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

Question

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?

Answer 1

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

and this should gove you what you need.

Answer 2

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 $\mathbb{R_+}.$