# $\lim_{x\to +\infty} \sin \left[2\pi \sqrt{ \lfloor x\rfloor^{2}+\lfloor \frac{x}{3}\rfloor} \right]$ [closed]

I would like to calculate this limit I tried geogebra but it's show me that there is no limit but we can't consider machine as prove

$$\lim_{x\to +\infty} \sin \left[2\pi \sqrt{ \lfloor x\rfloor^{2}+\lfloor \frac{x}{3}\rfloor} \right]$$

Could someone calculate this limit ?

## closed as off-topic by Winther, Steven Stadnicki, user99914, Delta-u, zoliMay 15 '18 at 17:32

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Winther, Steven Stadnicki, Community, Delta-u, zoli
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• Write $\sqrt{a^2+b} = a\sqrt{1 + b/a^2}$ and use $\sqrt{1+t} \approx 1 + t/2$ as $t\to 0$. Also useful to notice that $[kx]/[x] \to k$ as $x\to \infty$ – Winther May 13 '18 at 14:05
• desmos.com/calculator/kfzvndrzuh . The limit seems to exist and $\approx 0.86$ – Tony Ma May 13 '18 at 14:06
• Cloud you write an answer please – Educ May 13 '18 at 14:12
• @Educ I'm not doing your homework for you. I gave you some hints. Now it's time to do some work and try yourself. – Winther May 13 '18 at 14:12

Let us write simply $n$ for the floor of $x$. Then the floor of $x/3$ is the same as the one for $n/3$. We then replace the factor of $2\pi$ under the sine with: $$\sqrt{n^2+[n/3]}-n = \frac{n^2+[n/3]-n^2} {\sqrt{n^2+[n/3]}+n} = \frac{[n/3]}{\sqrt{n^2+[n/3]}+n} \overset{n\to\infty}\longrightarrow \frac{1/3}2=\frac 16\ .$$ The limit thus exists and is $$\sin\left(2\pi\cdot\frac 16\right) =\frac{\sqrt 3}2\approx0.866025403784439 \dots\ .$$ $\blacksquare$

Let us make the situation clear at the point where $[x/3]$, resp. $[n/3]$ appears. We can put a sandwich on the expression, using something like $n/3-1<[n/3]<n/3+1$.
(We can use also $n/3-2018<[n/3]<n/3+2018$ if in doubt here or with $[x/3]$)
Then the expression under the limit is made sandwich between the two expressions $$\frac{n/3\mp 1}{\sqrt{n^2+n/3\pm 1}+n}\ .$$ (We are lowering the numerator, while in the same time increasing the denominator. The fraction is replaced with a lower one, we get a lower bounding sequence. Analogously, increasing the numerator and decreasing the denominator, we get an upper bounding sequence.) To compute the limit of the above sequence, we divide by force with the "chief of the expression", which is $n$, mathematically spoken, the numerator and the denominator are of the shape $O(n)$, so we divide by $n$ to filter the limiting constant. Then we get: $$\frac{\frac n3\mp 1}{\sqrt{n^2+\frac n3\pm 1}+n}= \frac{\frac 13\mp \frac1n}{\sqrt{1+\frac 1{3n}\pm \frac 1{n^2}}+1} \longrightarrow \frac{\frac 13\mp 0}{\sqrt{1+0\pm 0}+1} =\frac{1/3}{1+1} \text{ for }n\to\infty \ .$$
• The limit depends only on the floor of $x$. So i put an $n$ instead to have a smooth typing. Use that floor of $x$ instead of all places where $n$ is used. – dan_fulea May 13 '18 at 14:18
• In doing so, you are using the fact that $\lfloor x/3\rfloor=\lfloor\lfloor x\rfloor/3\rfloor$, which, while true, is maybe not immediately obvious. – Barry Cipra May 13 '18 at 14:24
• If not doing so, then just type everywhere $[x]$ and $[x/3]$ where the corresponding signs appear. There is only more "white noise notation", the correpsonding limit $$\lim_{x\to\infty}\frac{[x/3]}{\sqrt{[x]^2+[x/3]}+[x]}$$ is the same. – dan_fulea May 13 '18 at 14:37
• @dan_fulea Could you please elaborate this limit $$\frac{[n/3]}{\sqrt{n^2+[n/3]}+n} \overset{n\to\infty}\longrightarrow \frac{1/3}2$$ – Educ May 13 '18 at 14:53