# Extending L' Hopital. (For my students. )

How can one deal with $\lim_{x \to 0^+} x^{x^2}$? It appears to be the indeterminate form $0^0$. The graph shows that it has a nice limit:

(Notice, how this function produces imaginary vales when  it is between -1 and 0. )

But, this is the wrong kind of indeterminate form for L'Hopital.  If we could somehow show that it was, in fact, the same as $\frac{0}{0}$ we could use the derivative to help us... Here is an extension of L'Hopital that will do just that:

$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)}$

See if you can apply it. We will talk more about this in class. If you simply said "we can't use L'Hopital" that is correct for this problem based on what you knew up until now. If you applied it anyway and tried to say that:  $\lim_{x \to c} f(x)^{g(x)} =\lim_{x \to c} f'(x)^{g'(x)}$ ... then that is incorrect! (And I believe it leads to the wrong answer here.)