evaluate
lim x -- infinity
of function (e ^ -x) ln x
using l'hopitals rule

Thanks to anyone who gives it a shot

Question

evaluate 
lim x --> infinity 
of function  (e ^ -x) ln x
 using l'hopitals rule

Thanks to anyone who gives it a shot

anonymous · Answer

$$\LARGE \lim_{x\to \infty}e^{-x}\cdot \ln x$$  is this what you mean ?

anonymous · Answer

yes that is correct

anonymous · Answer

$$\LARGE (u\cdot v)'=u'\cdot v+u\cdot v'$$  you're supposed to know this rule... do you know it ?

anonymous · Answer

absolutely..but hopitals rule doesnt apply it..at least i didnt think it did

anonymous · Answer

...now you got me confused to.. O_O , is L'Hopital's Rule about derivatives when you're $$\LARGE \frac{\infty}{\infty} \quad \quad , \quad \quad \frac00$$  ... ? :(

experimentx · Answer

attachment

anonymous · Answer

@experimentX yeah I know it's 0 .. http://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto+%5Cinfty%7De%5E%28-x%29*ln+x but how to do it !! ??

experimentx · Answer

put that e^x on down and use L'hospital rue

anonymous · Answer

i got that far experiment..but then i get (1/x)/(e^x) after taking the derivative...at that point im stuck

experimentx · Answer

now .. as x-> inf your function = 1/inf = 0

anonymous · Answer

doesnt numerator go to 0 and denom to inf?  (1/x --> 0 with x --> inf)

experimentx · Answer

yeah .. so it is your limit ...better than inf/inf isn't it??

anonymous · Answer

use drawing.. you can understand each other better !

experimentx · Answer

yeah .. just click on the link posted by @Kreshnik

anonymous · Answer

alright thanks alot..didnt see the link..take care

experimentx · Answer

oops sorry there wasn't graph!