot \pi cos(10\cdot \pi x -\frac{ \pi }{2})sin(10\cdot \pi x -\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{-162x}{6400\cdot \pi ^{2}sin(10\cdot \pi x -\frac{ \pi }{2})^{2} - 6400\cdot \pi ^{2}cos(10\cdot \pi x -\frac{ \pi }{2})^{2}}\rightarrow\frac{0}{6400\cdot \pi ^{2}}=0\\&lim_{x\rightarrow0-}\frac{-27x^{3}}{32cos(10\cdot \pi x -\frac{ \pi }{2})^{2}}=0\end{align*}$

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Example Question #427 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{11ln(x^{2})^{3}}{-28sin(4\cdot \pi x)^{3}}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{11}{(224\cdot \pi ^{3})}$

$\infty$

$\frac{11}{(224\cdot \pi ^{3})}$

Correct answer:

$-\frac{11}{(224\cdot \pi ^{3})}$

Explanation:

$\begin{align*}&\text{Numerator/denominator have 0 values at the limit specified.}\\&\text{L'Hopital's method may useful:}\\&\text{If both numerator/denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{11ln(x^{2})^{3}}{-28sin(4\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{\frac{(66ln(x^{2})^{2})}{x}}{-336\cdot \pi cos(4\cdot \pi x)sin(4\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{\frac{(264ln(x^{2}))}{x^{2}}-\frac{ (66ln(x^{2})^{2})}{x^{2}}}{1344\cdot \pi ^{2}sin(4\cdot \pi x)^{3} - 2688\cdot \pi ^{2}cos(4\cdot \pi x)^{2}sin(4\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{\frac{528}{x^{3}}-\frac{ (792ln(x^{2}))}{x^{3}}+\frac{ (132ln(x^{2})^{2})}{x^{3}}}{37632\cdot \pi ^{3}cos(4\cdot \pi x)sin(4\cdot \pi x)^{2} - 10752\cdot \pi ^{3}cos(4\cdot \pi x)^{3}}\rightarrow\frac{528}{-10752\cdot \pi ^{3}}\\&lim_{x\rightarrow1+}\frac{11ln(x^{2})^{3}}{-28sin(4\cdot \pi x)^{3}}=-\frac{11}{(224\cdot \pi ^{3})}\end{align*}$

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Example Question #428 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{30sin(15\cdot \pi x)}{31ln(x^{6})^{3}}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{(25\cdot 6696^{(\frac{2}{3})}\cdot \pi )}{372}$

$\infty$

$\frac{(25\cdot 6696^{(\frac{2}{3})}\cdot \pi )}{372}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{30sin(15\cdot \pi x)}{31ln(x^{6})^{3}}\rightarrow\frac{0}{0}\\&\frac{450\cdot \pi cos(15\cdot \pi x)}{\frac{(558ln(x^{6})^{2})}{x}}\rightarrow\frac{-450\cdot \pi}{0}=-\infty\\&lim_{x\rightarrow1+}\frac{30sin(15\cdot \pi x)}{31ln(x^{6})^{3}}=-\infty\end{align*}$

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Example Question #421 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{5tan(9\cdot \pi x)^{3}}{-4sin(4\cdot \pi x)}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{3645}{4096}$

$\infty$

$\frac{3645}{4096}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{5tan(9\cdot \pi x)^{3}}{-4sin(4\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{135\cdot \pi tan(9\cdot \pi x)^{2}\cdot (tan(9\cdot \pi x)^{2} + 1)}{-16\cdot \pi cos(4\cdot \pi x)}\rightarrow\frac{0}{-16\cdot \pi}=0\\&lim_{x\rightarrow1-}\frac{5tan(9\cdot \pi x)^{3}}{-4sin(4\cdot \pi x)}=0\end{align*}$

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Example Question #430 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{9 - 9x}{5sin(20\cdot \pi x)^{3}}\end{align*}$

Possible Answers:

$\frac{9}{(-40000\cdot \pi ^{3})^{(\frac{1}{3})}}$

$-\frac{9}{(-40000\cdot \pi ^{3})^{(\frac{1}{3})}}$

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{9 - 9x}{5sin(20\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{-9}{300\cdot \pi cos(20\cdot \pi x)sin(20\cdot \pi x)^{2}}\rightarrow\frac{-9}{0}=-\infty\\&lim_{x\rightarrow1-}\frac{9 - 9x}{5sin(20\cdot \pi x)^{3}}=-\infty\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #421 : Calculus Ii

Example Question #422 : Calculus Ii

Example Question #423 : Calculus Ii

Example Question #424 : Calculus Ii

Example Question #425 : Calculus Ii

Example Question #426 : Calculus Ii

Example Question #427 : Calculus Ii

Example Question #428 : Calculus Ii

Example Question #421 : Limits

Example Question #430 : Calculus Ii

All Calculus 2 Resources

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