# Find $\lim_{x\to 0^+}{x^{x^x} -1}=?$

Question: $\lim_{x\to 0^+}{x^{x^x} -1}=?$

I guess L'Hôpital must be used. But where? $$\lim_{x\to 0^+} e^{x^x\ln x}-1=\lim_{x\to 0^+} e^\frac{\ln x}{\frac{1}{x^x}}-1$$ I mean $\ln x\to -\infty$ as $x\to 0^+$, this is OK. But $1/x^x\to \infty$??

If so, how to find this limit?

• $x^x$ is finite and nonzero around $0$ (it tends to $1$), so the limit is of the form $0^1-1$, which raises no problem.
– user65203
Nov 26 '17 at 19:17
• @AntonioVargas: where do you see a ratio ???
– user65203
Nov 26 '17 at 19:19
• so the answer is $-1$. Thank you. @YvesDaoust Nov 26 '17 at 19:22
• Are you sure about the question ? $x^{x^x-1}$ is more likely (and more interesting).
– user65203
Nov 26 '17 at 19:24
• Yes, I'm pretty sure.$x^{x^x-1}$ is definitely more interesting. Nov 26 '17 at 19:27

As pointed out in the comments, $$\lim_{x\to 0} x^x=1$$ So $$\lim_{x\to 0^+} x^{x^x}=\lim_{x\to 0^+} e^{x^x\ln x}=e^{\lim_{x\to 0^+} (x^x\ln x)}$$ $$\lim_{x\to 0^+} x^x\ln x=-\infty.$$ So $$\lim_{x\to 0^+} x^{x^x}=\lim_{t\to -\infty}e^t=0$$ This the required limit is $-1$.

Since $x^x\to 1,$ we have $x^x >1/2$ for small positive $x.$ Thus $0 \le x^{x^x} \le x^{1/2}$ for such $x.$ Thus the limit is $0$ by the squeeze theorem.

hint

$$x^x=e^{x\ln (x)}$$

and $$\lim_{x\to0^+}x\ln (x)=0$$

thus

$$\lim_{x\to 0^+}e^{x^x\ln (x)}=0$$

• LOL. Everybody has a right to ask you anything. Of course, you don't have to answer. I was just curious why you think your answer might be helpful, because it may be obvious for me (or you) that $x^{x^x}=e^{x^x\ln (x)}$, but it may be not obvious for everybody. So looking at related answers actually downvoted, though correct (but not all too helpful), you might consider making your answer helpful.
– user436658
Nov 26 '17 at 19:54
• I prefered a hint to a complete answer. You can ask me what you want. No problem Professor. Nov 26 '17 at 20:01
• Just a question, no problem, especially since the OP doesn't seem to be sure what their problem was.
– user436658
Nov 26 '17 at 20:08