\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Infinite Limits - Limits At Infinity

Limits

Using Limits


Derivatives

Graphing

Related Rates

Optimization

Other Applications

Integrals

Improper Integrals

Trig Integrals

Length-Area-Volume

Applications - Tools

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Calculus Tools

Learning Tools

Articles

This page discusses infinite Limits (or they may be called Limits At Infinity in your textbook) which refers to cases where the variable in question goes off to infinity. In limit notation, they look like \(\displaystyle{ \lim_{x \rightarrow \infty }{~f(x)} }\) or \(\displaystyle{ \lim_{x \rightarrow -\infty }{~f(x)} }\) .

If your limit looks like \(\displaystyle{ \lim_{x \rightarrow c }{~f(x)} }\) where \(c\) as a finite number, then you need to go to the finite limits page. (The following panel explains the terminology and how 17calculus defines finite and infinite limits.)

Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained

The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.

When we talk about limits, we are looking at the \(\displaystyle{ \lim_{x \to c}{f(x)} = L }\). The various terms apply to the description of \(c\) and \(L\) and are shown in the table below. The confusion lies with the terms finite limits and infinite limits. They can mean two different things.

\(\displaystyle{ \lim_{x \to c}{f(x)} = L }\)

when

term(s) used

\(c\) is finite

limits approaching a finite value or finite limits

\(c\) is infinite \(\pm \infty\)

limits at infinity or infinite limits

\(L\) is finite

finite limits

\(L\) is infinite \(\pm \infty\)

infinite limits

You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either \(c\) or \(L\). It is possible to have \(c = \infty\) and \(L\) be finite. So is this an infinite limit or a finite limit? It depends if you are talking about \(c\) or \(L\).

How 17calculus Uses These Terms
The pages on this site are constructed based on what \(c\) is, i.e. we use the terms finite limits and infinite limits based on the value of \(c\) only ( using the first two rows of the table above and ignoring the last two ). This seems to be the best way since, when we are given a problem, we can't tell what \(L\) is until we finish the problem, and therefore we are unable to determine what type of problem we have and know what techniques to use until we are done with the problem.

Important: Make sure and check with your instructor to see how they use these terms.

When evaluating \(\displaystyle{ \lim_{x \to \pm \infty}{~f(x)} }\) you need to determine if the graph of the function is leveling off at a value (and, if so, what that value is) or if it is going off to infinity (either \(+\infty\) or \(-\infty\)). You don't want to try to figure it out off a graph. You need to do it mathematically (from the equation).

This is the main theorem you will use.

Infinite Limits Theorem

\(\displaystyle{ \lim_{x \rightarrow \pm \infty}{\left[ \frac{1}{x} \right]} = 0 }\)

You can use the limit laws to apply this theorem to the case when you have \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{a}{x^k}\right]} }\) where \(k\) is a positive rational number and \(a\) is a real number. Here is an example. Try it on your own before looking at the solution.

Example

Evaluate \(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} }\).

Solution

\(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} = 3\lim_{x\to\infty}{\frac{1}{x^2}} }\)

Constant Multiple Rule

\(\displaystyle{ 3\left[ \lim_{x\to\infty}{\frac{1}{x}} \right] \left[ \lim_{x\to\infty}{\frac{1}{x}}\right] }\)

Multiplication Rule

\(\displaystyle{ 3(0)(0) = 0 }\)

Infinite Limits Theorem

Answer

\(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} = 0 }\)

Practice

Unless otherwise instructed, evaluate these limits, giving your answers in exact terms.

Basic

\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}=\infty}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\)

Solution

Direct substitution yields

\(\displaystyle{ \lim_{x \to \infty}{(x^4 + 7x^2 + 3)} = \infty + \infty + 3 = \infty }\)

This form is determinate.

Note: Some textbooks and websites will give the answer as DNE (Does Not Exist). However, according to the discusson about whether a limit exists or not, strictly this is not correct. Read your textbook carefully and ask your instructor what they expect (and make sure you understand why they expect a certain answer).

