Uniform convergence of function series Let $f_n(x)=n(\sqrt{x^2+\frac{1}{n}}-x)$.
I want to prove that $f_n(x)$ is uniformly converging in $[1,\infty)$. 
I found that the pointwise limit function is $f(x)=\frac{1}{2x}$, and looked for the supremum of $|f_n(x)-f(x)|$ to show that it is tending to $0$.
However, I couldnt find a supremum such as this. Not an extremum and not a supremum. What should I do next?

Observe that
$$f_n(x) = n\left(\sqrt{x^2 + \frac{1}{n}} - x\right)
\left(\frac{\sqrt{x^2 + \frac{1}{n}} + x}{\sqrt{x^2 + \frac{1}{n}} + x}\right)
= \frac{1}{\sqrt{x^2 + \frac{1}{n}} + x}$$
Consequently, for $x \geq 1$,
$$\begin{aligned}
\left|f_n(x) - \frac{1}{2x}\right| &=
\left|\frac{1}{\sqrt{x^2 + \frac{1}{n}} + x} - \frac{1}{2x}\right| \\
&= \left|\frac{x - \sqrt{x^2 + \frac{1}{n}}}{2x(\sqrt{x^2 + \frac{1}{n}} + x)}\right| \\
& \leq \left| \frac{x - \sqrt{x^2 + \frac{1}{n}}}{4x^2}\right| \\
\end{aligned}$$
where the inequality follows because $\sqrt{x^2 + \frac{1}{n}} \geq x$ for positive $x$. Multiplying numerator and denominator of the last expression by $$x + \sqrt{x^2 + \frac{1}{n}},$$ we obtain the bound
$$\begin{aligned}
\left|f_n(x) - \frac{1}{2x}\right| &\leq
\left|\frac{1/n}{4x^2(x + \sqrt{x^2 + \frac{1}{n}})}\right| \\
&\leq \frac{1/n}{8x^3} \\
&\leq \frac{1}{8n}
\end{aligned}$$
since $x \geq 1$. Since the bound $1/(8n)$ is independent of $x$, we have established uniform convergence for $x \in [1,\infty)$.
