1. ## l'hospitals rule

I have a problem xe^(1/x) and i'm supposed to use l'hospital's rule to explain its behavior as x -> 0. But L'hospital's rule is for division only I thought? Any help would be great thanks.

2. Originally Posted by UMStudent
I have a problem xe^(1/x) and i'm supposed to use l'hospital's rule to explain its behavior as x -> 0. But L'hospital's rule is for division only I thought? Any help would be great thanks.
we spoke about a problem similar to this just tonight. I'll find the post for you. but the short answer is:

note that we can write x as 1/(1/x)

so xe^(1/x) = 1/(1/x) * e^(1/x) = [e^(1/x)]/(1/x) ........this is a quotient, however, this goes to 1/infinity, which is not a condition to use l'hopital's, i think you left out an x somewhere or something

see http://www.mathhelpforum.com/math-he...e-inf-inf.html

3. xe^(1/x) is the exact problem thats why I was having trouble and the problem says to specifically use l'hospital's rule

4. Originally Posted by UMStudent
xe^(1/x) is the exact problem thats why I was having trouble and the problem says to specifically use l'hospital's rule
ah yes, you can use l'hopital's on it. the lim is x-->0, i thought it was x-->infinity since that's what the problem we were doing was like. you can follow my original guidelines then. l'hopital's will work

5. lim{x-->oo}xe^(1/x)
= lim{x-->oo}(1/(1/x))e^(1/x)
= lim{x-->oo} [e^(1/x)]/(1/x) .......this goes to inf/inf as x-->0, we can use l'hopital's

Apply L'hopital's

=> lim{x-->oo}[e^(1/x)]/(1/x) = lim{x-->oo}[(-x^-2)e^(1/x)]/(-x^-2)
= lim{x-->oo} e^(1/x)
= infinity

6. ## Re:

UMStudent here is a good website I discovered earlier today. It goes through all of the possible scenarios. Just keep working with it and you will pick it up rather quickly.

http://spot.pcc.edu/~sperry/M252_S-4_5.pdf

7. Lol why didn't you say you were a genius. But thanks for all the help from both of you appreciate it.