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 $x\to a$

Question

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 $g(x)<+\infty$.

Answer 1

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.

Answer 2

You can't because it's false.

Counter-example:

Let $a=0$ and $f(x)=x$. define $g(x)$ as: $$\begin{cases} g(x)=n^2 &\text{if}\enspace\smash[t]{\dfrac1{n+1}<\lvert x\rvert \le \dfrac1n},\\ g(0)=0. \end{cases}$$ On $\Bigl(-\dfrac1n,\dfrac1n\Bigr)\smallsetminus\{0\}$, we have $f(x)g(x)>\dfrac{n^2}{n+1}$, hence it tends to $+\infty$ as $x$ tends to $0$.