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.