Limit of indeterminate in format 0^infinity.

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February 11th 2013, 12:12 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: $<br /> \lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+1))}_x$ . 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!(Nod)
February 11th 2013, 01:03 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: $<br /> \lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+1))}_x$ . 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.

Quote:

Any one got any ideas? Thanks y'all in advance!(Nod)
February 11th 2013, 01:21 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?
February 12th 2013, 07:35 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