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

Question

I'm trying to solve the limit $$\lim_{x\rightarrow 0^+} \frac{x^{-x}-1}{x}$$

I think we should use L'Hospital rule and the limit becomes

$$\lim_{x\rightarrow 0^+} -x^{-x}(\log x + 1)=\lim_{x\rightarrow 0^+} \frac{\log x + 1}{-x^{x}}= +\infty$$

Is it right?

I've tried to modify the form and not use L'Hospital's rule but without success.

Answer 1

Without the Hospital's rule: $$\frac{x^{-x}-1}{x}=-\frac{x^x-1}{x\cdot x^x}=-\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot\ln{x}\cdot\frac{1}{x^x}\rightarrow+\infty.$$