Q:

If $f$ and $g$ are finite valued on the interval $(a-1,a+1)$ and $f(x)\to0$ as $x\to a$, then $f(x)g(x)\to0$ as $x\to a$

How would I go about proving this? On my first attempt, I thought finite-valued was the same thing as bounded, but finite valued just means $g(x)<+\infty$.

A:

If $f$ and $g$ are not continuous, here is a counterexample.
Consider $(0-1,0+1) = (-1,1)$.
Let $f(x) = x$, and $g(x) = \begin{cases}\frac{1}{x^{2}} & x \neq 0 \\ 0 & x = 0 \end{cases}$.
Then $\lim \limits_{x \to 0} f(x) = 0$ and $\lim \limits_{x \to 0} f(x)g(x) = \lim \limits_{x \to 0} x \frac{1}{x^{2}} = \lim \limits_{x \to 0} \frac{1}{x} \neq 0$, so the claim is false.

