Verified answer

Because lim x-> a f(x) = L > 0, then taking epsilon = L/2 > 0, we can apply the definition of a limit to see that some d > 0 exists such that:

0 < |x - a| < d ===> |f(x) - L| < L/2

===> -L/2 < f(x) - L < L/2

===> L/2 < f(x) < 3L/2

===> f(x) > L/2

From there we know that, around this neighbourhood:

(fg)(x) = f(x)g(x) > (L/2)g(x)

which tends to infinity, by algebra of limits. Thus, by (infinite) squeeze theorem, we see that (fg)(x) ---> infinity.