what is the limit point of the sequence $\{\sqrt n \tan n\}$ what  is the limit point of the  sequence $\{\sqrt n \tan n\}$ ?
My attempt : i thinks  there  will be no limit point  because $\lim_{n \rightarrow \infty }{\sqrt n \tan n}= \infty$ 
Is its  true ?
 A: The sequence is dense in $\mathbb{R}$. Every real number is a limit point.
Let $\alpha$ be an irrational number, and consider the sequence $y_n=\tan(n\alpha\pi)$. Most of the time, these $y_n$ are large positive or negative. However, it's possible for $n\alpha\pi$ to be very close to a multiple of $\pi$, so then $\tan(n\alpha\pi)$ will be very close to zero, and multiplying by $\sqrt{n}$ will leave it close to zero. For example, with $\alpha=\frac1{\pi}$ as in the original problem, $355\cdot\alpha\cdot\pi \approx 113 + 3.0\cdot 10^{-5}$. Take the tangent of $355$ and multiply by $\sqrt{355}$, and we get about $5.7\cdot 10^{-4}$.
Specifically, there are infinitely many pairs $(p,q)$ such that $|q\alpha-p|<\frac1q$. How do we find them? We use the continued fraction of $\alpha$. If $\frac pq$ is a "convergent" of this continued fraction, the result of terminating it after some finite number of terms, then $|\alpha-\frac pq|<\frac1{q^2}$. Multiply by $q$, and we have $|q\alpha-p|<\frac1q$ as desired. The convergents alternate between being larger than $\alpha$ and smaller than $\alpha$, so the differences $q\alpha-p$ alternate between being positive and negative.
Choose such a pair $(p,q)$ with $q\ge 4$. Then $|q\alpha-p|<\frac1q\le\frac14$, and $|\tan(q\alpha\pi)|<\tan\frac {\pi}{q}\le \frac{4}{q}$ (using $\tan x\le\frac{4}{\pi}x$ on $[0,\frac{\pi}{4}]$). Multiply by $\sqrt{q}$, and we have $|\sqrt{q}\tan(q\alpha\pi)|<\frac{4}{\pi\sqrt{q}}$. There are infinitely many such pairs, and this sequence immediately gets us zero as a limit point.
All right, how do we get nonzero values as limit points? We modify these $q$ values. Multiply by a positive integer $k$, and $\sqrt{kq}\tan(kq\alpha\pi)\approx k\sqrt{k}\cdot \sqrt{q}\tan(q\alpha\pi)$ as long as the angle $|k(q\alpha-p)\pi|$ stays small. Using the same $\frac{\pi}{4}$ bound on the angle, the derivative of $\sqrt{ax}\tan(bx)$ is
$$b\sqrt{ax}\sec^2(bx)+\frac{\sqrt{a}}{2\sqrt{x}}\tan(bx) < 2b\sqrt{a}\sqrt{x}+\frac{4b}{\pi}\frac{\sqrt{a}}{2}\sqrt{x}=\left(2+\frac{2}{\pi}\right)b\sqrt{a}\cdot \sqrt{x}$$
and the difference between adjacent values of $\sqrt{kq}\tan(kq\alpha\pi)$ is
\begin{align*}\left|\sqrt{kq}\tan(k(q\alpha\pi-p\pi))-\sqrt{(k-1)q}\tan((k-1)(q\alpha\pi-p\pi))\right| &< \left(2+\frac{2}{\pi}\right)(q\alpha-p)\pi\sqrt{kq}\\
&< (2\pi+2)\sqrt{\frac{k}{q}}\end{align*}
These adjacent values are less than $\epsilon$ apart as long as $k<\frac{\epsilon^2 q}{(2\pi+2)^2}$. As long as the approximation of $\alpha$ isn't too good - say, $|q\alpha-p|\ge\frac{1}{4q}$ - we get $|\sqrt{kq}\tan(k(q\alpha-p)\pi)|\ge \sqrt{kq}\cdot k\cdot\frac{1}{4q}\cdot\pi$. At the upper limit of the range, that bound becomes $\frac{\pi\epsilon^3}{4(2\pi+2)^3}q$, which goes to $\infty$ as $q$ does. What if our approximation is "too good"? Double $p$ and $q$ repeatedly until it's in that range.
So then, a convergent $\frac{p}{q}$ leads to us being able to approximate everything between $0$ and $A\epsilon^3 q$ to within $\epsilon$. We only get one side of the origin, but moving to the next convergent flips the sign and gets us the other side. Letting $q\to\infty$, the sequence $n\tan(n\alpha\pi)$ takes values within $\epsilon$ of everything. Since $\epsilon$ was arbitrary, we then get that the values are dense in $\mathbb{R}$.
The details get complicated, but the basic premise is simple: we use the continued fraction to find places where the values are small, then scale them to find places where the value is close to whatever we want.
