I need to find the value of $a,b \in \mathbb R$ such that the given limit is true I am given that $\lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx=1$ need to find the value of $a,b \in \mathbb R$ such  that the given limit is true. I was able to work the whole thing out, but I have a question about one step in my work. There is a lot of rough work because I simplify by using the rule of the difference of cubes, so here is a condensed part of my work:
$$\begin{align} \lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx &=\lim_{x \to \infty} \frac{8x^3+ax^2-b^3x^3}{(\sqrt[3]{8x^3+ax^2})^2+bx\sqrt[3]{8x^3+ax^2}+b^2x^2} \\&= \lim_{x \to \infty} \frac{8+a\frac{1}{x}-b^3}{\frac{1}{x^3}(\sqrt[3]{8x^3+ax^2})^2+b\frac{1}{x^2}\sqrt[3]{8x^3+ax^2}+b^2\frac{1}{x}} \\&=\frac{\lim_{x \to \infty}8+\lim_{x \to \infty}a\frac{1}{x}-\lim_{x \to \infty}b^3}{\lim_{x \to \infty}\frac{1}{x^3}(\sqrt[3]{8x^3+ax^2})^2+\lim_{x \to \infty}b\frac{1}{x^2}\sqrt[3]{8x^3+ax^2}+\lim_{x \to \infty}b^2\frac{1}{x}} \\&= \frac{8-b^3}{0+0+0} \\&= \frac{8-b^3}{0}\end{align}$$ Thus $8-n^3$ must also equal $0$ which implies that $b=2$. (This is the part I am unsure about. Is what I said true? If $b=2$ then this would give me an indeterminate form, but other than that I'm not sure if what I said holds, and if it does hold why does it hold?) Regardless of my uncertainty, I went on and using this assumption I found that $a=12$ in a similar manner, and when I check $\lim_{x \to \infty} \sqrt[3]{8x^3+12x^2}-2x$ it does equal $1$ .
Any help as to why/why not my assumption is correct? 
Thanks in advance!
(If anyone wants me to post the method as to how i got 12 for $a$, let me know and then I'll type it up).

Your assumption is correct; if $b\neq 2$, then $8-b^3\neq 0$, and hence the limit would either not exist or be infinite.  But you know the limit is $1$.
A shorter way to do the first part is: $\sqrt[3]{8x^3+ax^2}-bx=x(\sqrt[3]{8+\frac{a}{x}}-b)$.  The cube root approaches 2 as $x\rightarrow \infty$, so if $b\neq 2$, the product approaches $\pm\infty$ (not $1$ as in the hypothesis).
