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Calculus 2 : Finding Limits and One-Sided Limits

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Example Questions

Example Question #381 : Finding Limits And One Sided Limits

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

Possible Answers:

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

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

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

Correct answer:

$\displaystyle 0$

Explanation:

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

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Example Question #382 : Finding Limits And One Sided Limits

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

Possible Answers:

$\displaystyle 0$

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

$-\infty$ $\displaystyle -\infty$

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

$\infty$ $\displaystyle \infty$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

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

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Example Question #383 : Finding Limits And One Sided Limits

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

Possible Answers:

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

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

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

Correct answer:

$\displaystyle 0$

Explanation:

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

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Example Question #384 : Finding Limits And One Sided Limits

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

Possible Answers:

$\displaystyle 0$

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

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

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

Correct answer:

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

Explanation:

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

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Example Question #385 : Finding Limits And One Sided Limits

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

Possible Answers:

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

$\displaystyle 0$

$\infty$ $\displaystyle \infty$

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

$-\infty$ $\displaystyle -\infty$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

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

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Example Question #386 : Finding Limits And One Sided Limits

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

Possible Answers:

$\displaystyle 0$

$\frac{3645}{4096}$ $\displaystyle \frac{3645}{4096}$

$-\frac{3645}{4096}$ $\displaystyle -\frac{3645}{4096}$

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

Correct answer:

$\displaystyle 0$

Explanation:

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

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Example Question #387 : Finding Limits And One Sided Limits

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

Possible Answers:

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

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

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

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

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

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Example Question #388 : Finding Limits And One Sided Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\end{align*}$

Possible Answers:

$-\infty$ $\displaystyle -\infty$

$\frac{(35\cdot \pi )}{2}$ $\displaystyle \frac{(35\cdot \pi )}{2}$

$\infty$ $\displaystyle \infty$

$-\frac{(35\cdot \pi )}{2}$ $\displaystyle -\frac{(35\cdot \pi )}{2}$

$\displaystyle 0$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-140\cdot \pi cos(7\cdot \pi x)}{\frac{(24ln(x^{8})^{2})}{x}}\rightarrow\frac{140\cdot \pi}{0}=\infty\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}=\infty\end{align*}$ $\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-140\cdot \pi cos(7\cdot \pi x)}{\frac{(24ln(x^{8})^{2})}{x}}\rightarrow\frac{140\cdot \pi}{0}=\infty\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}=\infty\end{align*}$

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Example Question #389 : Finding Limits And One Sided Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{10x}{19ln((x + 1)^{8})^{2}}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{10x}{19ln((x + 1)^{8})^{2}}\end{align*}$

Possible Answers:

$\frac{(5\cdot 1216^{(\frac{1}{2})})}{608}$ $\displaystyle \frac{(5\cdot 1216^{(\frac{1}{2})})}{608}$

$-\frac{(5\cdot 1216^{(\frac{1}{2})})}{608}$ $\displaystyle -\frac{(5\cdot 1216^{(\frac{1}{2})})}{608}$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

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Example Question #390 : Finding Limits And One Sided Limits

$\lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}$ $\displaystyle \lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}$

Possible Answers:

2xh $\displaystyle 2xh$

$\displaystyle 0$

$\displaystyle 2x$

$\displaystyle 2$

$\displaystyle 2h$

Correct answer:

$\displaystyle 2x$

Explanation:

This question is asking you to recall the limit definition of a derivative, which states:

$\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}=f'(x).$ $\displaystyle \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}=f'(x).$

Therefore, if we can define our function for our problem, we can simply evaluate this limit by taking its derivative.

$\lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}.$ $\displaystyle \lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}.$ In these types of questions, you look at the second term in the numerator. That is the negative of the function.

Since $\displaystyle f(x)=x^2$, $\displaystyle f'(x)=2x.$

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Calculus 2 : Finding Limits and One-Sided Limits

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #381 : Finding Limits And One Sided Limits

Example Question #382 : Finding Limits And One Sided Limits

Example Question #383 : Finding Limits And One Sided Limits

Example Question #384 : Finding Limits And One Sided Limits

Example Question #385 : Finding Limits And One Sided Limits

Example Question #386 : Finding Limits And One Sided Limits

Example Question #387 : Finding Limits And One Sided Limits

Example Question #388 : Finding Limits And One Sided Limits

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Example Question #390 : Finding Limits And One Sided Limits

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