A limit that doesn't work with l'Hôpital's rule

[daudaudaudau]

Homework Statement

Evaluate \lim_{x\rightarrow0}\frac{e^{-\frac{1}{x}}}{x}

The Attempt at a Solution

I tried taking logarithms and applying l'Hôpital's rule, but that just led to an expression which diverged, and as far as I understand, the limit in l'Hôpital's rule MUST exist.

 

[Cyosis]

You could also use the following hand waving argument. The numerator will go to zero much faster than the denominator will so the limit will be 0.

 

[Mark44]

How about rewriting the original expression as
\frac{1/x}{e^{1/x}}
and then using l'Hopital's rule on that?

 

[daudaudaudau]

Beautiful, Mark44, that works.

 

[Mark44]

How about rewriting the original expression as
\frac{1/x}{e^{1/x}}
and then using l'Hopital's rule on that?

Wouldn't it be nice if you could just cancel the 1/x's?

 

[zcd]

Wouldn't it be nice if you could just cancel the 1/x's?

You could, in a way. Replace 1/x = u and take u to infinity instead.

 

[daudaudaudau]

By the way, this limit does not exist when we approach zero from the negative numbers, because then the result is infinity/0. I guess both limits from the positive and the negative numbers have to be equal in order for the limit to exist?

 

[JG89]

Nevermind.

 

[daudaudaudau]

JG89: I don't get it. Why do you say that x=1 ?

 

[JG89]

Ah, I made a mistake. Please disregard that post.