Proof for a limit using epsilon-delta proof and squeeze theorem

Question

Suppose $f$ is a function that satisfies $\displaystyle \lim_{x \to 0} \frac{f(x)}{x} = 3$. And suppose $f(0) = 0$.

Show that $\displaystyle \lim_{x \to 0} f(x) = 0$

Hint: Start by taking $\varepsilon= 1$ in definition of limit for the function $\frac{f(x)}{x}$ and then try to use the squeeze theorem.

Thanks for the help!

Answer 1

If $\lim_{x \to 0} \frac{f(x)}{x} =c $, where $c$ is some positive real, then, for any $\epsilon > 0$, for small enough $x$, $|\frac{f(x)}{x}-c| < \epsilon $, or $-\epsilon < \frac{f(x)}{x}-c < \epsilon $, or $-\epsilon+c < \frac{f(x)}{x} < \epsilon+c $. Multiplying by $x$, $x(-\epsilon+c) < f(x) < x(\epsilon+c) $.

Choosing $x$ small enough, we can make $f(x)$ as small as we want. This implies that $\lim_{x \to 0} f(x) =0 $.

You can make my "small enough"s more rigorous, but this is good enough for me.