Spivak's Calculus (Chapter 5, Problem 41): Proof that $\lim_{x \to a} x^2 = a^2$ 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?
 A: 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!
A: It is wrong to call the proof "fallacious". The comments in the solution book, 

"41. How do we know that $\sqrt{a^2 - \epsilon}$ and $\sqrt{a^2 + \epsilon}$ exist?  In Chapter 7 we prove (Theorem 8) that every positive number has a square root, but the proof of this theorem uses the fact that $f(x)=x^2$ is continuous, which is essentially what we are trying to prove.  In fact the existence of square roots is essentially equivalent to the continuity of $f$ --- compare Problem 8-8"

contain several errors.


*

*The proof and calculations are correct; the stated value of $\delta$ exists and suffices for the $\epsilon$ argument.

*The supposed problem with assuming the existence of the square roots does not derive from the proof, but from the sequence chosen for the material in the book.  The quantities with square roots do exist as real numbers, and using those values for $\delta$ makes the proof work.  

*The existence of square roots does not logically depend on proving continuity of $f(x)=x^2$.  Other books give the existence statement as an exercise on least upper bounds (of the positive rational solutions of $x^2 < a$) or as convergence of the Babylonian square root iteration.

*In epsilon-delta proofs, including the one Spivak wrote down for this exercise, equations like $\delta = \min(A,B,C,\dots, K)$ do not necessarily assume that the bounds $A,B,C...$ all exist.  It is only required that all inequalities $\delta \leq A$ and  $\delta \leq B$ can be satisfied at once.  Some of the upper bounds might not exist for all values of $a$ and $\epsilon$ (in this problem, take $\epsilon > a^2$ and one of the square roots becomes imaginary).  

*Completing the proof without taking existence of roots as an assumption, does not require an existence proof for the root.  All we need is solutions to conditions like $\delta \leq \sqrt{a^2+\epsilon} - a$, which can be done by squaring,

$(\delta + a)^2 < a^2 + \epsilon$  

an inequality that is easy to achieve with pure algebraic epsilon-delta calculations.  No properties of real numbers such as convergence or completeness are necessary.
A: This may just be tom - foolery that I'm pointing out, but if $\epsilon > a^2$ then I'm not too sure what $\delta$ you would pick...
