Spivak's Calculus (Chapter 5, Problem 41): Proof that $\lim_{x \to a} x^2 = a^2$

Question

In Chapter 5, Problem 41, Spivak provides an alternative way to prove that

$$\lim_{x \rightarrow a} x^2 = a^2\,\,,\,\,a > 0$$

Given $\,\epsilon > 0\,$ let

$$\delta = \min\left\{\sqrt{a^2 + \epsilon} - a, a - \sqrt{a^2 - \epsilon}\right\}$$

Then

$$|x - a| < \delta\Longrightarrow \sqrt{a^2 - \epsilon} < x < \sqrt{a^2 + \epsilon}\Longrightarrow a^2 - \epsilon < x^2 < a^2 + \epsilon\,\,,\, |x^2 - a^2| < \epsilon$$

Then he goes on to claim that this proof is fallacious. But wherein lies the fallacy?

Answer 1

In Spivak's book, this limit fact (later stated as: function $x^2$ is continuous) is proved quite early. Before the existence of square-roots is known. Indeed, continuity of the function $x^2$ will later be used to prove existence of square-roots. So an argument with square-roots here would be circular reasoning!