Is zero a limit point of the sequence $n \sin \sqrt n$? 
Is zero a limit point of the sequence $(n \sin \sqrt n)$?

My attempt: 
Considering @MikhailKatz' answer to the question "Is zero a limit point of the sequence $(\sqrt{n}\sin n)$:

By Dirichlet's approximation theorem, $\pi$ can be approximated by rationals $p/n$ so that $|n\pi-p|<\frac{1}{n}$. Since sine is 1-Lipschitz, we get $\left|\sin(p)\right|<\frac{1}{n}$ from which the result is immediate.

is it then valid to write the following here?

By Dirichlet's approximation theorem, $\pi$ can be approximated by rationals $p/n$ so that $|n\pi-p|<\frac{1}{\sqrt n}$. Since sine is 1-Lipschitz,  we get $\left|\sin(p)\right|<\frac{1}{\sqrt n}$ from which the result is immediate.

Can I conclude that $0$ is a limit point?
 A: Partial result only (too long for a comment and I will go to bed now):
I will use the following heavy machinery:
Theorem. (Weyl/equidistribution of polynomial sequences) Let $$P(n)=a_dn^d+a_{d-1}+\dots+a_0$$ be a polynomial where $a_d$ is irrational. Then the sequence $(\{P(n)\})_{n\in\Bbb N}$ is equidistributed in $[0,1[$. (Here, $\{\cdot\}$ denotes the fractional part.)

In particular, we know that the squence $(\{m^2\pi^2\})_{m\in\mathbb N}$ is equidistributed in $[0,1[$, so for all $\varepsilon>0$, (here, $|\cdot|$ denotes cardinality) $$\lim_{m\to\infty} \frac{|\big\{\{\pi^2\},\{2^2\pi^2\},\dots,\{m^2\pi^2\}\big\}|
\cap[0,\varepsilon]
}{m}
= \varepsilon.$$
This means that, "on average", one out of $\frac1\varepsilon$ members of $m^2\pi^2$ is at most $\varepsilon$ away from the greatest integer from below. (Note that, heuristically, we want $n\cdot\sin(\sqrt n)\approx 0$, so $\sqrt n\approx m\cdot\pi$ for some integer $m$ and thus $m^2\cdot\pi^2$ should be close to an integer.)
Lemma. Let $\{m^2\pi^2\}<\varepsilon$ and $n=\lfloor m^2\pi^2\rfloor$. Then $n\cdot\sin(\sqrt n)\in[-n\sqrt \varepsilon, n\sqrt \varepsilon]$.
Proof. We have (note  that $\sqrt{m^2-\varepsilon}=\sqrt{m+\sqrt\varepsilon}\sqrt{m-\sqrt\varepsilon}>m-\sqrt\varepsilon$) $$\sqrt n = \sqrt{m^2\pi^2-\{m^2\pi^2\}}\in[\sqrt{m^2-\varepsilon},m\pi]\subset[m-\sqrt \varepsilon,m\pi].$$
Since $\sin$ is $1$-Lipschitz, we have $$\sin(\sqrt n)\in[-\sqrt \varepsilon,\sqrt\varepsilon].$$ This achieves a proof. $\square$

Now I would like to conclude saying that $\varepsilon$ is arbitrarily small. But, $n$ depends on $m$ and thus on $\varepsilon$ and can get really large before $\{m^2\pi^2\}<\varepsilon$. It seems that some further quantification of "how fast the series becomes equidistributed is needed."
Remark. I am not even sure if $0$ actually is a limit point. A converse result may also be provable using a similar approach as mine from above.
