# $\lim_{x\to 0}\left\lfloor\frac{n\sin(x)} x \right\rfloor,n\in\mathbb N$

If $\lfloor\,\cdot\,\rfloor$ denotes the greatest integer function $n\in\mathbb N$, then what is the value of $$\lim_{x\to 0}\left\lfloor\frac{n\sin(x)} x \right\rfloor \text{ ?}$$

My try:

\begin{align*} \lim_{x\to 0} \left\lfloor\frac{n\sin(x)} x \right\rfloor &= \lim_{x\to 0} \left\lfloor \frac{n(x-x^3/3!+x^5/5!-\cdots)} x \right\rfloor \\[10pt] &= \lim_{x\to 0} \left\lfloor \frac{n(1-x^2/3!+x^4/5!-\cdots)} 1 \right\rfloor \\[10pt] &= \lfloor n\rfloor \end{align*}

But the answer is $n-1$.

How am I committing a mistake? Why can't limits enter into the greatest integer function? What is the correct solution for this problem?

• Limits can't enter into the greatest integer function because the greatest integer function isn't continuous. Commented Jul 28, 2017 at 20:31
• Your last equality doesn't hold.
– user65203
Commented Jul 28, 2017 at 20:47
• @YvesDaoust:Yeah it is wrong,by mistake i'd assumed that greatest integer function is a continuous function. Commented Jul 28, 2017 at 20:51
• If the domain ($\mathbb{R}$) is connected and the image ($\mathbb{Z}$) is not connected, then the function cannot be continuous. Commented Jul 28, 2017 at 20:52
• @P.Styles, you don't have to write what $\lfloor\rfloor$ means because it is part of a normed notation. Commented Jan 27, 2020 at 20:27

Consider that $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$ and that $\frac{\sin(x)}{x} < 1$ for all $x\in \mathbb{R}\setminus \{0\}$. Therefore, for any $\epsilon > 0$, there is some $\delta > 0$ such that $0 < \lvert x\rvert < \delta$ implies $1-\epsilon < \frac{\sin(x)}{x} < 1$. From this, we can see that for any $\epsilon > 0$, we can choose the same $\delta > 0$ to find that $0 < \lvert x\rvert < \delta$ implies $$(1-\epsilon)n < \frac{n\sin(x)}{x} < n$$ Namely, this holds true for $\epsilon < \frac{1}{n}$ (which gives $(1-\epsilon)n > n-1$), so the result follows, i.e. $$\lim_{x\to 0} \left\lfloor \frac{n\sin(x)}{x}\right\rfloor = n-1$$

• As $n$ is independent from $x$ I ask the reason why I can't "export" $n$ out of the limit and conclude that the limit is $0$ Commented Jul 29, 2017 at 17:53
• Generally speaking, $\lfloor nr\rfloor\neq n\lfloor r\rfloor$ for real $r$. Commented Jul 29, 2017 at 17:56

For $x\ne0$, $$\dfrac{\sin x}x<1$$ so that $$\dfrac{n\sin x}x<n.$$

For all $x$ sufficiently close to $0$, the floor is $n-1$* (but never $n$), and this is the requested limit.

*By continuity of the function you will always find an interval where this holds.

• :$\frac{sin x}{x}<1$ also holds for $x<0$? Commented Jul 28, 2017 at 20:56
• It does! Both $x$ and $\sin(x)$ are odd functions, so for $x > 0$, $$\frac{\sin(-x)}{-x} = \frac{-\sin(x)}{-x} = \frac{\sin(x)}{x} < 1$$ Commented Jul 28, 2017 at 20:59
• @PKStyles : Note this MathJax code: $$\mathbb R \setminus \{0\} \text{ or } \mathbb R \smallsetminus \{0\}$$ Commented Jul 28, 2017 at 21:04
• I forgot to mention that $\text{sinc } x$ is an even function. Updating...
– user65203
Commented Jul 28, 2017 at 21:05
• @MichaelHardy:got it.Thanks!! Commented Jul 28, 2017 at 21:06

Maybe lots of students recall that $\dfrac{\sin x} x \to 1$ as $x\to0,$ but if you recall the proof of that fact you see that $\dfrac{\sin x} x \uparrow 1$ as $x\to0,$ i.e. it approaches $1$ from below. Therefore $\dfrac{n\sin x} x \uparrow n,$ i.e. this approaches $n\vphantom{\frac {\sum^\sum}{}}$ from below.