Q:

Find limit $\lim_{x \rightarrow 0^+} \frac{x}{x^x-1}$

Find $$\lim_{x \rightarrow 0^+}(x^{x-1}-x^{-1})^{-1}$$
my approach
Firstly I should represent factor in more intuitive form
$$\lim_{x \rightarrow 0^+}\frac{x}{x^x-1} $$
I know that $$\lim_{x \rightarrow 0^+} x^x = 1$$
so I suspect that I have expression of type $ \frac{0}{0} $ 
Ok. Now I am goind to find $$\lim_{x \rightarrow 0^+} \frac{f'(x)}{g'(x)}= \lim_{x \rightarrow 0^+}\frac{1}{(\ln (x) +1)\cdot x^x}$$
Ok but know I have no idea how I can deal with that because $ln x\rightarrow -\infty$ when $x\rightarrow 0^-$ and from other hand $x^x$ is going to $\infty$ and I can't use there Hospital rule again..

A:

$x^x$ does not approach $+\infty$ as $x \to 0^+$. 
The limit $\lim_{x \to 0+} x^x$ is in fact equal to $1$, as we see by expressing $x^x = e^{x\ln{x}}$. 
Thus, your limit $$\lim_{x \to 0+} \frac{1}{(\ln(x) + 1)x^x}$$ 
is in fact equal to $$\lim_{x \to 0+} \frac{1}{\ln(x) + 1} \cdot \lim_{x \to 0+} \frac{1}{x^x},$$ which is equal to $0 \cdot 1 = 0$.

