# Limits: factoring out $x$ from $\lim_\limits{x\to +\infty}\left(\frac{5-x^3}{8x+2}\right)$

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?

• What you do there is completely correct. Your teacher was wrong. Commented Mar 26, 2019 at 4:28
• As far as "not rigorous" mathematics concerned, your calculation is valid. Can your teacher point out precisely at which part you are not allowed arithmetic operations? If he or she cannot, then you are definitely not wrong. Commented Mar 26, 2019 at 4:35

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.
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 $$\lim_{x\rightarrow\infty}x^2=+\infty.$$