I'm puzzled by the answer to a problem for Spivak's Calculus (4E) provided in his Combined Answer Book.

Problem 5-3(iv) (p. 108) asks the reader to prove that $\mathop{\lim}\limits_{x \to a} x^{4} =a^{4}$ (for arbitrary $a$) by using some techniques in the text to find a $\delta$ such that $\lvert x^{4} - a^{4} \rvert<\varepsilon$ for all $x$ satisfying $0<\lvert x-a\rvert<\delta$.

The answer book begins (p. 67) by using one of these techniques (p. 93) to show that $$\lvert x^{4} - a^{4} \rvert = \lvert (x^{2})^{2} - (a^{2})^{2} \rvert<\varepsilon$$ for $$\lvert x^{2} - a^{2} \rvert <\min \left({\frac{\varepsilon}{2\lvert a^{2}\rvert+1},1}\right) = \delta_{2} .$$

In my answer, I use the same approach to show that $$\lvert x^{2} - a^{2} \rvert <\delta_{2}$$ for $$\lvert x - a \rvert <\min \left({\frac{\delta_{2}}{2\lvert a\rvert+1},1}\right) = \delta_{1} ,$$ so that $$\lvert x^{4} - a^{4} \rvert<\varepsilon$$ when $$\delta = \delta_{1}=\min \left({\frac{\delta_{2}}{2\lvert a\rvert+1},1}\right). \Box$$

But Spivak's answer book has $$\delta =\min \left({\frac{\delta_{1}}{2\lvert a\rvert+1},1}\right),$$ which I believe is an error.

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    $\begingroup$ If it is incorrect, perhaps you can find a particular value of $a$ and a particular value of $\epsilon$ where his formula fails. Can you? $\endgroup$
    – GEdgar
    Sep 18, 2011 at 0:31
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    $\begingroup$ @GEdgar: That may be worth determining, but the question at hand is, first, should it be obvious that $\delta_{1}$ was intended, and if so, what step am I missing, since applying the techniques of the chapter, as well as all the steps explicitly worked out in the answer key, leads to $\delta_{2}$ where it ends up with $\delta_{1}$. $\endgroup$
    – orome
    Sep 18, 2011 at 3:14
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    $\begingroup$ Are you sure the first $\delta$ that Spivak introduces is $\delta_2$? Seems a bit strange to me to name the first $\delta$ "$\delta_2$"... $\endgroup$ Sep 18, 2011 at 4:54
  • 3
    $\begingroup$ Since $\delta_2<\varepsilon$, $\delta_1<\delta_2$ and Spivak's $\delta$ (based on $\delta_1$) is smaller than yours (based on $\delta_2$). Thus, if your $\delta$ is correct, Spivak's $\delta$ is correct as well. So much for counterexamples. $\endgroup$
    – Did
    Sep 18, 2011 at 9:30
  • $\begingroup$ @Arturo: Yes, they are introduced on the opposite order on the key. $\endgroup$
    – orome
    Sep 18, 2011 at 14:12

1 Answer 1


Where you (correctly) iterated the bound twice it seems that Spivak iterated three times. This particular $\delta$ is shrinking at each iteration, because it satisfies $\delta(\epsilon,a) < \epsilon$ for all $a$. Given that two iterations are enough, three are more than needed, but still logically correct.

Without seeing the answer book, it is impossible to determine whether Spivak's extra layer of work is consistent with the methods he gives for this and other problems.


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