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Calculus 2 : Finding Limits and One-Sided Limits

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Example Question #444 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-38\cdot (x - 1)^{3}}{13sin(3\cdot \pi x)^{3}}\end{align*}$

Possible Answers:

$\infty$

$-\frac{38}{(351\cdot \pi ^{3})}$

$-\infty$

$\frac{38}{(351\cdot \pi ^{3})}$

Correct answer:

$\frac{38}{(351\cdot \pi ^{3})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-38\cdot (x - 1)^{3}}{13sin(3\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{-114\cdot (x - 1)^{2}}{117\cdot \pi cos(3\cdot \pi x)sin(3\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{228 - 228x}{702\cdot \pi ^{2}cos(3\cdot \pi x)^{2}sin(3\cdot \pi x) - 351\cdot \pi ^{2}sin(3\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{-228}{2106\cdot \pi ^{3}cos(3\cdot \pi x)^{3} - 7371\cdot \pi ^{3}cos(3\cdot \pi x)sin(3\cdot \pi x)^{2}}\rightarrow\frac{-228}{-2106\cdot \pi ^{3}}\\&lim_{x\rightarrow1+}\frac{-38\cdot (x - 1)^{3}}{13sin(3\cdot \pi x)^{3}}=\frac{38}{(351\cdot \pi ^{3})}\end{align*}$

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Example Question #445 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{-13\cdot (x - 1)^{2}}{-41cos(\frac{(7\cdot \pi x)}{2})}\end{align*}$

Possible Answers:

$-\infty$

$\frac{52}{(82369\cdot \pi ^{2})}$

$\infty$

$-\frac{52}{(82369\cdot \pi ^{2})}$

Correct answer:

Explanation:

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Example Question #446 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{7tan(11x)^{2}}{29x^{3}}\end{align*}$

Possible Answers:

$\frac{(847\cdot (-29)^{(\frac{1}{3})})}{29}$

$-\frac{(847\cdot (-29)^{(\frac{1}{3})})}{29}$

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{7tan(11x)^{2}}{29x^{3}}\rightarrow\frac{0}{0}\\&\frac{14tan(11x)\cdot (11tan(11x)^{2} + 11)}{87x^{2}}\rightarrow\frac{0}{0}\\&\frac{308tan(11x)^{2}\cdot (11tan(11x)^{2} + 11) + 14\cdot (11tan(11x)^{2} + 11)^{2}}{174x}\rightarrow\frac{1694}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{7tan(11x)^{2}}{29x^{3}}=-\infty\end{align*}$

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Example Question #447 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{33sin(17\cdot \pi x)}{35ln(x + 1)^{3}}\end{align*}$

Possible Answers:

$\frac{(561\cdot 35^{(\frac{2}{3})}\cdot \pi )}{35}$

$\infty$

$-\infty$

$-\frac{(561\cdot 35^{(\frac{2}{3})}\cdot \pi )}{35}$

Correct answer:

$\infty$

Explanation:

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Example Question #448 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{23x}{-5sin(18\cdot \pi x)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{23}{(90\cdot \pi )}$

$-\frac{23}{(90\cdot \pi )}$

Correct answer:

$-\frac{23}{(90\cdot \pi )}$

Explanation:

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Example Question #449 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{26sin(19\cdot \pi x)}{21cos(5\cdot \pi x -\frac{ \pi }{2})^{2}}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{(494\cdot \pi )}{(525\cdot \pi ^{2})^{(\frac{1}{3})}}$

$-\frac{(494\cdot \pi )}{(525\cdot \pi ^{2})^{(\frac{1}{3})}}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{26sin(19\cdot \pi x)}{21cos(5\cdot \pi x -\frac{ \pi }{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{494\cdot \pi cos(19\cdot \pi x)}{-210\cdot \pi cos(5\cdot \pi x -\frac{ \pi }{2})sin(5\cdot \pi x -\frac{ \pi }{2})}\rightarrow\frac{494\cdot \pi}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{26sin(19\cdot \pi x)}{21cos(5\cdot \pi x -\frac{ \pi }{2})^{2}}=-\infty\end{align*}$

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Example Question #450 : Calculus Ii

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{46sin(19\cdot \pi x)^{2}}{19ln(x^{5})}\end{align*}$

Possible Answers:

$\infty$

$-\frac{(46\cdot \pi ^{2})}{25}$

$-\infty$

$\frac{(46\cdot \pi ^{2})}{25}$

Correct answer:

Explanation:

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Example Question #401 : Finding Limits And One Sided Limits

Evaluate the 1-sided limits

$\small \lim_{x->4^+}\ln(4x^2-16x)$

Possible Answers:

$\small -\infty$

$\small \infty$

Correct answer:

$\small -\infty$

Explanation:

$\small \lim_{x->4^+}\ln(4x^2-16x)$

If we were to look at a plot of the function, the limit would be obvious. Drawing the graph would be difficult without a graphing device. We can overcome this by first defining a new variable and assessing how it behaves as the original variable approaches $\small x=4$ from the right.

Let,

$\small u = 4x^2 -16x$

$\small \lim_{x->4^+}u(x)=\lim_{x-4^+}(4x^2-16x)=0$

Therefore, $\small \small u ->0^+$ as $\small x->4^+$

$\small \small \small \lim_{x->4^+}\ln(4x^2-16x)=\lim_{u->0^+}\ln(u)$

We notice that the function in terms of $\small u$ has a graph that is much easier to draw from recollection since we readily know how the natural logarithm behaves. In the plot of the natural log, the y-axis is a horizontal asympote, and the function becomes infinite in the negative direction as becomes arbitrarily close to 0 from the right. Hence $\lim_{u->0^+}\ln (u) = -\infty$ .

$\small \small \small \lim_{x->4^+}\ln(4x^2-16x)=\lim_{u->0^+}\ln(u)= -\infty$

$\small \small \small \lim_{x->4^+}\ln(4x^2-16x)= -\infty$

Calc problem 4 plot

The graph above is more familiar. This is not the graph of the original function $\ln(4x^2-16x)$ , which is in terms of The plot above represents the function defined in terms of the variable we defined, .

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Example Question #402 : Finding Limits And One Sided Limits

One-Sided Limits

Find

$\lim_{x \to 3^+}\frac{2x}{x-3}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

As gets closer to 3 (but remains larger than 3), then x-3 gets closer to 0 (but remains a small positive number).

The numerator, , gets closer to 6.

So, $\frac{2x}{x-3}$ is an arbitrarily large positive number.

Thus, we conclude that the limit is $\infty$ .

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Example Question #403 : Finding Limits And One Sided Limits

One-Sided Limits

Find

$\lim_{x \to 3^-}\frac{2x}{x-3}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

$-\infty$

Explanation:

As gets closer to 3 (but remains smaller than 3), then x-3 gets closer to 0 (but remains a small negative number).

The numerator, , gets closer to 6. So, $\frac{2x}{x-3}$ is an arbitrarily large negative number.

Thus, we conclude that the limit is $-\infty$ .

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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 #444 : Calculus Ii

Example Question #445 : Calculus Ii

Example Question #446 : Calculus Ii

Example Question #447 : Calculus Ii

Example Question #448 : Calculus Ii

Example Question #449 : Calculus Ii

Example Question #450 : Calculus Ii

Example Question #401 : Finding Limits And One Sided Limits

Example Question #402 : Finding Limits And One Sided Limits

Example Question #403 : Finding Limits And One Sided Limits

All Calculus 2 Resources

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