<s>[INST] calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$ So I multiplied by the conjugate and got $$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$
and this is where I got stuck.
 [/INST] Here's a completely elementary solution.
For $x > 1$
$\begin{array}\\
d(x)
&=x + \sqrt[3]{1-x^3}\\
&=x - \sqrt[3]{x^3-1}\\
&=x(1 - \sqrt[3]{1-1/x^3})\\
&\gt 0\\
\end{array}
$
Also,
since
$(1-z)^3 < 1-z$
for $0 < z < 1$,
$\sqrt[3]{1-z} > 1-z$
so that
$\sqrt[3]{1-1/x^3}
\gt 1-\frac1{x^3}
$
for $x > 1$.
Therefore
$\begin{array}\\
d(x)
&=x(1 - \sqrt[3]{1-1/x^3})\\
&<x(1 - (1-\frac1{x^3}))\\
&= x(\frac1{x^3})\\
&=\frac1{x^2}\\
\end{array}
$
Therefore
$0
\lt d(x)
\lt \frac1{x^2}
$
so
$\lim_{x \to \infty} d(x)
=0
$.
</s>