$\lim \frac{\sqrt{x}-1}{x-1}=\frac{1}{2}$ as $x \rightarrow 1$ Epsilon-Delta proof help.

Question

I'm trying to sort out how to define epsilon-delta proofs and this one is a tricky one. Any ideas?

Answer 1

$$\left|\frac{\sqrt x-1}{x-1}-\frac12\right|=\left|\frac1{\sqrt x+1}-\frac12\right|=\frac{|1-\sqrt x|}{2(\sqrt x+1)}=\frac1{2(\sqrt x+1)^2}|x-1|\;(*)$$

Now, decide arbitrarily that $\;\delta <1\;$ , then we get

$$|x-1|<\delta<1\implies 0<x<2\implies 0<\sqrt x<\sqrt 2\implies 1<(\sqrt x+1)^2<(\sqrt2+1)^2<9$$

so continuining at $\;(*)\;$ :

$$(*)\;\;\;\frac1{2(\sqrt x+1)^2}|x-1|\le\frac{|x-1|}2\;\left(\stackrel{\text{we want}}<\right)\epsilon $$

so if we choose now $\;\delta = \min\,\{1,2\epsilon\}\;$ we are done.