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Calculus 2 : Limits

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Example Questions

Example Question #381 : Calculus Ii

Evaluate:

$\lim_{n \to \infty} \frac{4n^3}{3n^3+n^2+1}$

Possible Answers:

$\frac{4}{3}$

$\infty$

Correct answer:

$\frac{4}{3}$

Explanation:

To evaluate the limit, we must factor out a term consisting of the highest power term divided by itself (which is the same as 1, so we aren't changing the original function):

$\lim_{{n \to \infty}}\frac{n^3}{n^3}(\frac{4}{3+n^{-1}+n^{-3}})$

After the term we factored out cancels to one, and the negative power terms in the denominator go to zero (as n goes to infinity, these terms contain infinity in their denominator, which makes the whole term go to zero), we are left with

$\lim_{n \to \infty}\frac{4}{3}=\frac{4}{3}$

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Example Question #382 : Calculus Ii

If it exists, find the following limit

$\small \small \lim_{x\to 2} \frac{1}{x+2}$

Possible Answers:

Does not exist

$\small \frac{1}{2}$

$\small \frac{1}{4}$

$\small -\frac{1}{4}$

Correct answer:

$\small \frac{1}{4}$

Explanation:

To find the following limit

$\small \small \lim_{x\to 2} \frac{1}{x+2}$

we can simply plug in $\small x=2$ because it's defined at that point and in infinitesimally small neighborhoods near $\small x=2$ , so we get

$\small \small \small \lim_{x\to 2} \frac{1}{x+2}=\frac{1}{2+2}=\frac{1}{4}$

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Example Question #383 : Calculus Ii

Find the following limit

$\small \small \small \lim_{x\to \infty} \frac{1}{2x+3}$

Possible Answers:

$\small \infty$

$\small \frac{1}{5}$

$\small \frac{1}{2}$

$\small 0$

Correct answer:

$\small 0$

Explanation:

To find the following limit

$\small \small \small \lim_{x\to \infty} \frac{1}{2x+3}$

we can see that $\small 2x+3$ approaches $\small \infty$ as $\small x\to \infty$ , so we get the fraction $\small \frac{1}{\infty}$ , which approaches , so we get

$\small \small \small \small \lim_{x\to \infty} \frac{1}{2x+3}=0$

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Example Question #341 : Finding Limits And One Sided Limits

Find the following limit

$\small \small \small \small \lim_{x\to \infty} \frac{2^x+3}{5^x+7}$

Possible Answers:

$\small \frac{2}{5}$

$\small \infty$

$\small 0$

$\small \frac{5}{2}$

Correct answer:

$\small 0$

Explanation:

To find the following limit

$\small \small \small \small \lim_{x\to \infty} \frac{2^x+3}{5^x+7}$

we can divide every term by $\small 2^x$ to get

$\small \small \small \small \small \small \small \lim_{x\to \infty} \frac{2^x+3}{5^x+7}=\lim_{x\to \infty} \frac{2^x/2^x+3/2^x}{5^x/2^x+7/2^x}=\lim_{x\to \infty} \frac{1+3/2^x}{(5/2)^x+7/2^x}$

It's obvious that $\small \frac{3}{2^x}$ and $\small \small \frac{7}{2^x}$ both approach $\small 0$ and that $\small \left(\frac{5}{2}\right)^x$ approaches $\small \infty$ as $\small x\to\infty$ , so we have

$\small \small \small \small \small \small \lim_{x\to \infty} \frac{2^x+3}{5^x+7}=\lim_{x\to \infty} \frac{2^x/2^x+3/2^x}{5^x/2^x+7/2^x}=\lim_{x\to \infty} \frac{1+3/2^x}{(5/2)^x+7/2^x}=\frac{1+0}{\infty+0}=\frac{1}{\infty}=0$

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Example Question #381 : Limits

Find the following limit

$\small \small \lim_{x\to 5}x^2$

Possible Answers:

$\small 25$

$\small \sqrt{5}$

$\small -25$

$\small 5$

Correct answer:

$\small 25$

Explanation:

To find the following limit

$\small \small \small \small \small \lim_{x\to 5}x^2$

we can simply plug in $\small x=5$ because it's defined there, so we get

$\small \small \small \small \small \small \lim_{x\to 5}x^2=5^2=25$

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Example Question #385 : Calculus Ii

