Find $\lim_{x\to 0}x^{\frac{1}{x}}$ I tried to solve $$\lim_{x\to 0} x^{\frac{1}{x}}$$

I tried doing it like this:

$$\lim_{x\to 0} x^{\frac{1}{x}}= \lim_{x\to 0} e^{\ln(x^{\frac{1}{x}})} = \lim_{x\to 0} e^{\frac{\ln(x)}{x}} = \exp\left(\lim_{x\to 0} \frac{\ln(x)}{x}\right)$$

Then solving $\lim_{x\to 0} \frac{\ln(x)}{x}$ with l'Hospital's Rule I get

$$\lim_{x\to 0} \frac{\ln(x)}{x}=\lim_{x\to 0} \frac{1}{x}$$

Which remains indeterminate as it has no two-sided limit.

So is that the answer, that it remains indeterminate?  Or would I say that it has two limits depending on the side you approach?  How do I phrase this result?
The expression $\lim_{x\to 0^-}x^{\frac{1}{x}}$ makes no sense, for in general $x^y$ is undefined if $x$ is negative. As a consequence, $\lim_{x\to 0}x^{\frac{1}{x}}$ does not exist.

As was pointed out by Lubin, L'Hospital's Rule is not appropriate here, since after taking the logarithm we are looking at $\frac{\ln x}{x}$. For $x$ near $0$, the numerator $\ln x$ is very large negative, but the denominator is close to $0$, so we do not have a suitable form. But luckily, the rule is not needed, since raising a number close to $0$ to a large power gives a number very close to $0$.
$\lim_{x\to 0} x^\frac{1}{x}$=$\lim_{x\to 0} (1+x-1)^{{\frac{1}{x-1}\cdot}{\frac{x-1}{x}}}$=$e^{{\lim_{x\to 0}}{\frac{x-1}{x}}}$=$e$.