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

This question is an exact duplicate of:

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 $\delta$)

## marked as duplicate by Travis, Rory Daulton, Paul, Rebecca J. Stones, user147263 Dec 15 '14 at 1:08

This question was marked as an exact duplicate of an existing question.

• Hmm ... actually, should the limit have been for $x\to 2/\pi$? That would be slightly more interesting. – Henning Makholm Dec 14 '14 at 22:56
• @HenningMakholm oh so sorry my mistake its indeed 2/pi was my mistake – The One Dec 14 '14 at 22:57

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 $(0,\pi)$.

• So what was the method to find out $\delta$? you tried the nearest range of $\pi$s? cause it may be harder with harder expressions like $\lim_{x\to \infty}{\sqrt{2x - \sin 3x}=\infty}$ – The One Dec 14 '14 at 22:54
• @TheOne: I guessed, and the guess worked. Since I know there's an entire interval of zeroes around the target point, it is just a matter of finding a small enough guess that works, and it only needs to be done once (there's no dependency on $\varepsilon$). – Henning Makholm Dec 14 '14 at 22:58
• I fixed my typo mistake its indeed 2 / pi how would you guess it now? – The One Dec 14 '14 at 22:59
• @TheOne: Some comments added – Henning Makholm Dec 14 '14 at 23:13
• Thank you I now got a better sense of it – The One Dec 14 '14 at 23:17

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 $\frac{\pi}{2}$.

• I don't have enough reputation, so there is my answer to Henning Makholm: The problem is that even if argument of sine takes values betweeen 0 and $\pi$ it does not mean that sine won't take 1 value and will have a limit at 0. – Mihail Dec 14 '14 at 22:57
• x @Mihail: You're right, and I fixed that right after by using a smaller interval. – Henning Makholm Dec 14 '14 at 22:59