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 $l$.


You want to compute


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


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



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

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.

  • $\begingroup$ To find b,we use the following formula $b = lim_{n\to\infty}(f(x)-kx)$, so in your case you didn't solve it by this formula, or you did but didn't show me how to. Could you please solve it again? I found that k = -1 but I don't know how to find b. If there is another way than Taylor, please show it to me. $\endgroup$ – Vala Ahmeti Apr 21 '17 at 13:25
  • $\begingroup$ @Vala Ahmeti, Just calculate $b=\lim\limits_{x\to\infty}(\sqrt[3]{x^2(2-x)}+x)=\lim\limits_{x\to\infty}\dfrac{(\sqrt[3]{\hspace{.2cm}})^3+x^3}{(\sqrt[3]{\hspace{.2cm}})^2-x(\sqrt[3]{\hspace{.2cm}})+x^2}=\ldots$ $\endgroup$ – Minz Apr 22 '17 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.