Limit of the sequence $\lim_\limits{n\to\infty}\sin(2\pi(n^2+n^{1/2})^{1/2})$. I have tried to solve this limit : $\lim_\limits{n\to\infty}\sin(2\pi(n^2+n^{1/2})^{1/2})$. 
Where n $\in\mathbb{N}$.
I have understood that the limit exists and goes to 0 if the argument becomes like $k2\pi$ with k $\in$ $\mathbb{Z}$, so I have collected $n^2$ and I have taken it out of the root , highlighting  that $(n(1+\frac{1}{n^{3/2}})^{1/2})$ "tend to become an integer". 
I guess if this is enough to say that the limit exist and goes to 0.
 A: Assuming $n$ is an integer, as we've clarified already, we have
$$\sin(2\pi (n^2 + n^{1/2})^{1/2}) = \sin (2\pi n (1+n^{-3/2})^{1/2}). $$
Now you can rewrite that as 
$$ \sin \left (2\pi n + 2\pi n \left[ \sqrt{1 + n^{-3/2}} - 1 \right] \right) $$
and adding $2\pi n$ to the argument doesn't change the value, so this is just
$$ \sin \left( 2\pi n \left[ \sqrt{1 + n^{-3/2}} - 1 \right] \right). $$
Now what happens to the argument $2\pi n \left( \sqrt{1 + n^{-3/2}} - 1 \right)$ as $n \to \infty$?  I leave this to you.
A: As $n$ approaches infinity, $(1+\frac{1}{n^{3/2}})^{1/2}$ approaches zero, whilst $n$ approaches infinity. Their product approaches infinity however, because n is of a higher degree than the bracketed quantity. This is evident also from the expansion of the square root. 
Therefore, your "$k$" value actually diverges to infinity and does not approach an integer. So the sine of that does not approach any quantity and the limit does not exist. 
EDIT: Note that this behavior also becomes obvious when you plot the expression whose limit you are trying to take. It is periodic and does not actually move towards any given quantity. 
