1. ## Limits and Logs

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.

2. 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

$\chi$ $\sigma$

3. Thanks but can you explain something else to me? How did the ln(x) manage to get on the bottom part?

4. Originally Posted by mei
Thanks but can you explain something else to me? How did the ln(x) manage to get on the bottom part?
Here's an example - $\left(\frac{1}{2}\right)^3 = \frac{1}{2^3}$ .

5. I see... because 1 to the power of anything will always be 1!
Thank you so much! =)