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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?

  • 0
    I don't see any fallacy.2012-09-14
  • 0
    Spivak's Calculus1994 page 1002012-09-14

3 Answers 3