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Calculus 2 : Finding Limits and One-Sided Limits

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Example Questions

Example Question #401 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow0+}\frac{19tan(2x)}{33ln((x + 1)^{3})^{3}}\end{align*}$

Possible Answers:

$\frac{(38\cdot 891^{(\frac{2}{3})})}{891}$

$\infty$

$-\frac{(38\cdot 891^{(\frac{2}{3})})}{891}$

$-\infty$

Correct answer:

$\infty$

Explanation:

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Example Question #402 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1+}\frac{9sin(6\cdot \pi x)^{2}}{7ln(x^{2})^{2}}\end{align*}$

Possible Answers:

$-\frac{(81\cdot \pi ^{2})}{7}$

$\infty$

$\frac{(81\cdot \pi ^{2})}{7}$

$-\infty$

Correct answer:

$\frac{(81\cdot \pi ^{2})}{7}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{9sin(6\cdot \pi x)^{2}}{7ln(x^{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{108\cdot \pi cos(6\cdot \pi x)sin(6\cdot \pi x)}{\frac{(28ln(x^{2}))}{x}}\rightarrow\frac{0}{0}\\&\frac{648\cdot \pi ^{2}cos(6\cdot \pi x)^{2} - 648\cdot \pi ^{2}sin(6\cdot \pi x)^{2}}{\frac{56}{x^{2}}-\frac{ (28ln(x^{2}))}{x^{2}}}\rightarrow\frac{648\cdot \pi ^{2}}{56}\\&\lim_{x\rightarrow1+}\frac{9sin(6\cdot \pi x)^{2}}{7ln(x^{2})^{2}}=\frac{(81\cdot \pi ^{2})}{7}\end{align*}$

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Example Question #403 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1-}\frac{-9ln(x^{5})^{2}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}\end{align*}$

Possible Answers:

$-\infty$

$\frac{450}{(539\cdot \pi ^{2})}$

$-\frac{450}{(539\cdot \pi ^{2})}$

$\infty$

Correct answer:

$-\frac{450}{(539\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-9ln(x^{5})^{2}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{-\frac{(90ln(x^{5}))}{x}}{-154\cdot \pi cos(\frac{(7\cdot \pi x)}{2})sin(\frac{(7\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{\frac{(90ln(x^{5}))}{x^{2}}-\frac{ 450}{x^{2}}}{539\cdot \pi ^{2}sin(\frac{(7\cdot \pi x)}{2})^{2} - 539\cdot \pi ^{2}cos(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{-450}{539\cdot \pi ^{2}}\\&\lim_{x\rightarrow1-}\frac{-9ln(x^{5})^{2}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}=-\frac{450}{(539\cdot \pi ^{2})}\end{align*}$

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Example Question #404 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow0-}\frac{-17x^{2}}{14sin(12\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{17}{(2016\cdot \pi ^{2})}$

$-\frac{17}{(2016\cdot \pi ^{2})}$

Correct answer:

$-\frac{17}{(2016\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-17x^{2}}{14sin(12\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{-34x}{336\cdot \pi cos(12\cdot \pi x)sin(12\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{-34}{4032\cdot \pi ^{2}cos(12\cdot \pi x)^{2} - 4032\cdot \pi ^{2}sin(12\cdot \pi x)^{2}}\rightarrow\frac{-34}{4032\cdot \pi ^{2}}\\&\lim_{x\rightarrow0-}\frac{-17x^{2}}{14sin(12\cdot \pi x)^{2}}=-\frac{17}{(2016\cdot \pi ^{2})}\end{align*}$

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Example Question #405 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow0+}\frac{-17x^{2}}{48cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi }{2})}\end{align*}$

Possible Answers:

$-\infty$

$\frac{17}{(207936\cdot \pi ^{2})}$

$\infty$

$-\frac{17}{(207936\cdot \pi ^{2})}$

Correct answer:

Explanation:

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Example Question #406 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1+}\frac{19tan(13\cdot \pi x)^{3}}{sin(19\cdot \pi x)^{3}}\end{align*}$

Possible Answers:

$\frac{2197}{361}$

$-\frac{2197}{361}$

$-\infty$

$\infty$

Correct answer:

$-\frac{2197}{361}$

Explanation:

