So my teacher said that I cannot use arithmetic operation to factor out $x$ from this type of equation, saying that it's because it's composed only by addition and subtraction. But I don't understand clearly, because I get the right answer (according to the book):
$$\lim_{x\to+\infty}\left(\frac{5-x^3}{8x+2}\right) =\lim_{x\to\infty}\frac{x\times\left(\frac{5}{x}-x^2\right)}{x\times\left(8+\frac{2}{x}\right)} =\lim_{x\to\infty}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}} =\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}-x^2\right)}{\lim_\limits{x\to\infty}\left(8+\frac{2}{x}\right)} =\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}\right)-\lim_\limits{x\to\infty}\left(x^2\right)}{\lim_\limits{x\to\infty}\left(8\right)+\lim_\limits{x\to\infty}\left(\frac{2}{x}\right)} =\frac{0-\infty}{8+0} =\frac{-\infty}{8}$$
Applying the infinity property: $\frac{-\infty}{-c}=\infty$
$=-\infty$
Can someone explain to me why I can't factor $x$ out?
The writing $$ \lim_{x\to\infty}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}} =\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}-x^2\right)}{\lim_\limits{x\to\infty}\left(8+\frac{2}{x}\right)} $$ is not so good because $\lim\limits_{x\rightarrow\infty}\left(\frac{5}{x}-x^2\right)$ is not number.
We can say the same words about your next steps.
I think it's better to write the following. $$\lim_{x\rightarrow\infty}\frac{5-x^3}{8x+2}=\lim_{x\rightarrow\infty}\left(x^2\cdot\frac{\frac{5}{x^2}-x}{8x+2}\right)=-\infty$$ because $$\lim_{x\rightarrow\infty}\frac{\frac{5}{x^2}-x}{8x+2}=\lim_{x\rightarrow\infty}\frac{\frac{5}{x^3}-1}{8+\frac{2}{x}}=-\frac{1}{8}$$ and $$\lim_{x\rightarrow\infty}x^2=+\infty.$$