Convergence In L2 norm Given that
\begin{align*}
&u_n(x) = \frac{x}{\sqrt{x^2+1/n}}&\\
&\text{and }&\\
&v(x) =  \left\{ \begin{array}{cc} 
                \frac{|x|}{x} & \hspace{5mm} x\neq 0 \\
                0 & \hspace{5mm} x=0 \\
                \end{array} \right.&
\end{align*}
Show that $u_n \to v$ with respect to $L^2$ norm as $n \to \infty$.
My Attempt!
Since $u^{}_n(-x)=-u^{}_n(x)$ and $v(-x)=-v(x)$, and  Define 
\begin{align*}
&g_n(x) := |u_n^{}(x)-v(x)|&.\\
&\text{Noticing that $g_n(x)$ is even.}&\\
\end{align*}
And
\begin{align*}
&\|u_n(x)-v(x)\|^2 = \|g_n(x)\|^2=\int_{-1}^{1}|g_n(x)|^2dx&\\
&=2\int_{0}^{1}|g_n(x)|^2dx=2\int_{0}^{1}\left|\frac{x}{\sqrt{x^2+(1/n)}}-1\right|^2dx&\\
%&=2\int_{0}^{1}|g_n(x)|^2dx=2\int_{0}^{1}\left|\frac{x}{\sqrt{x^2+(1/n)}}-1\right|^2dx&\\
&=2\int_{0}^{1}\left|\frac{x-\sqrt{x^2+(1/n)}}{\sqrt{x^2+(1/n)}}\right|^2dx&\\
&= 2\int_{0}^{1}\left|\frac{-1/n}{\sqrt{(x^2+1/n)}(x+\sqrt{x^2+(1/n)})}\right|^2dx&\\
&\leq 2\int_{0}^{1}\left|\frac{-1/n}{\sqrt{(x^2+1/n)}}\right|^2dx&\\
&= 2\int_{0}^{1}\frac{1}{n^2} \frac{1}{{(x^2+1/n)}}dx&\\
\end{align*}
The inequality is obtained by dropping the term $x+\sqrt{x^2+1/n}$ from the denominator of preceding term. 
I am little hesitant to drop this term because it depends on $n$.
Can any body tell me if I am doing it right so far?

No, since
$$ \frac{1}{x + \sqrt{x^2 + 1/n}} > 1 $$ 
for $n$ large and $x \in [0, 1]$ small. I think the easiest proof is probably by the dominated convergence theorem, as $u_n \rightarrow v$ pointwise and $|u_n| \leq 1$. But if you want to proceed in this way, you can. Since
$$ \frac{1}{x + \sqrt{x^2 + 1/n}} \leq \frac{1}{\sqrt{x^2 + 1/n}}, $$
using Mathematica I find: 
$$ ||g_n(x)||^2 \leq \frac{2}{n^2}\int_0^1 \frac{dx}{(x^2+1/n)^2} \xrightarrow{n\to\infty} 0.$$
Although that integral, which I have omitted if you want to evaluate it yourself, seems rather complicated.
