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.
 A: 
So I multiplied by the conjugate and got

What conjugate expression was that exactly...?
You want to get rid of the cube root by using:
$$a+b=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}=\frac{a^3+b^3}{a^2-ab+b^2}$$
with, in your case, $a=x$ and $b=\sqrt[3]{1-x^3}$.
A: HINT
By binomial expansion we have
$$\sqrt[3]{1-x^3}=-x \left(1-\frac1{x^3}\right)^{\frac13} =-x+\frac1{3x^2}+o\left(\frac1{x^2}\right)$$
A: The simplest strategy is to use $x=1/t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{\sqrt[3]{t^3-1}+1}{t}
$$
which coincides with the derivative at $0$ of $f(t)=\sqrt[3]{t^3-1}$; since
$$
f'(t)=\frac{3t^2}{3\sqrt[3]{(t^3-1)^2}}
$$
your limit is $f'(0)=0$.
A: 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
$.
