Uniform convergence of $n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ I need to say whether or not $f_n(x)=n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ is uniformly convergent on $(0,\infty)$.

I've found that the function is locally convergent to $f(x)=\frac{1}{2\sqrt{x}}$ and was trying to find $\sup{|f_n(x)-f(x)|}$.

I got the derivative $f_n'(x)= \frac{2nx\left(x-\sqrt x\sqrt{x+\frac{1}{n}}\right)+\sqrt{x}\sqrt{x+\frac{1}{n}}}{...}$ and could not find $x$ so that $f_n'(x)=0$

Any ideas?
Note that each of the functions $\displaystyle f_n(x) = n\biggl(\sqrt{x+\frac{1}{n}}-\sqrt{x}\biggr)$ is bounded on $(0,\infty)$, with $f_n(x) \leq \sqrt{n}$.  Since $\displaystyle f(x) = \frac{1}{2\sqrt{x}}$ is unbounded on $(0,\infty)$, the sequence $\{f_n\}_{n=1}^\infty$ does not converge uniformly to $f$.
As already pointed out by several others, the convergence is not uniform on $(0,\infty)$ but it may be worth noting that it is uniform on every subinterval $[c,\infty)$ for $c &gt; 0$.

This follows from the following computation:

$$
\begin{split}
|f_n(x) - f  (x)| &amp;= \left|\frac{1}{\sqrt{x+\dfrac1n}+\sqrt{x}} - \frac1{2\sqrt{x}}\right| \\
&amp;= \left|\frac{\sqrt{x}-\sqrt{x+\dfrac1n}}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)} \right|\\
&amp;= \left|\frac{\dfrac1n}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)^2} \right|
\le \frac1n \cdot \frac{1}{8c\sqrt{c}}.
\end{split}
$$