$\begin{align*}\\&\frac{250458\cdot \pi ^{3}\cdot (tan(13\cdot \pi x)^{2} + 1)^{3} + 500916\cdot \pi ^{3}tan(13\cdot \pi x)^{4}\cdot (tan(13\cdot \pi x)^{2} + 1) + 1753206\cdot \pi ^{3}tan(13\cdot \pi x)^{2}\cdot (tan(13\cdot \pi x)^{2} + 1)^{2}}{41154\cdot \pi ^{3}cos(19\cdot \pi x)^{3} - 144039\cdot \pi ^{3}cos(19\cdot \pi x)sin(19\cdot \pi x)^{2}}\rightarrow\frac{250458\cdot \pi ^{3}}{-41154\cdot \pi ^{3}}\\&\lim_{x\rightarrow1+}\frac{19tan(13\cdot \pi x)^{3}}{sin(19\cdot \pi x)^{3}}=-\frac{2197}{361}\end{align*}$ $\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{19tan(13\cdot \pi x)^{3}}{sin(19\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{741\cdot \pi tan(13\cdot \pi x)^{2}\cdot (tan(13\cdot \pi x)^{2} + 1)}{57\cdot \pi cos(19\cdot \pi x)sin(19\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{19266\cdot \pi ^{2}tan(13\cdot \pi x)\cdot (tan(13\cdot \pi x)^{2} + 1)^{2} + 19266\cdot \pi ^{2}tan(13\cdot \pi x)^{3}\cdot (tan(13\cdot \pi x)^{2} + 1)}{2166\cdot \pi ^{2}cos(19\cdot \pi x)^{2}sin(19\cdot \pi x) - 1083\cdot \pi ^{2}sin(19\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\end{align*}$

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Example Question #407 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow0+}\frac{-5tan(7x)^{3}}{26sin(14\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$\frac{1715}{(5096\cdot \pi ^{2})^{(\frac{3}{2})}}$

$-\frac{1715}{(5096\cdot \pi ^{2})^{(\frac{3}{2})}}$

$\infty$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-5tan(7x)^{3}}{26sin(14\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{-15tan(7x)^{2}\cdot (7tan(7x)^{2} + 7)}{728\cdot \pi cos(14\cdot \pi x)sin(14\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{- 30tan(7x)\cdot (7tan(7x)^{2} + 7)^{2} - 210tan(7x)^{3}\cdot (7tan(7x)^{2} + 7)}{10192\cdot \pi ^{2}cos(14\cdot \pi x)^{2} - 10192\cdot \pi ^{2}sin(14\cdot \pi x)^{2}}\rightarrow\frac{0}{10192\cdot \pi ^{2}}=0\\&\lim_{x\rightarrow0+}\frac{-5tan(7x)^{3}}{26sin(14\cdot \pi x)^{2}}=0\end{align*}$

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Example Question #411 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1-}\frac{11x - 11}{-9sin(17\cdot \pi x)}\end{align*}$

Possible Answers:

$\frac{11}{(153\cdot \pi )}$

$\infty$

$-\frac{11}{(153\cdot \pi )}$

$-\infty$

Correct answer:

$\frac{11}{(153\cdot \pi )}$

Explanation:

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Example Question #412 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1-}\frac{-17sin(17\cdot \pi x)^{3}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}\end{align*}$

Possible Answers:

$\infty$

$-\frac{(83521\cdot \pi ^{3})}{(\frac{(539\cdot \pi ^{2})}{2})^{(\frac{3}{2})}}$

$\frac{(83521\cdot \pi ^{3})}{(\frac{(539\cdot \pi ^{2})}{2})^{(\frac{3}{2})}}$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-17sin(17\cdot \pi x)^{3}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{-867\cdot \pi cos(17\cdot \pi x)sin(17\cdot \pi x)^{2}}{-154\cdot \pi cos(\frac{(7\cdot \pi x)}{2})sin(\frac{(7\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{14739\cdot \pi ^{2}sin(17\cdot \pi x)^{3} - 29478\cdot \pi ^{2}cos(17\cdot \pi x)^{2}sin(17\cdot \pi x)}{539\cdot \pi ^{2}sin(\frac{(7\cdot \pi x)}{2})^{2} - 539\cdot \pi ^{2}cos(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{539\cdot \pi ^{2}}=0\\&\lim_{x\rightarrow1-}\frac{-17sin(17\cdot \pi x)^{3}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}=0\end{align*}$

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Example Question #413 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1+}\frac{49ln(x^{3})}{27x - 27}\end{align*}$

Possible Answers:

$-\frac{49}{9}$

$\frac{49}{9}$

$-\infty$

$\infty$

Correct answer:

$\frac{49}{9}$

Explanation:

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Calculus 2 : Finding Limits and One-Sided Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #401 : Calculus Ii

Example Question #402 : Calculus Ii

Example Question #403 : Calculus Ii

Example Question #404 : Calculus Ii

Example Question #405 : Calculus Ii

Example Question #406 : Calculus Ii

Example Question #407 : Calculus Ii

Example Question #411 : Calculus Ii

Example Question #412 : Calculus Ii

Example Question #413 : Calculus Ii

All Calculus 2 Resources

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