$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$ - show that $a_n$ is convergent sequence Problem:
Show that $a_n$ is convergent sequence and find a limit of $a_n$.
$$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$$
I tried to look at this as normal limit problem so I wrote this:
$$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\lim_{n \to \infty}(\frac{1}{\sqrt{1+1}})=\frac{1}{2}$$
But I didn't get anything which can help me to solve a problem.

You can invert the function $y = {x \over \sqrt{x^2 + 1}}$ as follows.
$$y^2 = {x^2  \over x^2 + 1} = 1 - {1 \over x^2 + 1}$$
$$1 - y^2 = {1 \over x^2 + 1}$$
$${1 \over 1 - y^2} = x^2 + 1$$
$${1 \over 1 - y^2} - 1 = x^2$$
So we have
$$x^2 = {y^2 \over 1 - y^2}$$
Seeing that $x$ and $y$ must have the same sign, we have
$$x = {y \over \sqrt{1 - y^2}}$$
Hence if for your sequence $x_n$ you write $y_n = {x_n \over \sqrt{1 + x^2}}$, then you have
$$x_n = {y_n \over \sqrt{1 - y_n^2}}$$
Since $\lim_{n \rightarrow \infty} y_n = {1 \over 2}$, by the continuity of 
${y \over \sqrt{1 - y^2}}$ you have
$$\lim_{n \rightarrow \infty} x_n = {{1 \over 2} \over \sqrt{1 - {1 \over 4}}}$$
$$= {1 \over \sqrt{3}}$$
