Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$ 
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$

My attempt: 
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\ &=
\lim _{n\to \infty}\frac{n^2\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\&= 
\lim_{n\to \infty}\frac{\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{\sqrt{3+\frac{1}{n^2}}}\end{align}$
Which looks like is equal to $\infty$ because of the $n^3\cdot 1$ but it's wrong. 
What am I doing wrong here?
 A: $$\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}\cdot\frac{\frac1{\sqrt n}}{\frac1{\sqrt n}}=\frac{\sqrt{1+\sqrt{1+\frac1{n^2}}}}{\sqrt{3+\frac1n}}\xrightarrow[n\to\infty]{}\sqrt\frac23$$
A: Hint: 
$$
\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}=\lim _{n\to \infty}\sqrt{\dfrac{n+\sqrt{n^2+1}}{3n+1}}.
$$
So you only need to determine the limit of the term inside the square root.
A: When you take a factor out of the square-root, it still gets square-rooted, but you have been squaring them.  For example, $\sqrt{n^2+1}=\sqrt{n^2(1+1/n^2)}=\sqrt{n^2}\sqrt{1+1/n^2}=n\sqrt{1+1/n^2}$
A: $$\lim_{n\to\infty}\frac{n+\sqrt{n^2+1}}{3n+1}=\lim_{n\to\infty}\frac{1+\sqrt{1+1/n^2}}{3+1/n}$$
$$=\frac{1+\sqrt1}3$$
A: Here are the steps
$$\lim\limits _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}
$$
$$= \lim\limits _{n\to \infty}\sqrt{\frac{n+\sqrt{n^2+1}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+\sqrt{n^2+1}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+\sqrt{n^2\left(1+\frac{1}{n^2}\right)}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+|n|\sqrt{1+\frac{1}{n^2}}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{1+\sqrt{1+\frac{1}{n^2}}}{3+\frac{1}{n}}} $$
$$= \sqrt{ \frac{1+\sqrt{1+0}}{3+0}} $$
$$= \sqrt{ \frac{2}{3}} $$
