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?