The problem. Prove $\lim \sqrt{x} = \sqrt{a}$ as $x \to a$ given $a > 0$. The problem suggests using the identity $|\sqrt{x} - \sqrt{a}| = |x - a|/(\sqrt{x} + \sqrt{a})$.
This problem is tricky for me because I must be careful with numbers $a$ that are close to zero, after all I cannot take the square root of a negative number. So I tried: given $a > 0$, let's restrict $a/2 < |x - a| < (a + 1)/2$. This restriction works, but this is the first time that I find myself in need of a non-symmetric restriction. I suppose there must be a way to specify a non-symmetric delta. How is that usually written? I suppose I can describe it in words. Let $d > 0$ be such that $a/2 < |x - a| < (a + 1)/2$ or let $d$ be $e(\sqrt{4a/3} - \sqrt{a})$, whichever $d$elta is smaller.
And then I can finish the proof saying that the restriction implies $$\frac{1}{\sqrt{x} - \sqrt{a}} < \frac{1}{\sqrt{4a/3} - \sqrt{a}}$$
and then using the problem's suggested identity we end up with $|\sqrt{x} - \sqrt{a}| < e$.
Reference. This is problem 37 in section 2.4 of James Stewart's Calculus, 6th edition. The book suggests (with a red icon on the number of the problem) that there's a visual help for the problem, but this help would be in some website --- I guess --- and I could not find any help in Stewart's website. I suppose this edition is too old and there's no online help anymore?
I think there is a typo in the expression of the proposed suggestion.
Indeed, if let $|x - a| < \delta_{\varepsilon}$, it yields that: \begin{align*} |\sqrt{x} - \sqrt{a}| & = \left|\frac{x - a}{\sqrt{x} + \sqrt{a}}\right| = \frac{|x - a|}{\sqrt{x} + \sqrt{a}} \leq \frac{|x - a|}{\sqrt{a}} < \frac{\delta_{\varepsilon}}{\sqrt{a}} < \varepsilon \end{align*}
Hopefully this helps!