Limit of indeterminate in format 0^infinity.

**EliteAndoy** · February 11th 2013, 12:12 PM

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: $<br /> \lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+1))}_x$ . 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!

**HallsofIvy** · February 11th 2013, 01:03 PM

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: $<br /> \lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+1))}_x$ . 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.

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

**ILikeSerena** · February 11th 2013, 01:21 PM

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?

**EliteAndoy** · February 12th 2013, 07:35 AM

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.

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