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: $$\frac{x^{-x}-1}{x}=-\frac{x^x-1}{x\cdot x^x}=-\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot\ln{x}\cdot\frac{1}{x^x}\rightarrow+\infty.$$

Answer 2

You can use $x^{-x}=e^{-x\ln(x)}$ and $e^u\approx 1+u$ for $u$ near $0$. Then the expression becomes $\approx \frac{-x\ln(x)}{x}=-\ln(x) \to \infty$ as $x\to 0$.

Answer 3

Use equivalence: near $u=0$, $\;\mathrm e^u=1+u+o(u)$, so $$\mathrm e^{-x\ln x}-1=-x\ln x+o(-x\ln x),\quad\text{which means }\;x^{-x}-1\sim_0-x\ln x$$ and finally $$\frac{x^{-x}-1}{x}\sim_0 -\ln x\xrightarrow[x\to 0^+]{}+\infty$$