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.


  • $\begingroup$ Now use, than limit of $g$ is zero. $\endgroup$
    – zkutch
    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.

  • $\begingroup$ I didn't think to do it in that direction! Thank you so much! $\endgroup$ May 4 at 23:45

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