Can anybody help me finding asymptotes of a function?

Question

The function given is $f(x) = \sqrt[3]{{x}^2(2-x)}$.

Can anybody help me to find all asymptotes of this function. I know it doesn't have a vertical asymptote and I know that it's horizontal asymptote is $\sqrt[3]{-1}$, but I don't know how to find asymptote of the slope.

I'd prefer if someone could help me solving it using the formula given below: $y = kx + l$ where $k = lim_{n\to\infty} \dfrac{f(x)}{x}$ and $l=lim_{n\to\infty}[f(x)-kx]$. I found $k$ that is $k=-1$ but I don't know how to find $l$.

Answer 1

You want to compute

$$k=\lim_{x\to\infty}\left(\sqrt[3]{{x}^2(2-x)}+x\right).$$

To get rid of the cubic root, you can multiply by the conjugate trinomial and get

$$k=\lim_{x\to\infty}\left(\frac{{x}^2(2-x)+x^3}{\sqrt[3]{{x}^2(2-x)}^2-\sqrt[3]{{x}^2(2-x)}x+x^2}\right).$$

The numerator simplifies to $2x^2$ and by factoring out $x^2$ at the denominator, the expression tends to

$$\frac2{(-1)^2-(-1)+1}.$$

Answer 2

$\sqrt[3]{x^2(2-x)}=-x\sqrt[3]{1-\frac2x}=\boxed{\text{via Taylor}}=-x(1-\frac2{3x}+o(\frac1x))=-x+\frac23+o(1)$

As $\sqrt[3]{x^2(2-x)}- (-x+\frac23)$ tends to $0$ the asymptotes at $-\infty$ and $+\infty$ are $y=-x+\frac23$ according to definition.

Alternative approach is standart: firstly search $k=\lim\limits_{x\to\pm\infty}\frac {f(x)}{x}$ and secondly define $b=\lim\limits_{x\to\pm\infty}(f(x)-kx)$. If both calculations would successful then the asymptotes would be $y=kx+b.$ They may be different at $-\infty$ and $+\infty$ or exist only at one of the infinities.