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 ?
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 $$ \sin\left(2\pi\cdot\frac 16\right) =\frac{\sqrt 3}2\approx0.866025403784439 \dots\ . $$ $\blacksquare$
Later edit, after comments:
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: $$ \frac{\frac n3\mp 1}{\sqrt{n^2+\frac n3\pm 1}+n}= \frac{\frac 13\mp \frac1n}{\sqrt{1+\frac 1{3n}\pm \frac 1{n^2}}+1} \longrightarrow \frac{\frac 13\mp 0}{\sqrt{1+0\pm 0}+1} =\frac{1/3}{1+1} \text{ for }n\to\infty \ . $$