An indeterminate power is
what we call the form of a limit limx→af(x)g(x) where the limit has the form 00 or 1∞ or
∞0 when you plug in the value a (which might be
∞).
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 limx→af(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 limx→af(x)g(x)≠limx→aln(f(x)g(x)).
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 limx→1sin(π2x)ln(2−x)cos(π2x).
DO: Can you find this
error? Hint: The
limit mentioned above should be 2π instead of 1, so
the answer to the problem is e2π.