Proof that $\lim_{ x\to\frac{2}{\pi}} {\sin {(\frac{1}{x})}} = 1$

Question

I'm new to epsilon-delta proofs and and I want to know how to prove this using only the function limit definition at a point.

Answer 1

The function $\sin (1/x)$ is an elementary function defined at $x=2 / \pi$. Since all elementary functions are continuous in their domain it holds that $$\lim _{x\to 2/\pi} \sin (1/x) = \sin \frac{1}{\left (\frac{2}{\pi}\right )} = \sin \frac{\pi}{2} = 1. $$

If you want the limit to be equal to $0$ it suffices to let $x\to 1/\pi$.

Answer 2

Hint: $\lim_{ x\to2/\pi} \sin(1/x)= \lim_{ t\to \frac{\pi}{2}} \sin(t)$.

Answer 3

You want to show that $| \sin(1/x) - 1| <\epsilon.$ Write it as $| \sin(1/x) - \sin(\pi/2)| <\epsilon,$ and then use the trig identity

$$\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$

to get

$$\left| 2\cos\left(\frac{2+\pi x}{4x}\right)\sin\left(\frac{2-\pi x}{4x}\right)\right|< \epsilon.$$

Since $\cos t\leq 1$, it suffices to show

$$\left| \sin\left(\frac{2-\pi x}{4x}\right)\right|< \frac{\epsilon}{2},$$

or

$$\left| \sin\left(\frac{1}{2x}-\frac{\pi}{4}\right)\right| <\frac{\epsilon}{2}.$$

Using the inequality $\sin(t)\leq t$ for all $t$, it suffices to show that

$$\left|\frac{1}{2x}-\frac{\pi}{4} \right| <\frac{\epsilon}{2}$$

which equivalent to

$$\left| \frac{\pi(2/\pi-x)}{2x}\right|<\frac{\epsilon}{2}$$

or

$$\left|\frac{2/\pi-x}{x}\right| < \frac{2\epsilon}{\pi}.$$

So we'll need to restrict $\delta$ to keep $x$ away from $0$. So assume $\delta <1/10$. Then $|x|>1/2$ when $|2/\pi -x|<\delta.$ With this restriction, it now suffices to have

$$\left|\frac{2}{\pi}-x\right|<\frac{\epsilon}{\pi}$$,

so take $\delta = \min(1/10, \epsilon/\pi)$ and work backward through this mess.