A calculus problem by Parth Sankhe

As it turns out, it is not necessary to approach from the positive direction. We find first find $\ln \lim\limits_{x\to 0} x^x = \lim\limits_{x\to 0}\ln x^x$ . $\begin{aligned} \lim_{x\to 0} \ln x^x &= \lim_{x\to 0} x\ln x \\ &=\lim_{x\to 0} \frac{\ln x}{^1\!/\!_x} && \color{#3D99F6}\text{Apply L'H}\hat{\text{o}}\text{pital's rule}\\ &= \lim_{x\to 0} \frac{^1\!/\!_x}{-^1\!/\!_{x^2}} \\ &= \lim_{x\to 0} -x = 0 \end{aligned}$

Thus, $\ln \lim\limits_{x\to 0} x^x = 0$ , so $\lim\limits_{x\to 0} x^x = \boxed{1}$ .

This is a classic example of the limit-finding technique "it looks like an indeterminate form, so make it a fraction however you can."