Q: Where is the fallacy in this "proof" of $\lim_{x \to a} x^2 = a^2$? I just started reading Michael Spivak's Calculus and I am at Chapter 5 - Limits now. I tried doing the problems, and at the very end is an interesting and hard problem (problem 41) which I really can't see the solution:
(This problem is in Spivak's personal view.)

41. After sending the manuscript for the first edition of this book off to the printer, I thought of a much simpler way to prove that $\lim_{x \to a} x^2 = a^2$ and $\lim_{x \to a} x^3 = a^3$ without going through all the factoring tricks on page 95. Suppose, for example, that we want to prove that $\lim_{x \to a} x^2 = a^2$, where $a > 0.$ Given $\epsilon > 0$, we simply let $\delta$ be the minimum of $\sqrt{a^2 + \epsilon}  -a$ and $a - \sqrt{a^2 - \epsilon}$ (see Figure 19); then $|x - a| < \delta$ implies that $\sqrt{a^2 - \epsilon} < x < \sqrt{a^2 + \epsilon}$, so $a^2 - \epsilon < x^2 < a^2 + \epsilon$, or $|x^2 - a^2| < \epsilon$. It is fortunate these pages had already been set, so I couldn't make these changes, because this "proof" is completely fallacious. Wherein lies the fallacy?

I don't know where the fallacy is, but either way, the "proof" that he gave was very promising and for me, it's hard to find the error.

So what is the fallacy in Spivak's "proof"?


A: Having taught out of that book something like 15 times, let me give the answer. We do not yet know — rigorously — that $\sqrt x$ exists. That will be proved as a consequence of the Three Hard Theorems, several chapters later.

A: What is $\sqrt{x}$? Typically one defines it by appealing to the intermediate value theorem, which requires continuity of $x\mapsto x^2.$
