# Find a limit in order to find an asymptote

I need to draw the graph of the function $$y=\sqrt[3]{(x^2)(x+9)}$$. Suppose $$y=k\cdot x + b$$ is the equation for the asymptote.

I managed to find $$k=\lim_{x\to\infty}\frac{y(x)}{x}=\lim_{x\to\infty}\sqrt[3]{\frac{x+9}{x}}=1$$

But then I got stuck finding $$b=\lim_{x\to\infty}y(x)-k\cdot x$$

Any ideas on how to deal with it?

You can write $$y(x)-x = \frac{y(x)^3-x^3}{y(x)^2+y(x)x+x^2}$$. Let us work with this expression \begin{align*} y(x)-x &= \frac{x^2(x+9) - x^3}{(x^2(x+9))^{2/3} + x\sqrt[3]{x^2(x+9)} + x^2} \\ &= \frac{9x^2}{(x^2(x+9))^{2/3} + x\sqrt[3]{x^2(x+9)} + x^2} \\ &= \frac{9}{\frac{1}{x^2}(x^2(x+9))^{2/3} + \frac{1}{x}\sqrt[3]{x^2(x+9)} + 1} \\ &= \frac{9}{(\frac{x+9}{x})^{3/2} + \sqrt[3]{\frac{x+9}{x}} + 1 } \end{align*}

Taking a limit yields $$\frac{9}{1+1+1}=3$$.

You can write it like this:

$$b=\lim_{x \to \infty}\sqrt[3]{x^2(x+9)}-x= \frac{x^2(x+9)-x^3}{(\sqrt[3]{x^2(x+9)})^2+\sqrt[3]{x^2(x+9)}\sqrt[3]{x}+\sqrt[3]{x^2}}= \frac{9}{3}=3$$

I have used:

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$

So the task is to find the following limit: $$b=\lim_{x\to\infty}\left(y(x)-kx\right)=\lim_{x\to\infty}\left(\sqrt[3]{x^3+9x^2}-x\right).$$ To rationalize a cube root, we can take advantage of the formula $$A^3-B^3=(A-B)(A^2+AB+B^2).$$ In this example we'll apply it to $$A=\sqrt[3]{x^3+9x^2}$$ and $$B=x$$. So let's multiply and divide out expression by $$(A^2+AB+B^2)$$: $$\lim_{x\to\infty}\left(\sqrt[3]{x^3+9x^2}-x\right)=\lim_{x\to\infty}\frac{\left(\sqrt[3]{x^3+9x^2}-x\right)\left(\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2\right)}{\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2}=\lim_{x\to\infty}\frac{(x^3+9x^2)-x^3}{\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2}=\lim_{x\to\infty}\frac{9x^2}{\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2}=\frac{9}{3}=3,$$ where the last step follows by simultaneously dividing the numerator and denominator by $$x^2$$ (or, equivalently, by comparing the highest degree terms and their coefficients, so to speak).

Rather than rationalizing the denominator, you could use binomial series:

$$b=\lim_{x\to\infty}y(x)-k\cdot x=\lim_{x\to\infty}\left[\sqrt[3]{x^3+9x^2}-x\right]=\lim_{x\to\infty}\left[x\left(1+\frac9x\right)^{1/3}-x\right]$$

$$=\lim_{x\to\infty}\left[x\left(1+\frac13\frac9x+o\left(\frac1{x}\right)\right)-x\right]=3.$$