Does $ \lim_{x \to 0^+} (1 + x)^{\ln x}$ exist?

Question

Does $\displaystyle \lim_{x \to 0^+} (1 + x)^{\ln x}$ exist?

I do not have any attempts solving it because I do not know how to transform it into the form in which I use l'Hôpital's rule.

Answer 1

Suppose there's a limit, say $$ L = \lim_{x \to 0^+} (1 + x)^{\ln x}. $$ Apply the natural logarithm, and note that $u \mapsto \ln u$ is continuous, so we can pass the limit through the logarithm: \begin{align} \ln L &= \ln \biggl(\, \lim_{x \to 0^+} (1 + x)^{\ln x} \biggr) \\ &= \lim_{x \to 0^+} \ln \Bigl( (1 + x)^{\ln x} \Bigr) \\[4pt] &= \lim_{x \to 0^+} (\ln x) \ln (1 + x) \\ &= \lim_{x \to 0^+} \frac{\ln (1 + x)}{(\ln x)^{-1}} \end{align} which is indeterminate of the form $\frac{0}{0}$. You can now apply l'Hôpital's rule and assuming that this converges, then you can recover the original limit by exponentiating $$ L = e^{\ln L}. $$