I want to prove: Take a real $c$, a $\delta > 0$, and functions $f, g: \mathbb{R} \to \mathbb{R}$. If $$\lim_{x \to c}f(x) = 0$$ and $$|g(x)| \leq |f(x)|$$ for all $0 < |x-c| < \delta,$ then $$\lim_{x \to c}g(x) = 0.$$
The problem I have is that, given an $\epsilon > 0$, I do not see from the given conditions a way to choose a $0 < \delta' \leq \delta$ such that $$|x-c| < \delta' \Rightarrow |g(x)| < \epsilon.$$
since $$\lim_{x\rightarrow c} f(x) = 0$$ it is true, that $$\forall_{\epsilon > 0} \exists_{\delta' > 0} \, |x - c| < \delta' \implies |f(x)| < \epsilon$$ Now let $\epsilon > 0$ and choose some $\delta'$ such that the above implication is true and $\delta' < \delta$. Now we have $$\forall_{x, |x - c| < \delta'} |g(x)| \leq |f(x)| < \epsilon$$