Final Answer

\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}=\infty}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}=\infty}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}}\)

Solution

Direct substitution yields \(\displaystyle{ \lim_{x \to \infty}{(x^5 - 3x^2 + x - 21)} = \infty - \infty + \infty - 21 }\)
This is indeterminate because of the \(\infty - \infty\). So we need to do some algebra to see if we can get a determinate form. There are several ways to do this. In this case, we will factor out the highest power of x to get \(\displaystyle{ \lim_{x \to \infty}{(x^5 - 3x^2 + x - 21)} = }\) \(\displaystyle{\lim_{x \to \infty}{\left[ x^5( 1 - 3/x^3 + 1/x^4 - 21/x^5)\right]} }\)

We know from the Infinite Limits Theorem that

\(\displaystyle{ \lim_{x \to \infty}{1/x^3} = 0 }\)

\(\displaystyle{ \lim_{x \to \infty}{1/x^4} = 0 }\)

\(\displaystyle{ \lim_{x \to \infty}{1/x^5} = 0 }\)

That leaves us with

\(\displaystyle{ \lim_{x \to \infty}{\left[ x^5( 1 - 3/x^3 + 1/x^4 - 21/x^5)\right]} }\)

\( \infty(1 - 0 + 0 - 0) \)

\( (\infty)(1) = \infty \)

Note: Some textbooks and websites will give the answer as DNE (Does Not Exist). However, according to the discusson about whether a limit exists or not, strictly this is not correct. Read your textbook carefully and ask your instructor what they expect (make sure you understand why they expect what they do).

Final Answer

\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}=\infty}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\frac{3x^2+5x+4}{x^3+7x}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\frac{3x^2+5x+4}{x^3+7x}}}\)

Solution

1053 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)

Solution

Direct substitution yields \(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} = \frac{-\infty}{-\infty}}\)

This is indeterminate. There are a couple of ways to work this. We will use algebra and the Infinite Limits Theorem here. (L'Hôpital's Rule would work here also.) To do this, we divide the numerator and denominator by the highest power of \(x\), in this case, \(x^1\).

\(\displaystyle{ \lim_{x \to -\infty}{\left[ \left(\frac{x+5}{3x+7} \right) \cdot \frac{1/x}{1/x} \right]} = }\) \(\displaystyle{\lim_{x \to -\infty}{\left[ \frac{1+5/x}{3+7/x} \right]} = \frac{1+0}{3+0} = 1/3 }\)

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}}\)

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}=0}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}}\)

Solution

Using direct substitution, we get \(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{7}{x^3-16}\right]} = \frac{7}{-\infty} = 0 }\)

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}=0}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to-\infty}{\frac{x^4+x}{5x^3+7}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\frac{x^4+x}{5x^3+7}}}\)

Solution

1056 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}=-\infty}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}}\)

Solution

Direct substitution yields \( -\infty - \infty = -\infty \) which is determinate.

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}=-\infty}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}=\infty}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}}\)

Solution

Direct substitution yields \(\displaystyle{ \lim_{x \to \infty}{( 3x^3-17x^2 )} = \infty - \infty }\), which is indeterminate. So we need to do some algebra.

\(\displaystyle{ \lim_{x \to \infty}{( 3x^3-17x^2 )} = }\) \(\displaystyle{\lim_{x \to \infty}{[(x^2)(3x-17)]} = (\infty)(\infty) = \infty }\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}=\infty}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}=0}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\)

Solution

Using direct substitution, we get \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{3}{x^2+5}\right]} = \frac{3}{\infty} = 0 }\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}=0}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}=7/2}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)

Solution

Direct substitution yields \( \infty - \infty \) in the denominator, which is indeterminate. So we need to make some changes. Let's divide the numerator and denominator by the highest power of x, which in this case is \(x^3\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3 - 5x} \cdot \frac{1/x^3}{1/x^3}\right]} = }\) \(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7 + 1/x^2 + 12/x^3}{2 - 5/x^2} \right]} = }\) \(\displaystyle{\frac{7 + 0 + 0}{2 - 0} = 7/2 }\)
In the last step, we used the Infinite Limits Theorem.

Note: If you know L'Hôpital's Rule, you could have used it here, which would have been easier.

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}=7/2}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }\)

Final Answer

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]}=0 }\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }\)

Solution

Direct substitution yields \(\infty - \infty\) in the numerator, which is indeterminate. Let's divide the numerator and denominator by the highest power of x, in this case \(x^3\).

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{(7x^2 - 3x + 12)}{(x^3 + 4x + 127)} \cdot \frac{1/x^3}{1/x^3} \right]} }\)

Use the Infinite Limits Theorem.

