general topology - limit points of subset of real numbers

Let $$A=\{ \frac{\sqrt{m} -\sqrt{n}}{\sqrt{m}+\sqrt{n}} | m,n\in \Bbb{N} \}$$ I think that we must find sequence of $A$ and find limit of sequence,let $a_m =\frac{\sqrt{ k ^2 m^2} -\sqrt{ m^2}}{\sqrt{k ^ 2 m^2}+\sqrt{ m^2}}=\frac{(k-1)m}{(k+1)m}$ that $k\in\Bbb{N}$ the limit of $a_m$ is $\frac{k-1}{k+1}$ and let $B=\{\frac{k-1}{k+1} |k\in\Bbb{N}\}$ then $B\subseteq A^\prime$,($A^\prime$ is set of limit points of$ A$),the answer is interval $[-1,1]$,

Answer

Some ideas:

First: for all $\;m,n\in\Bbb N\;$ :

$$-1=\frac{-\sqrt n}{\sqrt n}\le\frac{-\sqrt n}{\sqrt m+\sqrt n}\le\frac{\sqrt m-\sqrt n}{\sqrt m+\sqrt n}\le\frac{\sqrt m}{\sqrt m}=1$$

so any limit point of $\;A\;$ indeed has to be in $\;[-1,1]\;$ .

Now, if $\;\alpha\in[-1,1]\;$ take a peek at

$$\frac{\sqrt m-\sqrt n}{\sqrt m+\sqrt n}-\alpha=\frac{\sqrt m(1-\alpha)-\sqrt n(1+\alpha)}{\sqrt m+\sqrt n}\le\frac{\sqrt m}{\sqrt n}(1-\alpha)$$

In order to make the last part above less than some predetermined $\;\epsilon >0\;$ it is then enough to take

$$\sqrt\frac nm>\frac{1-\alpha}\epsilon$$