Prove $f_n(x)=\sqrt{x+\frac{1}{n}}-\sqrt{x}$ converges uniformly on $\mathbb{R}$. From Spivak's Calculus 24-2:


  Prove $f_n(x)=\sqrt{x+\frac{1}{n}}-\sqrt{x}$  converges uniformly on
  $\mathbb{R}$.


My issue is whether this makes sense since square root of negative numbers are not yet defined.  In the book complex numbers have not been covered.  In any case can anyone offer up a solution?
Correct me if wrong:

Let 

$f_n (x):= \sqrt{x+1/n} -\sqrt{x}$, $x\ge 0.$

$|\sqrt{x+1/n}-\sqrt{x}| =$

$(1/n)|\dfrac{1}{\sqrt{x+1/n} +\sqrt{x}}| \le$

$(1/n)\sqrt{n} = \dfrac{1}{\sqrt{n}}.$

Let $\epsilon \gt 0$ be given.

Choose $n_0 \gt \dfrac{1}{\epsilon^2}.$

For $n \ge n_0 :$

$|f_n(x)| \lt \dfrac{1}{\sqrt{n}} \le \dfrac{1}{\sqrt{n_0}} \lt \epsilon,$ I.e.

$f_n(x)$ converges uniformly to $0$  on $\mathbb{R_+}.$
If you use
$(1+x/2)^2
=1+x+x^2/4
\gt 1+x
$,
you get
$\sqrt{1+x}
\lt 1+x/2
$.

Therefore

$\begin{array}\\
\sqrt{x+1/n}-\sqrt{x}
&amp;=\sqrt{x}\sqrt{1+1/(nx)}-\sqrt{x}\\
&amp;\lt\sqrt{x}(1+1/(2nx))-\sqrt{x}\\
&amp;\lt\sqrt{x}(1+1/(2nx))-\sqrt{x}\\
&amp;=\sqrt{x}+1/(2n\sqrt{x})-\sqrt{x}\\
&amp;\dfrac1{2n\sqrt{x}}\\
\end{array}
$

and this should gove you what you need.