Limit as x approaches 0+ of x^(1/x)

[Jalo]

Homework Statement

limit as x approaches 0+ of x^(1/x)

Homework Equations

The Attempt at a Solution

Usually I solve this limit by rewriting the limit as e^log(f(x)) and applying L'hopital rule. However:

(All limits approach 0+)

Lim x^(1/x) = exp( Lim log(x^(1/x)) ) = exp( Lim log(x) / x )

This is where I'd usually apply L'hopital rule and solve the problem. I can't tho, since x tends to 0 and log(x) tends to - infinity. I'm stuck here..

If anyone could point me in the right direction I'd appreciate.
Thanks!

 

[Dick]

Homework Statement

limit as x approaches 0+ of x^(1/x)

Homework Equations

The Attempt at a Solution

Usually I solve this limit by rewriting the limit as e^log(f(x)) and applying L'hopital rule. However:

(All limits approach 0+)

Lim x^(1/x) = exp( Lim log(x^(1/x)) ) = exp( Lim log(x) / x )

This is where I'd usually apply L'hopital rule and solve the problem. I can't tho, since x tends to 0 and log(x) tends to - infinity. I'm stuck here..

If anyone could point me in the right direction I'd appreciate.
Thanks!

If log(x) approaches -infinity and x approaches 0 and is positive then log(x)/x approaches -infinity. Good idea not to use l'Hopital on it. It's not indeterminant.

 

[Jalo]

If log(x) approaches -infinity and x approaches 0 and is positive then log(x)/x approaches -infinity. Good idea not to use l'Hopital on it. It's not indeterminant.

Oh lol... I did so many exercises with l'Hopital rule that I forgot to check that...
Thanks a lot!

 

[Dick]

Oh lol... I did so many exercises with l'Hopital rule that I forgot to check that...
Thanks a lot!

You did catch it. You realized you can't use l'Hopital. That would have given you the wrong answer! That's a good piece of work right there. If it's not indeterminant the answer should be easy to get. That's all you missed.