Calculus I - Limits At Infinity, Part II

Show General Notice Show Mobile Notice Show All Notes Hide All Notes

General Notice

I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.

Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.

Paul
February 18, 2026

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.8 : Limits at Infinity, Part II

Back to Problem List

10. Evaluate \(\mathop {\lim }\limits_{x \to - \infty } {\tan ^{ - 1}}\left( {7 - x + 3{x^5}} \right)\).

Show Solution

First notice that,

\[\mathop {\lim }\limits_{x \to - \infty } \left( {7 - x + 3{x^5}} \right) = - \infty \]

If you aren’t sure about this limit you should go back to the previous section and work some of the examples there to make sure that you can do these kinds of limits.

Now, recalling Example 7 from this section, we know that because the argument goes to negative infinity in the limit the answer is,

\[\mathop {\lim }\limits_{x \to - \infty } {\tan ^{ - 1}}\left( {7 - x + 3{x^5}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{\pi }{2}}}\]