# How do you find the slant asymptote of (x^2)/(x-1)?

Dec 21, 2015

Re-express as the sum of a polynomial and a term whose limit as $x \to \infty$ is $0$:

${x}^{2} / \left(x - 1\right) = x + 1 + \frac{1}{x - 1}$

so the oblique asymptote is $y = x + 1$

#### Explanation:

${x}^{2} / \left(x - 1\right) = \frac{{x}^{2} - x + x}{x - 1} = \frac{x \left(x - 1\right) + x}{x - 1} = x + \frac{x}{x - 1}$

$= x + \frac{x - 1 + 1}{x - 1} = x + 1 + \frac{1}{x - 1}$

Note that:

$\frac{1}{x - 1} \to 0$ as $x \to \infty$,

So the oblique asymptote is $y = x + 1$