How to find $\lim_{x\to\infty} {\sin x \over x}$ without squeeze theorem?

Question

How can I find the limit:

$$\lim_{x\to\infty} {\sin x \over x}$$

Using an epsilon-delta proof? I am only really familiar doing this with simple polynomials.

Answer 1

First, note that $$ \left|\sin x\right|\leq1. $$ Now, for each $\epsilon>0$, pick $X > 0$ large enough such that $$ \frac{1}{X}<\epsilon. $$ It follows then that for all $x\geq X$, $$ \left|\frac{\sin x}{x}-0\right|= \text{_________} <\epsilon, $$ as desired. Can you fill in the blanks?

Answer 2

First of all, note that it's a slightly weird kind of "epsilon-delta" proof, since your $x$ is going to $\infty$. Call the limit $l$. With that in mind: you want to prove that, for all (small) $\varepsilon > 0$, there exists (large!) $\delta > 0$ such that, whenever $x > \delta$, then $\left|\dfrac{\sin(x)}{x} - l\right| < \varepsilon$. (The limit should, of course, be $0$.)

But since $|\sin(x)| \leq 1$, we have $\left|\dfrac{\sin(x)}{x}\right| \leq \left|\dfrac{1}{x}\right|$, so it'll suffice to pick $\delta$ such that $\left|\dfrac{1}{x}\right|<\varepsilon$. Can you do this?