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.
Best Answer
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.