Prove that $\sqrt[3]{x}$ is continuous. There are many solutions for this available; this question is to verify and critique my proof, which uses an approach different and, IMO, simpler, than the solutions I've seen.
Proof: By definition, $\sqrt[3]x$ is continuous if $\lim_{h \to 0} \sqrt[3]{x+h} = \sqrt[3]x$. At $x = 0,\lim \sqrt[3]{x+h} = \sqrt[3]x$ by the $\epsilon$-$\delta$ definition of limit, with $\delta < \epsilon^3.$ To show this for $x \neq 0$, we first show that $\lim_{b \to 0} \sqrt[3]{1+b} = 1.$
Assume $b > 0$. Then $1 < \sqrt[3]{1 + b} < 1 +b$, since $1^3 < 1 + b < (1 + b)^3$. Similarly, if $-1 < b < 0$, then $1 > \sqrt[3]{1 + b} > 1 +b.$ By the Squeeze Theorem, $\lim_{b \to 0} \sqrt[3]{1+b} = 1.$
Write $b = \frac h x$, and $\sqrt[3]{x+h} = \sqrt[3] x \sqrt[3] {1 + b} \to \sqrt[3] x.$ QED.
Questions: Is this proof correct, rigorous, and well written? How can it be improved?
Update: Drawing on Thomas Andrews' very helpful critique, I revise the proof that $\lim_{b \to 0} \sqrt[3]{1+b} = 1$ as follows:
Restrict $|b| < 1$. Assume $b > 0$. Then $b^2 < b$, so $(1 - b)^3 = (1-b)(1 + b^2 - 2b) < 1 - b < 1 + b < (1 +b)^3.$ Likewise, if $b < 0$, we have by symmetry $(1-b)^3 > 1 + b > (1 + b)^3$. By the Squeeze Theorem, $\lim_{b \to 0} \sqrt[3]{1+b} = 1.$
Or better yet:
Observe that for all $a > 0, a^3 < a$ if and only if $a < 1$. Restrict $|b| < 1$. Assume $b > 0$. Then $(1 -b)^3 < 1 - b < 1 + b < (1 + b)^3.$ And, if $b < 0$, by symmetry we have a similar inequality: $(1 +b)^3 < 1 + b < 1 - b < (1 - b)^3.$ By the Squeeze Theorem, $\lim_{b \to 0} \sqrt[3]{1+b} = 1.$
Does this solve the deficiencies of the original proof?