Help solving the limit when $\lim_{x\to 1}{(\ln x)^{x-1}}$ I'm stuck with this one, so far I've tried taking logarithms to apply "The limit of the logarithm is the $ln$ of the limit" And then turning it into an infinite/infinite indeterminate but keep obtaining $0$ D:
 A: L'Hospital's also a possibility:
$$(\log x)^{x-1}=e^{(x-1)\log\log x}$$
and 
$$\lim_{x\to 1^+}(x-1)\log\log x=\lim_{x\to 1^+}\frac{\log\log x}{\frac{1}{x-1}}\stackrel{\text{L'H}}=\lim_{x\to 1^+}-\frac{(x-1)^2}{x\log x}\stackrel{\text{L'H}}=$$
$$=\lim_{x\to 1^+}-2\frac{x-1}{\log x+1}=-2\frac{0}{0+1}=0$$
Thus the limit is, by continuity of the exponential function, $\,e^0=1\,$
A: Your limit is of the form $0^0$. In this case, it is helpful to use the fact that $a^b = \exp(b \ln a)$. You then have that (exp is continuous)
$$\lim_{x\to 1} (\ln x )^{x-1} = \exp\left[ \lim_{x\to 1} (x-1) \ln\ln(x)\right]
=\exp(0) =1$$
because $\ln x$ grows slower than any polynomial.
A: Let us first set $u=x-1$.
Note that your limit requires $x-1=u>0$ to make sense.
So your limit is equal to
$$
\lim_{u\rightarrow 0^+} (\ln (1+u))^u.
$$
Now for all $u>0$, we have
$$
\ln(1+u)\leq u
$$
since the graph of $\ln (1+u)$ is below the tangent at $u=0$ (by a concavity argument).
Using the fact that $\ln$ is increasing on $]0,+\infty[$, we have
$$
1\leq (\ln (1+u))^u=\exp\left( u\ln(\ln (1+u))\right)\leq \exp(u\ln u)
$$
for all $u>0$.
It remains to invoke the Squeeze theorem and the fact that
$$
\lim_{u\rightarrow 0^+} u\ln u=0
$$
to deduce from the inequality above that your limit is equal to $\exp(0)=1$.
