Is that $\displaystyle \lim_{x\to 1^{+}}(\ln(x))^{x-1}$

or

$\displaystyle \lim_{x\to 1^{+}}(\ln (x))^{x}-1$

I am guessing the former.

If so, it is the form $\displaystyle 0^{0}$

Since the ol' hospital rule is the topic at hand, we can try this:

Rewrite:

$\displaystyle \lim_{x\to 1^{+}}e^{ \displaystyle (x-1) \ln (\ln(x))}$

L'Hopital:

$\displaystyle e^{\displaystyle \left( -\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{x \ln (x)}\right) }$

quotient rule:

$\displaystyle {\Large}{e^{\left(\frac{\displaystyle -\lim_{x \to 1^{+}}\frac{(x-1)^{2}}{\ln (x)}}{\displaystyle \lim_{x\to 1^{+}}x}\right)}}$

The limit in the denominator of the above is 1, so we have:

$\displaystyle e^{\displaystyle \left(-\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{\ln (x)}\right)}$

L'Hopital again:

$\displaystyle e^{\displaystyle \left(-2\lim_{x\to 1^{+}}x(x-1)\right)}$

Now, we could continue using sum rules and product rules and what not, but it is easy to see what the limit is now. See it?.

**Does anyone know the trick to get the limit to display nicely instead of all spread out like like?. Also, how do we make the font larger?. I can not get the displaystyle to work on this site.**