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?

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

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

  • $\begingroup$ Thanks. How did I know I needed to multiply by $\sqrt{1-x}+ \sqrt{1+x}$ to get that nice form, though? That comes from experience or there's some pattern I should recognize? $\endgroup$ – pdb Feb 9 '19 at 10:55
  • $\begingroup$ It is just something that works. You will get a better sense of what you need to do as time goes by $\endgroup$ – Locally unskillful Feb 9 '19 at 10:56
  • 1
    $\begingroup$ @pbd: Notice that $\left|\frac{\sqrt{1-x}}{\sqrt{1+x}}- 1\right| = \left|\frac{\sqrt{1-x} - \sqrt{1+x}}{\sqrt{1+x}}\right|$. Then, $\sqrt{1-x} + \sqrt{1+x}$ is the conjugate of the numerator. $\endgroup$ – JavaMan Feb 13 '19 at 4:27

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