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7/x - 3/x^2 + 12/x^3}{1 + 4/x^2 + 127/x^3} \right]} = 0/1 = 0 }\)

Note: We could have used L'Hôpital's Rule, which would have been easier.

Final Answer

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]}=0 }\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}=+\infty}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)

Solution

Direct substitution yields \(\infty - \infty\) in the numerator which is indeterminate. So let's divide the numerator and denominator by the highest power, which is \(x^2\).

\(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{(7x^2 + x + 21)}{(11-x)} \cdot \frac{1/x^2}{1/x^2}\right]} }\)

Use the Infinite Limits Theorem

\(\displaystyle{\lim_{x \to -\infty}{\left[ \frac{7 + 1/x + 21/x^2}{11/x^2-1/x} \right]} = 7/0 = \infty }\)

Now, the tendency here is to stop and say the answer is \(+\infty\). But since x goes to \(-\infty\), we need to stop and take the negative sign into account. In this case, since we divided the numerator and denominator by \(x^2\) which is always positive, then the sign doesn't change. However, if the highest power would have been odd, the sign would have changed on both the numerator and denominator leaving the limit \(-\infty\).

Note: We could have used L'Hôpital's Rule, which would have been easier.

Final Answer

\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}=+\infty}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{ \lim_{x\to\infty}{\frac{4x^{10}+10000x^9}{5x^{10}+4}} }\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{ \lim_{x\to\infty}{\frac{4x^{10}+10000x^9}{5x^{10}+4}} }\)

Solution

1063 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\frac{3x^7}{5x^8+10x+2}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\frac{3x^7}{5x^8+10x+2}}}\)

Solution

1064 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\frac{x^4}{x^3+5}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\frac{x^4}{x^3+5}}}\)

Solution

1065 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

Intermediate

\(\displaystyle{\lim_{x\to\infty}{\frac{x+3}{\sqrt{x^2+4}}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\frac{x+3}{\sqrt{x^2+4}}}}\)

Solution

1045 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\pm\infty}{\frac{3x^4+5x}{2x^4-3x^3+10}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pm\infty}{\frac{3x^4+5x}{2x^4-3x^3+10}}}\)

Solution

1046 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}=0}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}}\)

Solution

Direct substitution yields \(\infty - \infty\) which is indeterminate. The technique we will use here is called rationalizing. We multiply the numerator and denominator by the conjugate of \(x - \sqrt{x^2+9} \). The conjugate is the same expression with one of the signs changed (it doesn't matter which sign).

\(\displaystyle{ \lim_{x \to \infty}{\left[ (x - \sqrt{x^2+9}) \frac{(x + \sqrt{x^2+9})}{(x + \sqrt{x^2+9})} \right]} }\)

\(\displaystyle{\lim_{x \to \infty}{\left[ \frac{(x - \sqrt{x^2+9})(x+\sqrt{x^2+9})}{x + \sqrt{x^2+9}} \right]}}\)

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^2 + x\sqrt{x^2+9} - x\sqrt{x^2+9} - (x^2+9)}{x + \sqrt{x^2+9}} \right]} }\)

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^2 - (x^2+9)}{x + \sqrt{x^2+9}} \right]} }\)

\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{ -9}{x + \sqrt{x^2+9}} \right]} }\)

Now if we use direct substitution, we get \( -9/\infty = 0 \)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}=0}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}=1/2}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\)

Solution

Direct substitution yields \(\infty / \infty\) which is indeterminate. So we need to use some algebra. Let's divide the numerator and denominator by the highest power of x, in this case \(x^{3/2}\).

\(\displaystyle{ \lim_{x \to \infty}{\left[ \sqrt{ \frac{x^3 + 3x}{4x^3+7} } \cdot \frac{x^{3/2}}{x^{3/2}}\right]} }\)

\(\displaystyle{ \lim_{x \to \infty}{\sqrt{ \frac{(x^3 + 3x)}{(4x^3+7)} \cdot \frac{x^3}{x^3} }} }\)

\(\displaystyle{ \lim_{x \to \infty}{\sqrt{ \frac{(1 + 3/x^2)}{(4+7/x^3)} }} }\)

\(\displaystyle{ \sqrt{ \lim_{x \to \infty}{ \frac{(1 + 3/x^2)}{(4+7/x^3)} } } }\)

Use the Infinite Limits Theorem

\(\displaystyle{ \sqrt{ \frac{1}{4} } = 1/2 }\)

Final Answer

\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}=1/2}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\arctan(x)}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\arctan(x)}}\)

Final Answer

\(\displaystyle{ \lim_{x \to \infty}{\arctan(x)} = \pi/2 }\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\infty}{\arctan(x)}}\)

Solution

The question was to determine the limit \(\displaystyle{ \lim_{x \to \infty}{\arctan(x)} }\).
Now, without studying this, I wouldn't know off the top of my head what this limit is or if it even existed. In fact, the technique to determine this is a little unusual. So, I am going to take you step-by-step through a procedure that you need to start doing as you work through problems.

