<s>[INST] 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?
 [/INST] 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).
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