Why is $\lim_{x\rightarrow 0^{+}} x^{1/x}=0?$ Why is $\lim_{x\rightarrow 0^{+}} x^{1/x}=0?$
I believe it is from graphing it, but I'm not able to prove it.  Log w/ L'Hospitals isn't working for me since ln(x)/x isn't the right indeterminant form.

$x^{1/x} = e^{\ln x / x}$ so indeed all you need to show is that $\lim_{x \to 0} \ln x / x = -\infty$, which you can do without L'hopital's rule, just simply by noting that $\ln x \to -\infty$ and $1/x \to +\infty$ as you approach $0$ from the right-hand side.
