Prove using definition of a limit that $\lim_{x\to \infty}\frac{\sin x}{x^2} = 0$.
Proof:
Let $\epsilon > 0$. Note that $\left|\frac{\sin x}{x^2}\right| \leq \frac 1 {x^2}$ for $x\ne 0$. Then choose $M= \frac 1{\sqrt{\epsilon}}$. Then if $x> M$ implies that $\left|\frac{\sin x}{x^2}\right| \leq \frac 1 {x^2} < \epsilon$.
Am I allowed to give an $M$ the way I did?