Let $A \subseteq \Bbb{R}$, let $a$ be a limit point of $A$, and $f,g : A \to \Bbb{R}$ functions. If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = L \in (0,\infty)$, then $\lim_{x \to a} f(x)g(x) = \infty$.
I first tried to solve this problem "from the definitions", but I couldn't see how to do this. Then I realized that $fg = \frac{f^2 + g^2 - (f-g)^2}{2}$, and I thought to myself that it would suffice to show that $\lim_{x \to a} f(x) = \infty$ implies $\lim_{x \to a} f(x)^2 = \infty$, since I had already proven that $\lim_{x \to a}[f(x)+g(x)] = \infty$. But then I realized that the above expression involves a nasty, miserable negative sign...
Is there any way to "bypass" the negative to get this reduction to work? If not, how do I solve this problem?
"from the definitions"
Let $B>0$ be given. Since $\frac{1}{2}L>0$ we know there is $\delta_1>0$ so that \begin{equation}|g(x)-L|<\frac{L}{2} \text{ whenever } 0<|x-a|<\delta_1. \end{equation} Since $\frac{2B}{L}>0$ we know there is $\delta_2>0$ so that \begin{equation}|\,f(x)|>\frac{2B}{L} \text{ whenever } 0<|x-a|<\delta_2. \end{equation} We select $\delta=\min\{\delta_1,\delta_2\}$, and now we may observe that we have \begin{equation}\left|\,f(x)g(x)\right|>\left|\,f(x)\frac{L}{2}\right|>B \text{ whenever } 0<|x-a|<\delta. \end{equation} Therefore, $\lim_{x \to a} f(x)g(x)=\infty$.