Question

L'Hopital's Rule:

lim x-> -inf of (arctan x) / (2^x)?

Please help....:'(

Answer 1

pretty sure you do not use l'hopital for this

Answer 2

Really?....but this section is on L'Hopital's Rule..

Answer 3

\[\lim_{x\to -\infty}\tan^{-1}(x)=-\frac{\pi}{2}\] so it is not in the form \(\frac{\infty}{\infty}\)
it may be in that section, they are checking to see if you know when you can and cannot use l'hopital

Answer 4

Oh... you are right. Thanks!

Answer 5

you can only use it in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)
if you try to use it here you will get the wrong answer

Answer 6

yw

Answer 7

We have 0 * infinity too for L'Hopital's, but I didn't think that would be it for this problem.. thank you!

Answer 8

well not really
if it looks like \(0\times \infty\) to have to change it to make it look like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)