Finding a $\delta$ for the limit of $f(x) = \sqrt{x}$ near $a = 1$ is $1$

Question

This is an exercise on Michael Spivak's Calculus. I decided $\delta =\epsilon$ would work as $|\sqrt{x}-1|\leqslant|x-1|$ for all $x\geqslant0$ and since we need to show that for all $\epsilon>0$ there is some $\delta>0$ that if $0<|x-1|<\delta$ then $|\sqrt{x}-1|<\epsilon$. So choosing $\delta =\epsilon$ gets us $|\sqrt{x}-1|<|x-1|<\epsilon$. However the solution given in the book is $\delta = 2\epsilon - \epsilon^2$. Is my reasoning flawed or the solution provided is overkill?

Answer 1

Your approach is correct. Perhaps that you could justify the inequality $\bigl|\sqrt x-1\bigr|\leqslant|x-1|$, but that's all.