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?
 A: By the definition of uniform convergence, if they did uniformly converge to a limit function $f$ there would be some $N$ such that if $n \geq N$ then $|f_n(x) - f(x)| < 1$ for all $x$. In particular this would hold for $n = N$ itself. Since $f_N(x)$ is a bounded function, this means so is $f(x)$. So we can let $M$ be such that $|f(x)| < M$ for all $x$. Thus by the above,  for all $n > N$ and all $x \in (0,\infty)$ we have
$$|f_n(x)| \leq |f_n(x) - f(x)| + |f(x)|$$
$$ < M + 1$$
But $f_n({1 \over n}) = (\sqrt{2} - 1)\sqrt{n}$. For $n$ large enough this will be greater than $M + 1$, a contradiction. So the functions don't converge uniformly.
A: 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 $f$.
A: In fact, $f_n'(x)<0$ for all $x>0$.  Thus, each $f_n$ is continuous, positive, and decreasing on $[0,\infty)$.  It follows that 
$$\sup\{f(x):x>0\} = f(0) = \sqrt{n}.$$
As $f$ is unbounded, you can't have uniform convergence.
A: 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 $c > 0$.
This follows from the following computation:
$$
\begin{split}
|f_n(x) - f  (x)| &= \left|\frac{1}{\sqrt{x+\dfrac1n}+\sqrt{x}} - \frac1{2\sqrt{x}}\right| \\
&= \left|\frac{\sqrt{x}-\sqrt{x+\dfrac1n}}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)} \right|\\
&= \left|\frac{\dfrac1n}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)^2} \right|
\le \frac1n \cdot \frac{1}{8c\sqrt{c}}.
\end{split}
$$
