Prove $\lim_{x\to \infty}\frac{\sin x}{x^2} = 0$.

Question

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?

Answer 1

Yes. Your proof is correct. That's how one chooses $M$. Though "Let we have for $x>M$" looks strange for me. The writing can be slightly modified as:

Let $\epsilon>0$. Note that $\displaystyle |\frac{\sin x}{x^2}|\le \frac{1}{x^2}$ for $x\ne 0$. Let $M=\frac{1}{\sqrt{\epsilon}}$. Then for $x>M$, one has $$ |\frac{\sin x}{x^2}|\le\frac{1}{x^2}<\epsilon\;. $$ Q.E.D.