I am working my way through Spivak. I am not incredibly confident in my solution to the problem, and the solutions in the back of the book gives a subtlety that I don't have in my own proof. I'm wondering why.
Claim: $\lim_{x \to 0} f(x) = \lim_{x \to 0} f(x^3).$
My Proof: First, suppose $\lim_{x \to 0} f(x) = L$. Then, by definition, $\forall \epsilon, \exists \delta$ such that $\forall x, \lvert{x - 0}\rvert < \delta \implies \lvert{f(x) - L}\rvert < \epsilon$. Since we may choose any $\delta \in \mathbb{R}^+$, consider $\lvert{x}\rvert < \sqrt[3]\delta$. Then $\lvert{x^3}\rvert < \delta \implies \lvert{f(x^3) - L}\rvert < \epsilon$. Thus, $\lim_{x\to 0} f(x^3) = L$.
Conversely, suppose $\lim_{x \to 0} f(x^3) = L$. Then, again by definition, $\forall \epsilon, \exists \delta$ such that $\forall x, \lvert{x - 0}\rvert < \delta \implies \lvert{f(x^3) - L}\rvert < \epsilon$. If $\lvert{x}\rvert < \delta^3$, then $\lvert{\sqrt[3]{x}}\rvert < \delta \implies \lvert{f(\sqrt[3]{x}^3}) - L\rvert < \epsilon$. Thus, $\lim_{x\to 0} f(x) = L$.
$\blacksquare$
Since this is the chapter on limits, the proof is written very pedantically. However, the crux of the argument for each case is:
- Since we may choose any $\delta \in \mathbb{R}^+$, consider $\lvert{x}\rvert < \sqrt[3]\delta$.
- Since $\lvert{x}\rvert < \delta$, we can simply choose $\delta^3$ such that $\lvert{x}\rvert < \delta^3 \implies \lvert{\sqrt[3]{x}}\rvert < \delta$.
My Question
Spivak's Solutions has the following as justification for the first case:
Then if $0 < \lvert{x}\rvert < \min{(1, \delta)}$, we have $0 < \lvert{x^3}\rvert < \delta$.
I don't understand why Spivak shows that $\delta < \min{(1, \delta)}$, or why knowing this we can deduce $0 < \lvert{x^3}\rvert < \delta$.
$$|x|<\min (1,\delta) \iff |x|<1 \text{ and } |x|<\delta$$
If $\delta<1,$ then $|x|^3<\delta^3<\delta<1.$
If $1<\delta,$ then $|x|^3<1<\delta<\delta^3.$
Otherwise, if $|x|>1$, then $|x|<\delta$ does not imply $|x|^3<\delta.$