# Oblique asymptotes by long division

Due to my horrid ability with long division I have an extremely hard time finding the oblique asymptote for any function.

In the problem given I am required to find the oblique asymptote for$$y = \frac{x^3}{x^2-x}$$

I have found that the asymptote is $x+1$ from looking online. But I am required to show how I came to the answer, I have not been able to do this as I do not understand long division.

Note that $y=\frac{x^2-1+1}{x-1}=x+1+\frac{1}{x-1}$. So

$$\lim_{x\to\infty}(f(x)-(x+1))=\lim_{x\to\infty}\frac{1}{x-1}=0$$

$y=x+1$ is an oblique asymptote.

Here you can use a workaround. Notice that $$\begin{split} \frac{x^3}{x^2-x} &= \frac{x^3 - x^2}{x^2-x} + \frac{x^2}{x^2-x} \\ &= \frac{x^3 - x^2}{x^2-x} + \frac{x^2-x}{x^2-x} + \frac{x}{x^2-x}\\ &= \frac{x^2(x - 1)}{x(x-1)} + 1 + \frac{1}{x-1}\\ &= x + 1 + \frac{1}{x-1} \\ \end{split}$$

• What is the format that you used for determining the answer, as I think I understand how you got the answer. But I would like to make sure that I understand the formatting before applying it to other equations.
– Miki
Commented May 31, 2017 at 15:31
• @Miki i don't understand -- are you asking how did i know whcih simplifications to make? If so, you just fit the numerator to be $x^n$ times the denominator until it fits for $n \ge 0$ Commented Jun 2, 2017 at 5:25
• It took me a while to understand this so hopefully this helps anyone else googling: Split the equation up into parts, e.x. (x^2-3x+2)/(x+2) = (x^2)/(x+2) - (3x)/(x-2) + 2/(x+2) and then make the top part of the equation be a multiple of the bottom by adding. Ex. (x^2-4)/(x+2) - (3x+6)/(x+2) + (2+4-6)/(x+2) and then simplify. There's the asymptote (x-5, ignore the remainder). Thanks!
– njha
Commented Oct 12, 2017 at 22:31