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Calculus 2 : Finding Limits and One-Sided Limits

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Example Questions

Example Question #301 : Finding Limits And One Sided Limits

Evaluate the limit:

$\lim_{x \to 3}\log_{x}(27)$

Possible Answers:

$\infty$

DNE

$-\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to 3}\log_{x}(27)=\log_{3}(27)=3$

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Example Question #302 : Finding Limits And One Sided Limits

Evaluate the limit:

$\lim_{x \to \pi}\csc(x)\sec(x)$

Possible Answers:

$\infty$

$-\infty$

DNE

Correct answer:

DNE

Explanation:

The cosecant is the reciprical of the sine of an angle. The sine of pi is 0; so the reciprical of 0 is DNE. So when we insert x=pi into the equation, we get:

$\lim_{x \to \pi}\csc(x)\sec(x)=\csc(\pi)\sec(\pi)=\frac{1}{0}*1\Rightarrow DNE$

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Example Question #303 : Finding Limits And One Sided Limits

Evaluate the limit:

$\lim_{x \to 0}\frac{10x^2-20}{5x}$

Possible Answers:

$-\infty$

DNE

$\infty$

Correct answer:

DNE

Explanation:

We see that we cannot factor this to make the denominator not equal 0; hence this limit DNE because the denominator is zero.

$\lim_{x \to 0}\frac{10x^2-20}{5x}=\frac{10*0^2-20}{5*0}=\frac{-20}{0}\Rightarrow DNE$

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Example Question #304 : Finding Limits And One Sided Limits

Evaluate the limit:

$\lim_{x \to \frac{\pi }{2}}\sin(x)\cos (x)$

Possible Answers:

$\infty$

$-\infty$

DNE

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to \frac{\pi }{2}}\sin(x)\cos (x)=\sin (\frac{\pi }{2})\cos (\frac{\pi }{2})=1*0=0$

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Example Question #305 : Finding Limits And One Sided Limits

Find the limit, approaching from the right:

$\lim_{x \to 0^+} \frac{\cos (x)}{x}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

To find the limit of the whole thing we have to find the individual limits of the top and bottom of the fraction, coming from the right.

For the top:

$\cos(x) \rightarrow 1$ as $x \rightarrow 0$ , because $\cos(0)=1$ and cos(x) is a continuous function.

$x \rightarrow 0$ as you approach 0.

So, near 0 we're looking at what's basically $\frac{1}{x}$ . And we know that $\cos(x)$ is positive for positive values of x near 0, so with these two facts combined we have enough information to see that

$\lim_{x \to 0^+} \frac{\cos x}{x} = \infty$ .

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Example Question #306 : Finding Limits And One Sided Limits

Evaluate the limit:

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}$

Possible Answers:

$-\infty$

Does not exist

$\frac{1}{6}$

$\infty$

Correct answer:

$\frac{1}{6}$

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}=\frac{0^2+2*0+1}{0+6}=\frac{1}{6}$

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Example Question #307 : Finding Limits And One Sided Limits

Evaluate the following limit:

$\lim_{x\rightarrow \infty}\frac{x^4}{x^3+5x^2+x}$

Possible Answers:

$-\infty$

$\frac{1}{5}$

$\infty$

Correct answer:

$\infty$

Explanation:

To evaluate the limit, we must first pull out a factor consisting of the highest power term divided by itself (one, essentially):

$\lim_{x\rightarrow \infty} \frac{x^4}{x^4}(\frac{1}{x^{-1}+5x^{-2}+x^{-3}})$

After the factor we created becomes 1, the negative exponent terms go to zero as x approaches infinity, therefore we are left with $\frac{1}{0}=\infty$ .

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Example Question #308 : Finding Limits And One Sided Limits

Evaluate the following limit:

$\lim_{x\rightarrow 4^+}f(x), f(x)=\left\{\begin{matrix} x^2+1, x\leq 4\\ \ln(x-4), x> 4 \end{matrix}\right.$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

To evaluate the limit, we must determine which side of the limit we are approaching. The plus sign "exponent" means that we are approaching 4 from the right, using numbers larger than 4. Therefore, we must evaluate the limit using the second part of the piecewise function; when we evaluate the limit, we get the limit as the natural logarithm approaches zero, which gives us $-\infty$ .

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Example Question #309 : Finding Limits And One Sided Limits

Evaluate the following limit:

$\lim_{x\rightarrow 3} \frac{\ln(3-x)}{2}$

Possible Answers:

$\infty$

$-\frac{1}{3}$

$-\infty$

Correct answer:

$-\infty$

Explanation:

To evaluate the limit, we simply substitute the x value we are approaching - 3 - into the limit. As the natural logarithm gets closer to zero, we reach $-\infty$ .

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Example Question #310 : Finding Limits And One Sided Limits

Find $\lim_{x\rightarrow \infty} \frac{4x}{x^2+5x}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

$\lim_{x\rightarrow \infty} \frac{4x}{x^2+5x}$

By L'Hopital's rule:

$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}$

$\lim_{x\rightarrow \infty} \frac{4x}{x^2+5x}=\lim_{x\rightarrow \infty} \frac{4}{2x+5}=0$

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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 #301 : Finding Limits And One Sided Limits

Example Question #302 : Finding Limits And One Sided Limits

Example Question #303 : Finding Limits And One Sided Limits

Example Question #304 : Finding Limits And One Sided Limits

Example Question #305 : Finding Limits And One Sided Limits

Example Question #306 : Finding Limits And One Sided Limits

Example Question #307 : Finding Limits And One Sided Limits

Example Question #308 : Finding Limits And One Sided Limits

Example Question #309 : Finding Limits And One Sided Limits

Example Question #310 : Finding Limits And One Sided Limits

All Calculus 2 Resources

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