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

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}

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.
• 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? – pdb Feb 9 '19 at 10:55
• @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. – JavaMan Feb 13 '19 at 4:27