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?