I have the following function $f(x)=(x-2)^{1/3}(x+4)^{2/3}$. I'm asked to find all asymptotes of this function.
Clearly, there are no vertical asymptotes since there are no points of discontinuity. There are also no horizontal asymptotes since $\displaystyle{\lim_{x\to \infty}f(x)=\infty}$ and $\displaystyle{\lim_{x\to -\infty}f(x)=-\infty}$
So that means I just need to check for the slant asymptotes.
I know that $\displaystyle m=\lim_{x \to \infty}\frac{f(x)}{x}$ and $\displaystyle b=\lim_{x \to \infty}{(f(x)-mx)}$
$\displaystyle m=\lim_{x \to \infty}\frac{f(x)}{x}$
$\displaystyle m=\lim_{x \to \infty}\frac{(x-2)^{1/3}(x+4)^{2/3}}{x}$
$\displaystyle m=\lim_{x \to \infty}\frac{x(1-\frac{2}{x})^{1/3}(1+\frac{4}{x})^{2/3}}{x}$
$\displaystyle m=\lim_{x \to \infty}\left(1-\frac{2}{x}\right)^{1/3}\left(1+\frac{4}{x}\right)^{2/3}=1$
The problem is $b$.
$\displaystyle b=\lim_{x \to \infty}{(f(x)-mx)}$
$\displaystyle b=\lim_{x \to \infty}{((x-2)^{1/3}(x+4)^{2/3}-x)}$
The answer SHOULD be $2$ have no idea how to proceed however. I tried to manipulate this into L'hopital's rule but I don't think that's possible here… Can someone help out? Thanks!
Best Answer
Use Newton's generalized binomial theorem: \begin{align} (x-2)^{1/3}(x+4)^{2/3}&=x^{1/3}\left(1-\frac{2}{x}\right)^{1/3} x^{2/3}\left(1+\frac{4}{x}\right)^{2/3} \\ &=x\left[1-\frac{2}{3x}+O\left(\frac{1}{x^2}\right)\right] \left[1+\frac{8}{3x}+O\left(\frac{1}{x^2}\right)\right] \\ &=x\left[1+\frac{2}{x}+O\left(\frac{1}{x^2}\right)\right] \\ &=x+2+O\left(\frac{1}{x}\right). \tag{1} \end{align} Therefore, $$ b=\lim_{x \to \infty}\left((x-2)^{1/3}(x+4)^{2/3}-x\right)=2. \tag{2} $$