The first question I asked myself was, what technique that I already know, can I use to determine this? Well, I could graph it and get an idea of an answer. If you graph this, it looks like the graph is leveling off to the right and that a limit might be about 1.5. However, we can't tell if it is just rising very slowly or if a limit does actually exist. We need to find an exact answer.
Whenever I have an inverse trig function, one technique I try is to convert to regular trig functions since I am much more familiar with them. So we have
\(\displaystyle{ \begin{align} \theta &= \arctan(x) \\ \tan( \theta ) &= x \\ \frac{\sin( \theta)}{\cos( \theta)} &= x \end{align} }\)
It's a good idea to use \(\theta\) (or another Greek letter) to remind yourself that inverse trig functions return angles. Also remember for inverse trig functions to make sense, we must restrict the domain of the corresponding trig function. For this case, we restrict the domain of the tangent function to \( -\pi/2 < \theta < \pi/2 \). [This is something you need to keep in mind whenever you work with inverse trig functions.]

So now we have enough information to determine the limit. The last equation is \( \displaystyle{ x=\frac{\sin( \theta )}{\cos( \theta )} }\) and we know that \( x \to \infty \). So we need to know the angle that would cause x to go off to infinity. Do you see it now? It would be \( \theta=\pi/2 \) since the numerator is 1 and the denominator is zero. (Why not \( \theta=-\pi/2 \)?)
Therefore our limit is \(\displaystyle{ \lim_{x \to \infty}{\arctan(x)} = \pi/2. }\)

Final Answer

\(\displaystyle{ \lim_{x \to \infty}{\arctan(x)} = \pi/2 }\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x \to \pm \infty}{ \ln \left[ \frac{2x^2-2x+1}{2x^2+2x+1} \right] }}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x \to \pm \infty}{ \ln \left[ \frac{2x^2-2x+1}{2x^2+2x+1} \right] }}\)

Final Answer

\( 0 \)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x \to \pm \infty}{ \ln \left[ \frac{2x^2-2x+1}{2x^2+2x+1} \right] }}\)

Solution

This limit is part of a improper integral problem. You do not need to know how to work improper integrals to understand how to evaluate this limit.
This solution uses the concept that we mentioned on the introduction to limits page stating that, when you are taking the limit of a continuous function, you can move the limit inside the function and then evaluate it.

3855 video

video by Michael Penn

Final Answer

\( 0 \)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to\infty}{\left[\sqrt{x^2+4x+1}-x\right]}}\)

Problem Statement

\(\displaystyle{\lim_{x\to\infty}{\left[\sqrt{x^2+4x+1}-x\right]}}\)

Solution

1047 video

video by Khan Academy

close solution

Log in to rate this practice problem and to see it's current rating.

Advanced

\(\displaystyle{\lim_{x\to-\infty}{\frac{x^3}{\sqrt{x^6+4}}}}\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to-\infty}{\frac{x^3}{\sqrt{x^6+4}}}}\)

Solution

1044 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

infinite limits 17calculus youtube playlist

Here is a playlist of the videos on this page.

You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

WikiBooks - Calculus/Infinite Limits

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on bags & supplies

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

Practice

How to Read and Do Proofs: An Introduction to Mathematical Thought Processes

Under Armour Clothing - Just Launched at eBags.com!

Shop Amazon - Rent Textbooks - Save up to 80%

As an Amazon Associate I earn from qualifying purchases.

Page Sections

How to Ace the Rest of Calculus: The Streetwise Guide, Including MultiVariable Calculus

Under Armour Clothing - Just Launched at eBags.com!

Shop Amazon - Rent eTextbooks - Save up to 80%

As an Amazon Associate I earn from qualifying purchases.

Practice Instructions

Unless otherwise instructed, evaluate these limits, giving your answers in exact terms.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.