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Calculus 2 : Limit Concepts

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Example Questions

Example Question #2 : Calculus Ii

Evaluate the limit:

$\lim_{y \rightarrow 0} \frac{\tan2y}{\sin 5y}$

Possible Answers:

$\frac{5}{2}$

$\frac{2}{5}$

Does Not Exist

$\sin \frac{2}{5}$

Correct answer:

$\frac{2}{5}$

Explanation:

Directly evaluating the limit will produce an indeterminant answer of $\frac{0}{0}$ .

Rewriting the limit in terms of sine and cosine, $\lim_{y \rightarrow 0} \frac{\sin 2y}{\cos 2y} \cdot \frac{1}{\sin 5y}$ , we can try to manipulate the function in order to utilize the property $\lim_{x \rightarrow 0} \frac{\sin x}{x}=1$ .

Multiplying the function by the arguments of the sine functions, $\lim_{y \rightarrow 0} \frac{\sin 2y}{2y} \cdot \frac{1}{\cos 2y} \cdot \frac{5y}{\sin 5y} \cdot \frac{2y}{5y}$ , we can see that the limit will be $(1)(1)(1)\left(\frac{2}{5}\right) = \left(\frac{2}{5} \right )$ .

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Example Question #1 : Limit Concepts

Find the limit of $\frac{sin^5(x)}{x^5}$ as approaches infinity.

Possible Answers:

Inconclusive

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

The expression $\frac{sin^5(x)}{x^5}$ can be rewritten as $\frac{1}{x^5}sin^5(x)$ .

Recall the Squeeze theorem can be used to solve for the limit. The sine function has a range from [0,1] , which means that the range must be inside this boundary.

$0\leq sin^5(x) \leq 1$

Multiply the $\frac{1}{x^5}$ term through.

$0\left(\frac{1}{x^5}\right)\leq \left(\frac{1}{x^5}\right)sin^5(x) \leq 1\left(\frac{1}{x^5}\right)$

Take the limit as approaches infinity for all terms.

$\lim_{x \to \infty}0\leq \lim_{x \to \infty}\frac{sin^5(x)}{x^5} \leq \lim_{x \to \infty}\frac{1}{x^5}$

$0\leq \lim_{x \to \infty}\frac{sin^5(x)}{x^5} \leq 0$

Since the left and right ends of this interval are zero, it can be concluded that $\lim_{x \to \infty}\frac{sin^5(x)}{x^5}$ must also approach to zero.

The correct answer is 0.

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Example Question #1 : Limit Concepts

Determine the limit. $\lim_{x \to 0}-\frac{1}{x^2}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

$-\infty$

Explanation:

To determine, $\lim_{x \to 0}-\frac{1}{x^2}$ , graph the function $y=-\frac{1}{x^2}$ and notice the direction from the left and right of the curve as it approaches x=0 .

Both the left and right direction goes to negative infinity.

The answer is: $-\infty$

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Example Question #2 : Limit Concepts

Which of the following is true?

Possible Answers:

If $\small \small \lim _{x\to a} f(x)+g(x)$ exists, then $\small \lim _{x\to a} f(x)$ and $\small \lim _{x\to a} g(x)$ both exist.

$\small \lim _{x\to a} f(x)$ and $\small \lim _{x\to a} g(x)$ exist if and only if $\small \small \lim _{x\to a} f(x)+g(x)$ exists.

If $\small \lim _{x\to a} f(x)$ and $\small \lim _{x\to a} g(x)$ , then $\small \lim _{x\to a} f(x)+g(x)$ exists.

If neither $\small \lim _{x\to a} f(x)$ nor $\small \small \lim _{x\to a} g(x)$ exist, then $\small \lim _{x\to a} f(x)+g(x)$ also doesn't exist.

Correct answer:

If $\small \lim _{x\to a} f(x)$ and $\small \lim _{x\to a} g(x)$ , then $\small \lim _{x\to a} f(x)+g(x)$ exists.

