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?
 A: 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 $x \in [1,\infty)$.
A: Let $g_n(x)$ be the function $f_n(x)-f(x)$.  
Then the first derivative if $g_n(x)$ with respect to $x$ is 
$$g_n'(x) =f'_n(x)-f'(x)=n\left(\frac{x}{\sqrt{x^2+1/n}}-1\right)+\frac{1}{2x^2}\tag 1$$
while the second derivative is given by
$$g_n''(x)=(x^2+1/n)^{-3/2}-x^{-3}$$
Obviously $g_n''<0$.  Thus, $g_n$ attains a local maximum when $g_n'=0$.   
To find the local maximum, we set the right-hand side of $(1)$ equal to zero and solve for $x$.  We find that the local maximum of $g_n$ occurs at $x=\sqrt{8n}$.  However, since $n\ge 1$ and $x\ge 1$, $g_n$ does not attain a local maximum.  
Therefore, the absolute maximum of $g_n(x)$ occurs at the boundary $x=1$.  It is easy to see that $g_n(1)<0$.  Thus, $|g_n(x)|$ has its absolute minimum at $x=1$.  
We can now write
$$\begin{align}
\left|f_n(x)-f(x)\right|&\le \left| n\left(\sqrt{1+\frac1 n}-1\right)-\frac12  \right|\\\\
&=\left| n\left(1+\frac{1}{2n}-\frac{1}{8n^2}+O(1/n^3) -1\right)-\frac12  \right|\\\\
&\le\frac{1}{8n}
\end{align}$$
Thus, given $\epsilon >0$, $|f_n(x)-f(x)|<\epsilon$ whenever $n\ge N=\frac{1}{8\epsilon}$.  And we are done!
