Proving continuity of $f(x)=\sqrt{1−x^3}$ on $[0,1]$ I have to prove the continuity of the function with $\epsilon, \delta$. For this I have to prove that the function is continuous at every $x \in [0, 1]$ by proving:
$$\forall 0 < \epsilon, \exists 0 < \delta, \forall a \in [0, 1]: |x - a| < \delta \implies \left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right| < \epsilon.$$
This is where I am right now:
\begin{align}
\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|&= \frac{\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{x^3 - a^3}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{(x - a)(x^2 + ax + a^2)}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3 + 1 - a^3}\right|}
\end{align}
Any hint is welcome.

If $a\not=1$, then you can continue with
$${3\delta\over\sqrt{1-x^3}+\sqrt{1-a^3}}\lt{3\delta\over\sqrt{1-a^3}}$$
so it suffices to let $\delta=\epsilon\sqrt{1-a^3}/3$ to make the epsilon-delta proof work for $a\in[0,1)$. But for $a=1$ you need a separate argument to show that there is a $\delta$ for which $|x-1|\lt\delta\implies|\sqrt{1-x^3}|\lt\epsilon$.  I'll let you think this over first; please leave a comment if you'd like more help.
