Let $f(x)$ and $g(x)$ continuous functions at $x = 0$ such that $f(0) = 0 = g(0)$. Show that limit as $x$ approaches zero of $f(x)^{g(x)} = 1$ The test is very simple for the case $f(x) = x = g(x)$ since
$$
\lim_{x\to 0}{x^x} = 1
$$
But in other cases?
Note that they do not specify that the functions are differentiable and neither that they are continuous throughout their domains. We can assume that the functions are defined in an open interval containing zero.
 A: Note that
$$
\log (f(x)^{g(x)}) = g(x)\log f(x)
$$
Now the result holds as soon as $ g(x)\log f(x)
\to 0$, but is is not always the case (if $g(x)$ goes to $0$ slow enough).
For instance:
$$
f(x) = e^{-1/|x|}\implies g(x)\log f(x) = -\frac {g(x)}{|x|}
$$
Now take $g(x) = \sqrt{|x|}$ and this is a counterexample.
A: As a counter-example, consider $f(x)=0$ and $g(x)=x^2$ which are each continuous and differentiable
A: To prove that
$$
\lim_{x\to 0}f(x)^{g(x)}=1
$$
We can instead prove
$$
\lim_{x\to 0} {g(x)}\ln(f(x))=0\\\\
$$
We can use L'Hopital's rule to prove this if we rearrange the limit first.
$$
\begin{align}
&\lim_{x\to 0} g(x)\ln(f(x))\\\\
=&\lim_{x\to 0}\frac{\ln(f(x))}{1/g(x)}\\\\
=&\lim_{x\to 0}\frac{\frac{1}{f(x)}f^\prime(x)}{-\frac{1}{g^2(x)}g^\prime(x)}\\\\
=&\lim_{x\to 0}\frac{g^2(x)}{f(x)}\frac{f^\prime(x)}{g^\prime(x)}\\\\
=&\lim_{x\to 0}g(x)=0
\\\\
\end{align}
$$
We can use L'Hopital's rule above since the limit $x\to 0$ of $\frac{f(x)}{g(x)}$ is of the indeterminate form $\frac00$. Since we've proven that the limit of $g(x)\ln(f(x))$ as $x\to 0$ is $0$, we can see that:
$$
\lim_{x\to 0}\ln f(x)^{g(x)}=0\implies\lim_{x\to 0}e^{\ln f(x)^{g(x)}}=1\implies\lim_{x\to 0}f(x)^{g(x)}=1
$$
Hopefully this helps answer your question.
