# Limit of indeterminate in format 0^infinity.

• Feb 11th 2013, 01:12 PM
EliteAndoy
Limit of indeterminate in format 0^infinity.
Hi everyone, we are asked to evaluate $\lim _{x \to 0^+}\(ln(x+1))^{1/x}$. Now what I did is I set $y=(ln(x+1))^{1/x}$ then took the natural log of both sides to bring down the exponent 1/x: $
\lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+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)
• Feb 11th 2013, 02:03 PM
HallsofIvy
Re: Limit of indeterminate in format 0^infinity.
Quote:

Originally Posted by EliteAndoy
Hi everyone, we are asked to evaluate $\lim _{x \to 0^+}\(ln(x+1))^{1/x}$. Now what I did is I set $y=(ln(x+1))^{1/x}$ then took the natural log of both sides to bring down the exponent 1/x: $
\lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+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)
• Feb 11th 2013, 02:21 PM
ILikeSerena
Re: Limit of indeterminate in format 0^infinity.
Hi EliteAndoy! :)

Alternatively, you can write

$\lim_{x \to 0+} (\ln(1+x))^{1/x} = \lim_{k \to \infty} (\ln(1+10^{-k})^{1/10^{-k}}$

Furthermore, we have:

$\frac x 2 \le \ln(1+x) \le x \qquad \text{if } 0\le x \le 1$

Can you find an upper and lower bound for the limit using this?
• Feb 12th 2013, 08:35 AM
EliteAndoy
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