A step in finding the slant asymptote I'm curious why this textbook shows an extra step when finding the slant asymptote in this function: $$f(x) = \frac {x^3}{x^2 + 1}.$$
So long division gives
$$f(x) = \frac{x^3}{x^2+1} = x - \frac{x}{x^2+1},$$
so
$$f(x) - x = - \frac{x}{x^2 + 1} = -\frac{\frac{1}{x}}{1 + \frac{1}{x^2}} \rightarrow 0$$ as $x \rightarrow \pm \infty$.
Why did they show that last step? Isn't it enough to have the $- \dfrac{x}{x^2 + 1}$?

Because $-\frac{x}{x^2+1}$ is in the indeterminate form $\frac{\infty}{\infty}$ and thus the additional step is to factor out an $x^2$ term before to take the limit in order to have a form $\frac{0}{1}$.
