Find the lim as x tends to 0+ of (ln x)^x.
$\displaystyle (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. So you should first consider $\displaystyle \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ which has the indeterminant form $\displaystyle \frac{\infty}{\infty}$. Using l'Hopitals rule is an obvious thing to do.
$\displaystyle \lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} }$$\displaystyle = \lim_{k \rightarrow \infty } \ (-\ln ( 2k+1 ))^{ \frac{1}{2k+1} }$$\displaystyle = -\lim_{k \rightarrow \infty } \ (\ln ( 2k+1 ))^{ \frac{1}{2k+1} }$$\displaystyle = -\lim_{x \rightarrow \infty } \ (\ln ( x ))^{ \frac{1}{x} }$$\displaystyle = -\lim_{x \rightarrow \infty} \ e^{ \frac{1}{x} \ln ( \ln x)}$$\displaystyle = -e^{ \lim_{x \rightarrow \infty} \ \frac{\ln ( \ln x)}{x} }$. Now use L'Hopital.
The problem with that is that the limit you get is 1, when the actual limit is -1$\displaystyle (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. So you should first consider $\displaystyle \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ which has the indeterminant form $\displaystyle \frac{\infty}{\infty}$. Using l'Hopitals rule is an obvious thing to do.
The problem is, Wolfram Alpha is calculating the limit in the complex plane with the definition I first gave, but by what I understood the limit that was asked ignores this and uses only the fact negative numbers have odd roots (ie. we want the limit of a sequence, not the limit of the function)