Finding linear asymptote

Question

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.

Answer 1

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 $x\to\infty$.

Answer 2

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} $