Limit of indeterminate in format 0^infinity.

Hi everyone, we are asked to evaluate . Now what I did is I set then took the natural log of both sides to bring down the exponent 1/x: . From there on I can't seem to figure out how to make the function in the form of 0/0 or infinity/infinity to use L'Hopital's rule. Any one got any ideas? Thanks y'all in advance!(Nod)

Re: Limit of indeterminate in format 0^infinity.

Quote:

Originally Posted by

**EliteAndoy** Hi everyone, we are asked to evaluate

. Now what I did is I set

then took the natural log of both sides to bring down the exponent 1/x:

. From there on I can't seem to figure out how to make the function in the form of 0/0 or infinity/infinity to use L'Hopital's rule.

Well, that's the whole point isn't it? The denominator goes to 0 but the numerator doesn't! What does the numerator go to? Dividing by smaller and smaller values in the denominator just makes the numerator larger.

Quote:

Any one got any ideas? Thanks y'all in advance!(Nod)

Re: Limit of indeterminate in format 0^infinity.

Hi EliteAndoy! :)

Alternatively, you can write

Furthermore, we have:

Can you find an upper and lower bound for the limit using this?

Re: Limit of indeterminate in format 0^infinity.

Now I get it. By the way I forgot to say that we are not jsut asked to evaluate the limit, but to rather prove it.Thanks everyone for the reply, really helped me out understanding this problem. :D