Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Limits / Computing Limits
Show Mobile Notice Show All Notes Hide All Notes
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 2.5 : Computing Limits

4. Evaluate \(\displaystyle \mathop {\lim }\limits_{z \to 8} \frac{{2{z^2} - 17z + 8}}{{8 - z}}\), if it exists.

Show Solution

There is not really a lot to this problem. Simply recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Recall that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

\[\mathop {\lim }\limits_{z \to 8} \frac{{2{z^2} - 17z + 8}}{{8 - z}} = \mathop {\lim }\limits_{z \to 8} \frac{{\left( {2z - 1} \right)\left( {z - 8} \right)}}{{ - \left( {z - 8} \right)}} = \mathop {\lim }\limits_{z \to 8} \frac{{2z - 1}}{{ - 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 15}}\]