An indeterminate power is
what we call the form of a limit $\displaystyle\lim_{x\rightarrow
a}f(x)^{g(x)}$ where the limit has the form $0^0$ or $1^\infty$ or
$\infty^0$ when you plug in the value $a$ (which might be
$\infty$).
Recall that when we have $f(x)^{g(x)}$ and we wish to
differentiate it, we use logarithmic differentiation, since taking
the log of $f(x)^{g(x)}$ allows us to pull the power $g(x)$ down
so it is easier to deal with.
Similarly, when looking at $\displaystyle\lim_{x\rightarrow
a}f(x)^{g(x)}$, we can take the log of $f(x)^{g(x)}$, and turn
this into an indeterminate product, which we can then tackle with
L'Hospital's rule. We have to be careful, as you will see in
the video, since
$$\lim_{x\rightarrow a}f(x)^{g(x)}\not=\lim_{x\rightarrow
a}\ln\left(f(x)^{g(x)}\right).$$
The video below will explain in detail how we compute such
limits.
Warning: There is an
error in the last step of the last example on the video, when she
uses L'Hospital's rule on $$\lim_{x\to
1}\frac{\sin\left(\frac{\pi}{2}x\right)\ln(2-x)}{\cos\left(\frac{\pi}{2}x\right)}.$$
DO: Can you find this
error? Hint: The
limit mentioned above should be $\frac{2}{\pi}$ instead of $1$, so
the answer to the problem is $\displaystyle e^{\frac{2}{\pi}}$.