Skip Navigation

8.3: Infinite Limits

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
Estimated5 minsto complete
Practice Infinite Limits
This indicates how strong in your memory this concept is
Estimated5 minsto complete
Estimated5 minsto complete
Practice Now
This indicates how strong in your memory this concept is
Turn In

Geeks-R-Us sells titanium mechanical pencils to computer algorithm designers. In an effort to attract more business, they decide to run a rather unusual promotion:

"SALE!! The more you buy, the more you save! Pencils are now \begin{align*}$ \frac{12x}{x - 3}\end{align*}$12xx3 per dozen!"

If the zillionaire Spug Dense comes in and says he wants to buy as many pencils as Geeks-R-Us can turn out, what will the cost of the pencils approach as the order gets bigger and bigger?

Watch This

Embedded Video:

- James Sousa: Limits at Infinity


Sometimes, a function may not be defined at a particular number, but as values are input closer and closer to the undefined number, a limit on the output may not exist. For example, for the function f(x) = 1/x (shown in the figures below), as x values are taken closer and closer to 0 from the right, the function increases indefinitely. Also, as x values are taken closer and closer to 0 from the left, the function decreases indefinitely.

We describe these limiting behaviors by writing

\begin{align*}\lim_{x \rightarrow 0^+} \frac{1} {x} = + \infty\end{align*}limx0+1x=+
\begin{align*}\lim_{x \rightarrow 0^-} \frac{1} {x} = - \infty\end{align*}limx01x=

Sometimes we want to know the behavior of f(x) as x increases or decreases without bound. In this case we are interested in the end behavior of the function, a concept you have likely explored before. For example, what is the value of f(x) = 1/x as x increases or decreases without bound? That is,

\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = ?\end{align*}limx+1x=?
\begin{align*}\lim_{x \to -\infty} \frac{1} {x} = ?\end{align*}limx1x=?

As you can see from the graphs (shown below), as x decreases without bound, the values of f(x) = 1/x are negative and get closer and closer to 0. On the other hand, as x increases without bound, the values of f(x) = 1/x are positive and still get closer and closer to 0.

That is,

\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = 0\end{align*}limx+1x=0
\begin{align*}\lim_{x \to -\infty} \frac{1} {x} = 0\end{align*}limx1x=0

Example A

Evaluate the limit by making a graph:

\begin{align*}\lim_{x\to 3^+} \frac{x + 6}{x - 3}\end{align*}limx3+x+6x3


By looking at the graph:

We can see that as x gets closer and closer to 3 from the positive side, the output increases right out the top of the image, on its way to \begin{align*}\infty\end{align*}

Example B

Evaluate the limit \begin{align*}\lim_{x\to \infty} \frac{11x^3 - 14x^2 +8x +16} {9x - 3}\end{align*}limx11x314x2+8x+169x3


To evaluate polynomial function limits, a little bit of intuition helps. Let's think this one through.

First, note that since we are looking at what happens as \begin{align*}x \to \infty\end{align*}x most of the interesting stuff will happen as x gets really big.

On the top part of the fraction, as x gets truly massive, the 11x3 part will get bigger much faster than either of the other terms. In fact, it increases so much faster that the other terms completely cease to matter at all once x gets really monstrous. That means that the important part of the top of the fraction is just the 11x3.

On the bottom, a similar situation develops. As x gets really, really big, the -3 matters less and less. So the bottom may as well be just 9x.

That gives us \begin{align*}\frac{11x^3}{9x}\end{align*}11x39x which reduces to \begin{align*}\frac{11x^2}{9}\end{align*}11x29

Now we can more easily see what happens at the "ends." As x gets bigger and bigger, the numerator continues to get bigger faster than the denominator, so the overall output also increases.

\begin{align*}\therefore \lim_{x\to +\infty} \frac{11x^3 - 14x^2 +8x +16} {9x - 3} = +\infty\end{align*}

Example C

Evaluate \begin{align*}\lim_{x\rightarrow 0} \frac{x + 2} {x + 3} \end{align*}


This one is easier than it looks! As x --> 0, it disappears, leaving just the fraction: 2/3

Concept question wrap-up

As Spug buys more and more pencils, the cost of each dozen will drop quickly at first, and level out after a while, approaching $12 per dozen.

You can see the effect on the graph here:


A function with an infinite limit continues to output greater and greater +/- values.

Guided Practice


1) Make a graph to evaluate the limit \begin{align*}\lim_{x\rightarrow \infty} \frac {1} {\sqrt x} \end{align*}

2) From problem 1, evaluate \begin{align*}\lim_{x\rightarrow 0^{+}} \frac {1} {\sqrt x} \end{align*}

3) Graph and evaluate the limit:

\begin{align*}\lim_{x\to 2^{+}} \frac {1} {x - 2}\end{align*}


1) By looking at the image, we see that as x gets huge, so does \begin{align*}\sqrt{x}\end{align*} which means that 1 is being divided by an ever-larger number, and the result is getting smaller and smaller.

The limit is 0

2) On the same image, we can see that as x gets closer and closer to zero, so does \begin{align*}\sqrt{x}\end{align*} which means that 1 is being divided by an ever smaller number, and the result gets bigger and bigger.

The limit is \begin{align*}+\infty\end{align*}

3) By looking at the image, we can see that as x gets closer and closer to 2 from the positive direction, 1 gets divided by smaller and smaller numbers, so the result gets larger and larger.


Evaluate the limits, you may graph if you wish:

  1. \begin{align*}\lim_{x\to 3^{-}} \frac {1} {x - 3}\end{align*}
  2. \begin{align*}\lim_{x\to -4^{+}} \frac {1} {x + 4}\end{align*}
  3. \begin{align*}\lim_{x\to -\left(\frac{8}{3}\right)^{+}} \frac {1} {3x + 8}\end{align*}
  4. \begin{align*}\lim_{x\to -5^{+}} \frac{\left(x^2+11x+30\right)}{x+5}\end{align*}
  5. \begin{align*}\lim_{x\to -\infty} \frac{\left(x^2+11x+30\right)}{x+5}\end{align*}

Evaluate the limits

  1. \begin{align*}\lim{x \to \infty}\end{align*} \begin{align*}\frac{-11x^3 + 20x^2 + 15x - 17}{-9x^3 + 5x^2 - x - 17}=\end{align*}
  2. \begin{align*}\lim{x \to \infty} 13 =\end{align*}
  3. \begin{align*}\lim{x \to \infty} \frac{-2x + 18}{17x - 3}=\end{align*}
  4. \begin{align*}\lim{x \to \infty} 15=\end{align*}
  5. \begin{align*}\lim{x \to \infty} -5x^2 + 5x + 14=\end{align*}
  6. \begin{align*}\lim{x \to \infty}7x + 12=\end{align*}
  7. \begin{align*}\lim{x \to \infty} -3x + 13=\end{align*}
  8. \begin{align*}\lim{x \to \infty} \frac{13x - 8}{19x^3 - 11x^2 + x + 4}=\end{align*}
  9. \begin{align*}\lim{x \to \infty} -17x + 14=\end{align*}
  10. \begin{align*}\lim{x \to \infty}-7x^2 - 2x - 13 =\end{align*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Show More


limit A limit is the value that the output of a function approaches as the input of the function approaches a given value.

Image Attributions

Show Hide Details
Difficulty Level:
At Grade
Date Created:
Nov 01, 2012
Last Modified:
Mar 23, 2016
Files can only be attached to the latest version of Modality
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original