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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.

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  • $\begingroup$ It looks fine to me. $\endgroup$ – José Carlos Santos May 27 at 21:13
  • $\begingroup$ It's not $\frac00$ to use L'hospital. $\endgroup$ – Nosrati May 27 at 21:14
  • $\begingroup$ 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 $\endgroup$ – Jake Freeman May 27 at 21:18
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Without the Hospital's rule: $$\frac{x^{-x}-1}{x}=-\frac{x^x-1}{x\cdot x^x}=-\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot\ln{x}\cdot\frac{1}{x^x}\rightarrow+\infty.$$

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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 $x\to 0$.

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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 $$\frac{x^{-x}-1}{x}\sim_0 -\ln x\xrightarrow[x\to 0^+]{}+\infty$$

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