# Proof using the $\left( \epsilon, \delta \right)$ definition of limit

Let $$f, g \colon \mathbb{R} \to \mathbb{R}$$ be functions such that f is bounded, i.e. $$\exists M \in \mathbb{R}^+ \colon \vert f(x) \vert \leq M$$ and $$\lim \limits_{x \to x_0} g(x) = 0$$.

I need to proof that $$\lim \limits_{x \to x_0} f(x) g(x) = 0$$ using the $$\left( \epsilon, \delta \right)$$ definition of limit.

This is what I've got so far ($$h(x) := f(x)g(x)$$):

$$\forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathrm{Dom}(h), 0 < \vert x - x_0 \vert < \delta \implies \vert h(x) - 0 \vert = \vert h(x) \vert< \epsilon$$.

Using the definition of $$h(x)$$ and the fact that $$f$$ is bounded I got:

$$\vert h(x) \vert = \vert f(x) g(x) \vert = \vert f(x) \vert \vert g(x) \vert \leq M \vert g(x) \vert = \epsilon$$.

I'm stuck there.

Thanks!

• Now use, than limit of $g$ is zero. May 4 at 23:37

You shall not start with $$h$$ with $$\epsilon$$-argument, because that is your aim.
Since $$\lim g(x)=0$$, for any $$\epsilon>0$$, consider the number $$\epsilon/M>0$$, one can find some $$\delta>0$$ such that $$0<|x-x_{0}|<\delta$$ implies that $$|g(x)|<\epsilon/M$$, with all such $$x$$, one has $$|f(x)g(x)|=|f(x)|\cdot|g(x)|\leq M\cdot |g(x)|\leq M\cdot(\epsilon/M)=\epsilon$$, done.