If $\lim\limits_{x\to a}f(x)=0$ and $g(x)$ is bounded in a neighborhood of $a$, then $\lim\limits_{x\to a}f(x)g(x)=0$

Question

If $\lim\limits_{x\to a}f(x)=0$ and $g(x)$ is a bounded function at $a$'s neighborhood, then $\lim\limits_{x\to a}f(x)g(x)=0$.

Thanks in advance for any help!

Answer 1

Hint: $|f(x)g(x)| = |f(x)|\cdot|g(x)| < \epsilon\cdot M$, where $|f(x)|< \epsilon$ and $|g(x)| < M$ when $x$ is in the appropriate neighborhood of $a$.

Answer 2

To show that $lim_{x \to a} f(x)g(x) = 0$, one needs to show that for any $\epsilon \gt 0$, $\exists$ a $\delta \gt 0$ such that for any point $x$ with $|x-a| \lt \delta$, we have $|f(x)g(x)-f(a)g(a)| \lt \epsilon$.

Fact 1: As $g$ is a bounded function in a neighborhood of $a$, $|g(x)| \lt M$ for some positive real M for all $x$ within some $\delta_1$ of $a$.

Fact 2: As $lim_{x \to a} f(x) =0$, $|f(x)| \lt \frac{\epsilon}{M}$ for all $x$ within some $\delta_2$ of $a$.

Can you put these 2 facts together to arrive at the desired conclusion?

Answer 3

$\lim\limits_{x\to a}f(x) = 0$ ie $\forall \epsilon > 0, \exists \alpha>0, |x-a|<\alpha \Rightarrow | f( x ) | < \epsilon$

$\exists \delta > 0, \exists M, \forall x, |x-a| < \delta \Rightarrow | g( x ) | < M$

So you have $\forall x, |x-a| < \min(\epsilon,\delta) \Rightarrow |f(x)|<\epsilon \land |g(x)|<M$

Then you just need to see that $|f(x)g(x)| \le |f(x)||g(x)|\le\epsilon M$

So $\lim\limits_{x\to a}|f(x)g(x)| = 0$ ie $\lim\limits_{x\to a}f(x)g(x) = 0$

Answer 4

$f$ gets as small as you want, and $g$ doesn't get too large, so $f g$ also gets as small as you want, though not necessarily as small as $f$.

That's how I think about these kinds of things.