TITLE: Prove that if $f(x) > 0$ for all $x$, $\lim_{x\to a} f(x) = 0$ and $\lim_{x\to a} g(x) = \infty$ ...
QUESTION [5 upvotes]: Prove that if $f(x) > 0$ for all $x$, $\lim_{x\to a} f(x) = 0$ and $\lim_{x\to a} g(x) = \infty$, then $\lim_{x\to a} [f(x)]^{g(x)} = 0$
NOTE: This shows that $0^\infty$ is not an indeterminate form.
Would this solution be correct?
$$
\lim_{x\to a}[f(x)]^{g(x)}=\lim_{x\to a}e^{g(x)\ln(f(x))}=0
$$
Because
$$
\lim_{x\to a}[\ln(f(x)]]=\lim_{y\to 0}(\ln y)=-\infty
$$
and
$$
\lim_{x\to a}[g(x)]=\infty
$$
Then
$$
\lim_{x\to a}[g(x)\ln(f(x)]]=-\infty
$$
Finally
$$
\lim_{x\to a}[e^{g(x)\ln(f(x)}]=0
$$

REPLY [2 votes]: Your solution looks good to me. Here's another approach - not better, just different.
There exists $\varepsilon_1$ such that $0 < f(x) < 1$ whenever $0 < |x-a| < \varepsilon_1$. There exists $\varepsilon_2$ such that $g(x) > 1$ whenever $0 < |x-a| < \varepsilon_2$. (Both of these are consequences of the limits given in the hypotheses.)
Therefore whenever $0 < |x-a| < \min\{\varepsilon_1,\varepsilon_2\}$, we have $0 < f(x)^{g(x)} < f(x)^1$.
The desired $\lim_{x\to a} f(x)^{g(x)} = 0$ then follows from $\lim_{x\to a} f(x) = 0$ and the squeeze theorem.