Finding linear asymptote We have 
$$ f(x) = (2x^2-x^3)^{1/3} $$ 
A linear asymptote is like: $$y = px+q$$  $$p \not= 0$$
We had this definition in our lecture:$$ p = \lim\limits_{x \rightarrow \infty}{\frac{f(x)}{x}}
$$
so
$$ p = \lim\limits_{x \rightarrow \infty}{\frac{(2x^2-x^3)^\frac{1}{3}}{x}} $$ 
 $$  =\lim\limits_{x \rightarrow \infty}{(\frac{2x^2}{x^3} - \frac{x^3}{x^3} )^\frac{1}{3}} $$
$$ = (-1)^\frac{1}{3}$$
So, $$p = (-1)^\frac{1}{3}$$
And 
$$q =  \lim\limits_{x \rightarrow \infty}{f(x)-px} $$
This would be $$  \lim\limits_{x \rightarrow \infty}{(2x^2-x^3)^\frac{1}{3}} -  (-1)^\frac{1}{3}x$$
In Wolframalpha this lim has a complex solution. Does this function have a linear asymptote ? Can a real number function (from $$R -> R$$ ) can have a complex asymptote? I guess not, because then it wouldn't be linear anymore.
 A: Notice that
$$q=\lim_{x\to\infty}(2x^2-x^3)^{1/3}+x=\frac23$$
I'll leave you to show this with the tip that $a^{1/3}+b^{1/3}=\frac{a+b}{a^{2/3}-(ab)^{1/3}+b^{2/3}}$ because this is not actually the main part of my answer.  Instead, I want to provide you with a new method for such an expansion known as the binomial expansion.  First, we want the largest powers to be at the front, so,
$$f(x)=-(x^3-2x^2)^{1/3}$$
Now we apply the binomial expansion theorem:
$$(a-b)^{1/3}=a^{1/3}-\frac13ba^{-2/3}-\frac29b^2a^{-5/3}+\dots$$
Let $a=x^3$ and $b=2x^2$,
$$f(x)=-\left(x-\frac23-\frac89x^{-1}+\dots\right)=-x+\frac23+\frac89x^{-1}+\dots$$
which not only returns the linear asymptote but also better asymptotes as $x\to\infty$.
A: I would use the generalized
binomial theorem:
$\begin{array}\\
f(x) 
&= (2x^2-x^3)^{1/3}\\
&= (-x)(1-\frac{2}{x})^{1/3}\\
&= (-x)(1-\frac{2}{x}\frac13+(-\frac{2}{x})^2\frac13 (-\frac23)+O(\frac1{x^3}))\\
&= (-x)(1-\frac{2}{3x}-\frac{8}{9x^2}+O(\frac1{x^3}))\\
&= -x+\frac{2}{3}+\frac{8}{9x}+O(\frac1{x^2}))\\
\end{array}
$
