Prove $\lim_{x\to 0} \frac{\sqrt{1-x}}{\sqrt{1+x}} = 1$ using $\epsilon-\delta$ definition

Question

Another user (Paramanand Singh) suggested that: $$ \left|\frac{\sqrt{1-x}}{\sqrt{1+x}}- 1\right| = \frac{2|x|}{1+x+\sqrt{1-x^2}}. $$

Starting from there, let $\delta = \frac{1}{2}$, so that $$|x|<\delta \Rightarrow \frac{1}{2}<x+1<\frac{3}{2}\Rightarrow \frac{2}{3}<\frac{1}{x+1}<2. $$

Since $1+x>1+x+\sqrt{1-x^2}$, we have:

$$ \frac{2|x|}{1+x+\sqrt{1-x^2}} < \frac{2|x|}{1+x}<4|x|<4\delta. $$

Therefore, we need to pick $\delta = \min\left(\dfrac12, \dfrac{\epsilon}{4}\right).$

So, I have two questions:

1. How can I get $\frac{2|x|}{1+x+\sqrt{1-x^2}}$ from $\left|\frac{\sqrt{1-x}}{\sqrt{1+x}}- 1\right|$?

2. Is the $\delta$ I found correct?

Answer 1

1. You just need to multiply the numerator and denominator by $\sqrt{1-x}+ \sqrt{1+x}$.

2. Your $\delta$ seems okay to me.