Prove that if there is a number $B$ such that $|\frac{f(x)}{x}| \leq B$ for all $x$ not $0$, then $\lim_{x \to 0} f(x) =0$.
My Proof:
then for all $\epsilon > 0$ there is a $\delta>0$ such that $|x| < \delta$ implies $|f(x)| < \epsilon$
Aside: If $|\frac{f(x)}{x}| \leq B \rightarrow |f(x)|<B|x| < \epsilon$
Pick $\delta = \frac{\epsilon}{B}$ then $|x| < \frac{\epsilon}{B} \rightarrow |f(x)| \leq B|x| < \epsilon$ , Thus $|f(x)| < \epsilon$ QED The limit holds.
Is this a valid proof? If not, how do I correct it?