Prove that $\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0$ in the $\epsilon$-$\delta$ way

Question

Given: $$\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0$$

How to prove this limit using the $\epsilon$-$\delta$ way? (the biggest problem is to find $\delta$)

Answer 1

Original answer from when the question was for the limit as $x\to \pi/2$:

$\frac{1}{\pi/2}\approx 0.6$, which is in $(0,\pi/2)$, so there's an entire neighborhood of $\pi/2$ in which $\lfloor \sin(1/x)\rfloor = 0$. Therefore the limit is $0$ too.

For example, you can take $\delta=1/2$. When $x$ varies between $\frac{\pi}2-\frac12$ and $\frac\pi2+\frac12$, $1/x$ will vary between ~0.93 and ~0.48, and all of this range is in $(0,\pi/2)$ so $\lfloor\sin(1/x)\rfloor=0$ everywhere in this range.

For the edited question: This is not quite as simple because $\lfloor \sin(1/x)\rfloor$ is $1$ when $x=2/\pi$. But fortunately what the $\varepsilon$-$\delta$ definition says is just that we must have $|f(x)-L|<\varepsilon$ when $0<|x-x_0|<\delta$. So the inequality doesn't actually need to hold for $x=2/\pi$ itself. And everywhere else in the close vicinity of $2/\pi$ we still have $\lfloor \sin(1/x)\rfloor=0$.

So we can still just guess at an appropriate $\delta$. It turns out that $\delta=1$ or $\delta=1/2$ don't work, but $\delta=1/4$ does: When $x$ varies between $\frac2\pi-\frac14$ and $\frac2\pi+\frac14$, $1/x$ will vary between ~2.59 and ~1.12. An all the values in that interval, except for $\pi/2$ have sines strictly between 0 and 1, so we do indeed have $\lfloor \sin(1/x)\rfloor=0$ there.

If guessing is too undignified, the example shows that the first thing that goes wrong if we use a too large $\delta$ is that $\frac{1}{2/\pi-\delta}$ becomes larger than $\pi$, where our function becomes $-1$ instead of $0$. So if we want to find the largest $\delta$ that works, we can solve $$ \frac{1}{2/\pi-\delta} = \pi $$ for $\delta$, yielding $\delta=1/\pi$. But there's not real sense in which $\delta=1/\pi$ is a better answer than $\delta=1/4$.

One could even make the guessing less careful and just say $\delta=0.000001$ and compute the resulting range of $1/x$, which will then easily stay within $(0,\pi)$.

Answer 2

Take $\delta=\frac{\pi}{8}$. Because $\frac{\pi}{8}<\frac{2}{\pi}<\frac{\pi}{4}$ and so $\frac{1}{x}$ will be always between 0 and $\frac{\pi}{2}$.