Compute $\lim_{x\to 0^{+}}x^{x^{x}}=?$

Question

Problem: find $\lim_{x\to 0^{+}}x^{x^{x}}$.

Solution: $$\lim_{x\to 0^+}x^{x^{x}}=\lim_{x\to 0^+}e^{x^{x}\ln x}=\lim_{x\to 0^+}e^{e^{\ln x^{x}}\ln x}=\lim_{x\to 0^+}e^{e^{x\ln x}\ln x}$$

Now what should I do?

Answer 1

First, $x\ln(x)$ goes to $0$.

Hence, $e^{x\ln(x)}$ goes to $1$.

It follows that $e^{x\ln(x)}\ln(x)$ goes to $-\infty$.

Thus, $e^{e^{x\ln(x)}\ln(x)}$ goes to $0$.

Answer 2

You're doing good; consider $$ \lim_{x\to0^+}x^x\ln x=-\infty $$ because $\lim_{x\to0^+}x^x=1$, so it is bounded in a right neighborhood of $0$.

Hence $\lim_{x\to0^+}e^{x^x\ln x}=0$.

In a different fashion, since $\lim_{x\to0^+}x^x=1$, we can find $0<\delta<1$ such that, for $0<x<\delta$, $$ \frac{1}{2}\le x^x\le 2 $$ Note that, for $0<x<1$, if $a<b$ then $x^a<x^b$ (take logarithms), so, for $0<x<\delta$, $$ x^2\le x^{x^x}\le x^{1/2} $$ The squeeze theorem makes us end.

Answer 3

There is a general limit theorem that if $f(x)\ge0$ for all $x$ near $c$ and if $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$ both exist and -- what's most important here -- are not both equal to $0$, then

$$\lim_{x\to c}f(x)^{g(x)}=(\lim_{x\to c}f(x))^{\lim_{x\to c}g(x)}$$

The theorem also applies to one-sided limits. (Note, the condition $f(x)\ge0$ is mainly to guarantee that $f(x)^{g(x)}$ is well defined.)

Now this theorem does not apply to $\lim_{x\to0^+}x^x$, but by L'Hopital, or whatever, we have $\lim_{x\to0^+}x\ln x=0$, so

$$\lim_{x\to0^+}x^x=e^{\lim_{x\to0^+}x\ln x}=e^0=1$$

and now the general limit theorem does apply to $f(x)=x$ and $g(x)=x^x$ with $c=0$, since $\lim_{x\to0}x=0$ and $\lim_{x\to0^+}x^x=1$ are not both $0$. We have, simply,

$$\lim_{x\to0^+}x^{x^x}=(\lim_{x\to0^+}x)^{\lim_{x\to0^+}x^x}=0^1=0$$

It's worth noting that the general theorem is again useless (as noted for $x^x$) if you ask about the next level, $x^{x^{x^x}}$. But when the "not both $0$" condition is met, it's a handy theorem indeed.