Is there a way that limit $\lim_{x \to 0}0^{0}$ to have alternative results besides 1? That is, is there a way so that
$= \lim_{x \to 0} f(x)^{g(x)}$
where $= \lim_{x \to 0} f(x)=0$ and $ \lim_{x \to 0} g(x)=0 $ is anything but 1
So is there a way that the limit of $0^{0}$ is not 1 but say, 0?
Could it be .5?
Could it be any other numbers?
 A: By continuity of the logarithm,
$$\log\left(\lim_{x\to0}f(x)^{g(x)}\right)=\lim_{x\to0}g(x)\log(f(x)).$$
The last expression is an indeterminate form $0\cdot\infty$, which is well-known to be able to take an arbitrary value.

E.g.
$$(e^{-1/x^2})^{x^2}\to e^{-1},$$
$$(e^{-1/x^4})^{x^2}\to 0,$$
$$(e^{-1/x^4})^{-x^2}\to \infty.$$
A: Elaborating on Yves Daoust's answer, here is a way to get any real number $C$ you would like. We separate into three cases.


*

*If $C>0$, then set $f(x) = e^{-1/x^2}$ and $g(x) = -\log(C) x^2$. You can check that $f$ and $g$ approach $0$ as $x \rightarrow 0$, and we get:
$$
f(x)^{g(x)}
=
e^{(-1/x^2)(-\log(C)x^2)}
=
e^{\log(C)} = C
$$

*If $C<0$, we just need to adjust a little bit. Take $f(x) = -e^{-1/x^2}$ and $g(x) = -\log(-C)x^2$. This works out just as above, and $\log(-C)$ makes sense since $-C>0$.

*Finally, if $C=0$, just take $f(x) = e^{-1/x^4}$ and $g(x) = x^2$. The limit of these functions as $x \rightarrow 0$ is still $0$, and we get:
$$
f(x)^{g(x)} = e^{-1/x^2}
$$
which approaches $0$ as $x \rightarrow 0$.
