Prove that $\lim\limits_{x\to 0^+}(x^{x^x}-x^x)=-1$ Prove that $\lim\limits_{x\to 0^+}(x^{x^x}-x^x)=-1$
Here neither L Hospital rule nor series expansion is working here.By what method should it be proved?Thanks.
 A: Consider the limit $$\lim_{x\to 0^+}\ln(x)e^{x\ln(x)}$$ We know that $\lim_{x\to 0^+}e^{x\ln(x)} =1$ and $\lim_{x\to 0^+} \ln(x) = -\infty$, so $$\lim_{x\to 0^+}\ln(x)e^{x\ln(x)} =-\infty$$ Hence $$\lim_{x\to 0^+}e^{\ln(x)e^{x\ln(x)}} \equiv e^{-\infty} = 0$$ and you can verify algebraically that $e^{\ln(x)e^{x\ln(x)}} = x^{x^{x}}$.
A: Define $f(x)=x^x$.  We know that $\lim_{x\to0}f(x)=1$.  Now then what's the limit of $x^{x^x}-x^x=x^{f(x)}-f(x)$?
A: The limit $\lim_{x \to 0^+} x^x$ is indeed $1$ which you can show by writing $x^x = e^{x \ln x}$ as mentioned in another answer, and then noting the exponential is continuous and $x \ln x \to 0$ which can be seen by L'Hospital's rule after first rewriting as $(\ln x) / (1/x)$.
Once you know that, you can show $\lim_{x \to 0^+} x^{x^x} = 0$ by choosing $x > 0$ small enough so that $x < \epsilon$ and $x^x < 1 + \epsilon$ for a given small $\epsilon > 0$. Then all you have to show is that for this $x > 0$ or smaller, you have $0 < x^{x^x} < \epsilon^{1 + \epsilon} < 2 \epsilon$ for example (where the last inequality holds for $\epsilon$ small enough, since $\epsilon^\epsilon \to 1$). And then taking $\epsilon$ to 0 gives you your result.
