How can I go about proving whether or not $f: (0, \infty) \rightarrow \mathbb{R}, f(x) = x^{1/x}$ is uniformly continuous? I graphed the function and it seemed like it wasn't uniformly continuous, but I might be wrong.
The way I'm trying to go about the problem is to use contradiction and assume that the function is uniformly continuous. So for some $x,y$ in the interval, we can find a $\delta$ so that if $|x-y| < \delta$, then $|x^{1/x} - y^{1/y}| < 1$. My plan is to replace $y$ with a function in terms of $x$ and show that there is a contradiction. However, no matter how many different substitutions I play around, I can't get anything to simplify.
The function $$f(x):=x^{1/x}=e^{\log x/x}\qquad(x>0)$$ is continuous, one has $\lim_{x\to\infty}f(x)=1$, and from $$e^{\log x/x}<e^{-1/x}\qquad\left(0<x<{1\over e}\right)$$ it follows that $\lim_{x\to0} f(x)=0$. In order to judge the quality of continuity we compute the derivative $$f'(x)={1\over x^2}(1-\log x)\>e^{\log x/x}\ ,$$ which is continuous as well. Furthermore one has $$\lim_{x\to 0}f'(x)=\lim_{x\to\infty}f'(x)=0\ .$$ Therefore there is an $M>0$ such that $$|f'(x)|\leq1\qquad\left(0<x<{1\over M}\quad {\rm or}\quad x>M\right)\ .$$ In addition there is a $C\geq1$ with $$|f'(x)|\leq C\qquad\left({1\over M}\leq x\leq M\right)\ .$$ It follows that $C$ is a global Lipschitz constant for $f$, so that $f$ is even Lipschitz continuous.