Explanation:

If $\small \lim _{x\to a} f(x)$ and $\small \lim _{x\to a} g(x)$ , then $\small \lim _{x\to a} f(x)+g(x)$ exists.

This can be proven rigorously using the $\small \epsilon-\delta$ definition of a limit, but it is most likely beyond the scope of your class.

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Example Question #1 : Limits

Determine the limit: $\lim_{x \to \infty}\frac{3x^n}{9n!}$

Possible Answers:

$\frac{1}{3}$

$\infty$

Correct answer:

Explanation:

Isolate the constant in the limit.

$\lim_{x \to \infty}\frac{3x^n}{9n!} = \frac{1}{3}\lim_{x \to \infty}\frac{x^n}{n!}$

The limit property $\lim_{x \to \infty}\frac{x^n}{n!}=0$ .

Therefore:

$\frac{1}{3}\lim_{x \to \infty}\frac{x^n}{n!} = \frac{1}{3}(0)=0$

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Example Question #1 : Limit Concepts

Evaluate the limit, if possible: $\lim_{x\to \infty}tan^{-1}(x^2)$

Possible Answers:

$\infty$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{2}$

Explanation:

To evaluate $\lim_{x\to \infty}tan^{-1}(x^2)$ , notice that the inside term x^2 will approach infinity after substitution. The inverse tangent of a very large number approaches to $\frac{\pi}{2} \textup{ radians}$ .

The answer is $\frac{\pi}{2}$ .

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Example Question #1 : Limit Concepts

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{x^2+1}{3x^2}$

Possible Answers:

$\frac{1}{3}$

$-\infty$

$\infty$

$-\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

The first step is to factor out the highest degree term from the polynomial on top and bottom (essentially pulling out 1):

$\lim_{x\rightarrow \infty }\frac{x^2}{x^2}\left(\frac{1+\frac{1}{x^2}}{3}\right)$

which becomes

$\lim_{x\rightarrow \infty }\left(\frac{1+\frac{1}{x^2}}{3}\right)$

Evaluating the limit, we approach $\frac{1}{3}$ .

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Example Question #1 : Limit Concepts

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{x}{2x^2+3x+1}$

Possible Answers:

$\frac{1}{3}$

$\infty$

$-\infty$

Correct answer:

Explanation:

To evaluate the limit, first pull out the largest power term from top and bottom (so we are removing 1, in essence):

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

which becomes

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

Plugging in infinity, we find that the numerator approaches zero, which makes the entire limit approach 0.

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Example Question #9 : Limit Concepts

Evaluate the following limit:

$\lim_{r\rightarrow \infty }\frac{3r^3}{r^3+r+1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

To evaluate the limit, first pull out the highest power term out of the numerator and denominator (so essentially you are pulling 1):

$\lim_{r\rightarrow \infty }\frac{r^3}{r^3}\cdot \frac{3}{1+r^{-2}+r^{-3}}$

As you can see, the $r^{-2}$ and $r^{-3}$ terms as they approach infinity go to zero. What is left over is $\frac{3}{1}=3$ .

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Example Question #2 : Limit Concepts

Evaluate the following limit:

$\lim_{g\rightarrow \infty }\frac{g}{g^2-2g+1}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

To evaluate this limit easily, simply pull out a factor of the highest degree term over the highest degree term (1):

$\lim_{g\rightarrow \infty }\frac{g^2}{g^2}\left(\frac{\frac{1}{g}}{1-\frac{2}{g}+\frac{1}{g^2}}\right)$

As you can see, after the g^2 divides to 1, then the denominator becomes 1 and the numerator becomes 0.

The final answer is therefore .

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Calculus 2 : Limit Concepts

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Calculus Ii

Example Question #1 : Limit Concepts

Example Question #1 : Limit Concepts

Example Question #2 : Limit Concepts

Example Question #1 : Limits

Example Question #1 : Limit Concepts

Example Question #1 : Limit Concepts

Example Question #1 : Limit Concepts

Example Question #9 : Limit Concepts

Example Question #2 : Limit Concepts

All Calculus 2 Resources

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