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Calculus 2 : Limits

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Example Questions

Example Question #356 : Finding Limits And One Sided Limits

$\lim_{x\rightarrow 3^+}\frac{1}{x-3}$

Possible Answers:

$-\infty$

$\infty$

Does not exist.

Correct answer:

$\infty$

Explanation:

To check a one-sided limit, we need to plug numbers into the function close to the value in question. Let's start with a whole number to the right of our value. The closest whole number is

$\frac{1}{4-3}=\frac{1}{1}=1.$

Now, we need to work our way closer to the value in question. Let's try 3.5.

$\frac{1}{3.5-3}=\frac{1}{0.5}=2.$

Now, we are getting a larger value. That is still inconclusive. We must try one more closer. Let's do 3.25.

$\frac{1}{3.25-3}=\frac{1}{0.25}=4.$

We are even bigger still! Therefore, we must be increasing to positive infinity.

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Example Question #357 : Finding Limits And One Sided Limits

$\lim_{x\rightarrow 2^-} \frac{2}{x-2}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

To check a one-sided limit, we need to plug numbers into the function close to the value in question. Let's start with a whole number to the left of our value. The closest whole number is

$\frac{2}{1-2}=\frac{2}{-1}=-2.$

Now, we need to work our way closer to the value in question. Let's try 1.5.

$\frac{2}{1.5-2}=\frac{2}{-0.5}=-4.$

Now, we are getting a smaller value. That is still inconclusive. We must try one more closer. Let's do 1.75.

$\frac{2}{1.75-2}=\frac{2}{-.25}=-8.$

We are even smaller still! Therefore, we must be decreasing to negative infinity.

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Example Question #358 : Finding Limits And One Sided Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow0+}\frac{-45sin(x)^{2}}{26x^{2}}\end{align*}$

Possible Answers:

$-\frac{45}{26}$

$-\infty$

$\frac{45}{26}$

$\infty$

Correct answer:

$-\frac{45}{26}$

Explanation:

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Example Question #401 : Limits

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

Possible Answers:

$\infty$

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

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

$-\infty$

Correct answer:

$\infty$

Explanation:

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Example Question #402 : Limits

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

Possible Answers:

$-\infty$

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

$\infty$

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

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 : Limits

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

Possible Answers:

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

$-\infty$

$\infty$

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

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 : Limits

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

Possible Answers:

$\infty$

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

$-\infty$

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

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 : Limits

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

Possible Answers:

$\infty$

$-\infty$

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

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

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-17x^{2}}{48cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{-34x}{-456\cdot \pi sin(\frac{(19\cdot \pi x)}{2}-\frac{ \pi }{2})}\rightarrow\frac{0}{456\cdot \pi}=0\\&\lim_{x\rightarrow0+}\frac{-17x^{2}}{48cos(\frac{(19\cdot \pi x)}{2}-\frac{ \pi }{2})}=0\end{align*}$

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Example Question #406 : Limits

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

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 : Limits

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

Possible Answers:

$-\infty$

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

$\infty$

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

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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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #356 : Finding Limits And One Sided Limits

Example Question #357 : Finding Limits And One Sided Limits

Example Question #358 : Finding Limits And One Sided Limits

Example Question #401 : Limits

Example Question #402 : Limits

Example Question #403 : Limits

Example Question #404 : Limits

Example Question #405 : Limits

Example Question #406 : Limits

Example Question #407 : Limits

All Calculus 2 Resources

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