Can we calculate $\lim_{x \to 0+}{x^{(x^x)}}$ without using extended real numbers?

Question

Can we calculate $\lim_{x \to 0+}{x^{(x^x)}}$ without using extended real numbers?

Today I tried to calculate $$\lim_{x \to 0+}{x^{(x^x)}},$$

which can be written as$$\lim_{x \to 0+}{e^{(x^x \log x)}},$$

and since $$x^x \log x \to -\infty\;\;\text{ as }\;\;x\to 0+,$$ the limit becomes $$e^{-\infty}=0.$$ However, I haven't learned the extended real number system. That is, I can't treat $-\infty$ as a number. Therefore, I can't calculate $\lim_{x \to 0+}{x^{(x^x)}}$ using this approach.

I have read this post, but I found that there seems to be an error: the condition is $\alpha \in \mathbb{R_+^*}$, but $0 \not\in \mathbb{R_+^*}$.

Are there other methods to calculate the limit without using extended real numbers?

Thank you for your help.

Answer 1

I don't see why you think that the extended reals are needed.

Once you have $x^x \log x \to -\infty\;\;\text{ as }\;\;x\to 0+, $ just taking the exponential works fine.

You are not taking $e^{-\infty}$, you are doing $\lim_{x \to 0^+}e^{f(x)}$ where $\lim_{x \to 0^+} f(x) =-\infty$.

Answer 2

Actually, $e^{-\infty}$ is undefined as a real number.

However, we always use this to interpret the limit of exponent is $-\infty$. $-\infty\notin\mathbb R$.

However, we can show that $\lim_{x\to -\infty}e^x=0$ (It would be easy using definition of limit of function and limit of sequence plus some knowledge about natural logarithm $\ln$).

To conclude, $\lim_{x\to 0^+}x^{x^x}=0$.

Answer 3

Note that for all $x\gt0$, $x\log(x)\ge-\frac1e$. Therefore, $x^x\ge e^{-1/e}$. Thus, for $0\lt x\le1$, $$ 0\lt x^{x^x}\le x^{e^{-1/e}} $$ Now, the Squeeze Theorem says that $$ \lim_{x\to0^+}x^{x^x}=0 $$ No extended reals are necessary.