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

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.

• It looks fine to me. – José Carlos Santos May 27 '19 at 21:13
• It's not $\frac00$ to use L'hospital. – Nosrati May 27 '19 at 21:14
• x^x as it approaches 0 will be 1, and ln(x) as x approaches 0 will be negative infinity, however, it is to the negative sign so it will approach positive infinity – Jake Freeman May 27 '19 at 21:18

Without the Hospital's rule: $$\frac{x^{-x}-1}{x}=-\frac{x^x-1}{x\cdot x^x}=-\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot\ln{x}\cdot\frac{1}{x^x}\rightarrow+\infty.$$
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 $$x\to 0$$.
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 $$\frac{x^{-x}-1}{x}\sim_0 -\ln x\xrightarrow[x\to 0^+]{}+\infty$$