Find the following limit

$\small \small \small \lim_{x\to 1} \frac{1}{x-1}$

Possible Answers:

$\small -\infty$

$\small 0$

$\small \infty$

Does not exist

Correct answer:

Does not exist

Explanation:

To find the following limit

$\small \small \small \lim_{x\to 1} \frac{1}{x-1}$

we cannot plug in $\small x=1$ directly because then we get an undefined statement. We can find the limits as we approach the left and right to get

$\small \small \small \small \lim_{x\to 1^+} \frac{1}{x-1}=\infty$

$\small \small \small \small \small \lim_{x\to 1^-} \frac{1}{x-1}=-\infty$

The left side and right side limits do not coincide so the limit does not exist.

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Example Question #386 : Calculus Ii

Find the following limit

$\small \small \small \small \small \lim_{x\to -\infty} \frac{2}{5+x}$

Possible Answers:

$\small \frac{1}{3}$

$\small 0$

$\small -\frac{2}{5}$

$\small \frac{2}{5}$

Correct answer:

$\small 0$

Explanation:

To find the following limit

$\small \small \small \small \small \lim_{x\to -\infty} \frac{2}{5+x}$

We know that $\small 5+x$ approaches $\small -\infty$ as $\small \small x\to-\infty$ , so the fraction $\small \frac{2}{5+x}$ approaches $\small \small \small \frac{2}{-\infty}=0$ , so we have

$\small \small \small \small \small \small \lim_{x\to -\infty} \frac{2}{5+x}=0$

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Example Question #387 : Calculus Ii

Find the following limit

$\small \small \small \small \small \small \small \lim_{x\to \infty} \frac{2x+1}{3x+2}$

Possible Answers:

$\small \infty$

$\small \frac{2}{3}$

$\small 0$

$\small -\frac{2}{3}$

Correct answer:

$\small \frac{2}{3}$

Explanation:

To find the following limit

$\small \small \small \small \small \small \small \lim_{x\to \infty} \frac{2x+1}{3x+2}$

we can divide everything in the expression by $\small x$ to get

$\small \small \small \small \small \small \small \small \lim_{x\to \infty} \frac{2x+1}{3x+2}=\lim_{x\to \infty}\frac{2+\frac{1}{x}}{3+\frac{2}{x}}=\frac{2+0}{3+0}=\frac{2}{3}$

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Example Question #346 : Finding Limits And One Sided Limits

Find the following limit

$\small \small \lim_{x\to \infty} \frac{2x+3}{5x-1}$

Possible Answers:

$\small \frac{1}{2}$

$\small \frac{5}{4}$

$\small \frac{2}{5}$

$\small 1$

Correct answer:

$\small \frac{2}{5}$

Explanation:

To find the following limit

$\small \small \lim_{x\to \infty} \frac{2x+3}{5x-1}$

we can divide everything in the expression by $\small x$ to get

$\small \small \small \lim_{x\to \infty} \frac{2x+3}{5x-1}=\lim_{x\to \infty}\frac{2+\frac{3}{x}}{5-\frac{1}{x}}=\frac{2+0}{5-0}=\frac{2}{5}$

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Example Question #382 : Limits

Find the following limit

$\small \small \small \lim_{x\to \infty} e^{\frac{1}{x}}$

Possible Answers:

$\small e$

$\small 1$

$\small \infty$

$\small e^{-1}$

Correct answer:

$\small 1$

Explanation:

To find the following limit

$\small \small \small \lim_{x\to \infty} e^{\frac{1}{x}}$

we can use limit $\small \lim_{x\to\infty}\frac{1}{x}=0$ to get

$\small \small \small \small \lim_{x\to \infty} e^{\frac{1}{x}}=e^0=1$

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #381 : Calculus Ii

Example Question #382 : Calculus Ii

Example Question #383 : Calculus Ii

Example Question #341 : Finding Limits And One Sided Limits

Example Question #381 : Limits

Example Question #385 : Calculus Ii

Example Question #386 : Calculus Ii

Example Question #387 : Calculus Ii

Example Question #346 : Finding Limits And One Sided Limits

Example Question #382 : Limits

All Calculus 2 Resources

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