# Is $f(x)=x^x=e^{x\ln (x)}$ defined on $(0, +\infty)$ continuously extendable to $x=0$?

Question: Show whether $f(x)=x^x=e^{x\ln (x)}$ defined on $(0, +\infty)$ is continuously extendable to $x=0$.

In my attempt, I tried to prove that $f$ is uniformly continuous on $(0, 1)$, but to no avail. Because if a real-valued function $f$ on $(a,b)$ is uniformly continuous on $(a, b)$, $f$ can be extended to a continuous function on $[a, b]$.

I'm assuming that $f$ is continuously extendable, since $x^x$ seems to go to $1$ if $x$ tends to $0$. Is it possible to prove this using uniform continuity?

You don't need uniform continuity. If $\lim_{x \to 0^+} f(x)$ exists, setting $f(0)$ to this limit will give you continuity on $[0, \infty)$ –  anonymous Dec 16 '12 at 2:36
Begin by evaluating $\lim_{x\downarrow 0} f(x).$ This should tell all.