For any $a>0$ can you find $f(x)$ and $g(x)$ such that $\lim_{x\to 0} f(x)^{g(x)} = a$?

Question

I watched a video today about how one can get an idea about how $0^0$ should be $1$. I think that what is happening is that $$ \lim_{x\to 0} x^x = 1. $$

So if $f(x) = x$ and $g(x) = x$. Then $$ \lim_{x \to 0} f(x)^{g(x)} = 1 $$

Now if $f(x) = 0$ and $g(x) = x$ then $$ \lim_{x\to 0} f(x)^{g(x)} = 0 $$

My question is: Given an $a\in \mathbb{R}$ with $a > 0$ is it possible to find two functions $f(x)$ and $g(x)$ such that $$ \lim_{x\to 0} f(x) =\lim_{x\to 0} g(x) = 0 $$ and $$ \lim_{x\to 0} f(x)^{g(x)} = a $$ ?

Answer 1

$a x^x$ would have a limit of $a$ at $0$, so let's work with that a bit. Firstly note that $a = x^{\log_x(a)}$. Thus

$$a x^x = a x^x = x^{\log_x(a)} x^x = x^{\log_x(a)+x}.$$

This is of the desired form and has the limit property you wanted. The convergence to $0$ of $\log_x(a)+x$ is very slow for what it is worth.

Answer 2

Yes, let $f(x)=x$, $g(x)=\ln (a)/\ln(x)$. Both obviously converge to 0 and $f^g=a$ everywhere.

Answer 3

The answer is yes. For example: let's consider $f : x \mapsto e^{-\log(a)/x}$ and $g:x \mapsto -x$.

Then, $f(x)^{g(x)} = a$ and both $f,g\underset{x\to 0}{\longrightarrow} 0$ if $a>1$.

Now if $0<a<1$, let's take $f : x \mapsto e^{+\log(a)/x}$ and $g:x \mapsto +x$.

It works the same.