Asymptotes of a rational function

Statement

Find the asymptotes of $f(x) = \dfrac{x^2+1}{x-1}$.

Solution

Domain: $\mathbb{R}\setminus\{1\}$.

Vertical asymptote. At $x=1$ the denominator vanishes and the numerator equals $2\ne 0$: $\lim_{x\to 1^\pm}\frac{x^2+1}{x-1} = \pm\infty \ \Rightarrow\ \text{vertical asymptote } \boxed{x=1}.$

Horizontal asymptote. There is none, because $\lim_{x\to\pm\infty}f(x)=\pm\infty$ (the degree of the numerator is greater than that of the denominator).

Oblique asymptote $y=mx+q$. By polynomial division $x^2+1 = (x-1)(x+1) + 2$, hence $f(x) = x+1+\frac{2}{x-1}.$ From this $m=\lim_{x\to\pm\infty}\dfrac{f(x)}{x}=1$ and $q=\lim_{x\to\pm\infty}\bigl(f(x)-x\bigr)=1$. The oblique asymptote is $\boxed{y=x+1}$.

Links

Topics: Limits
Concepts: Asymptote · Oblique asymptote · Vertical asymptote
Skills: Calculating limits · Studying functions
Exercise type: Function study