Limits and Logs

mei · #1 $Old$ June 22nd, 2009, 06:45 AM

Hey there!

I'm at a loss on how to solve this particular limit:

lim (1/(x-1))^(ln x)
x-> 1+

I have tried doing ln y = ln (1/(x-1))^(ln x)

and then I got to

lim ln x * ln (1/(x-1))
x->1+

I'm not quite sure how to get out of there. I tried using L'Hopital's rule but I think I must have done something wrong. I would like to know what's the best approach to solve such limits.

Thanks in advance for your help!

chisigma · #2 $Old$ June 22nd, 2009, 07:25 AM

Setting $t=x-1$ we obtain...

$\lim_{x \rightarrow 1+} \frac{1}{(x-1)^{\ln x}} = \lim_{t \rightarrow 0+} e^{-\ln t \cdot \ln (1+t)} = \lim _{t \rightarrow 0+} e^{-\ln t \cdot \sum_{n=1}^{\infty} (-1)^{n+1} \frac{t^{n}}{n}} = e^{-0} = 1$

Kind regards

$\chi$ $\sigma$