Thread: limits with e^x

lim e^(-1/x)/x
x-> 0

lim e^(-1/x)/x
x-> infinity


i keep getting that they do not exist, and i need an actual value.
No idea what im doing wrong, any help much appreciated.

first one looks like..

approaching infinity / approaching zero, which just boosts the infinity into more infinitiness ;D

so its infinity..

it looks that way to me anyways without writing it down..

second one looks like...

top goes to 1/ bottom is x as x approaches infinity which would make it zero.

ok, thanks, so im not completely out of it!!

It turns out i reduced my equation incorrectly. So im actually getting an answer instead of infinity now! YAY!!

was I right?

If you let u= 1/x your limits become $\displaystyle \lim_{u\rightarrow \infty} ue^{-u}$ and $\displaystyle \lim_{u\rightarrow 0}ue^{-u}$. The second of those is easy: setting u= 0 we get 0(1)= 0. For the first, you need to know that an exponential "dominates" and polynomial. $\displaystyle e^{-x}$ goes to 0 "faster" than u goes to infinity.