How do you find vertical, horizontal and oblique asymptotes for f(x)=(3x^2 + 2x - 3 )/( x - 1)?

May 4, 2016

This function $f$ has:
a vertical asymptote $x = 1$,
no horizontal asymptotes,
and an oblique asymptote $y = 3 x + 5$.

Explanation:

I'm going to use these formulas:

1. Function $f$ has a vertical asymptote in $c$ (equation: $x = c$) iff $c \notin {D}_{f}$ ($c$ is not in the domain of the function) and ${\lim}_{x \to {c}^{\pm}} f \left(x\right) = \pm \infty$.

2. Function $f$ has a horizontal asymptote in $\pm \infty$ (equation: $y = c$) iff ${\lim}_{x \to \pm \infty} f \left(x\right) = c < \infty$ (this limit must be a finite number).

3. Function $f$ has an oblique asymptote in $\pm \infty$ (equation: $y = a x + b$) iff both of these limits are finite:
$a = {\lim}_{x \to \pm \infty} f \frac{x}{x}$
$b = {\lim}_{x \to \pm \infty} \left(f \left(x\right) - a x\right)$

Important note: A horizontal asymptote is a special case of an qblique one (where $a = 0$).

For $f \left(x\right) = \frac{3 {x}^{2} + 2 x - 3}{x - 1}$ we have:
${D}_{f} = m a t h \boldsymbol{R} \setminus \setminus \left\{1\right\}$

1.
${\lim}_{x \to {1}^{+}} f \left(x\right) = \left[\frac{2}{{0}^{+}}\right] = + \infty$
${\lim}_{x \to {1}^{-}} f \left(x\right) = \left[\frac{2}{{0}^{-}}\right] = - \infty$
which gives us a vertical asymptote $x = 1$.

2.
${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{3 x + 2 - \frac{3}{x}}{1 - \frac{1}{x}} = + \infty$
${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{3 x + 2 - \frac{3}{x}}{1 - \frac{1}{x}} = - \infty$
which gives us no horizontal asymptotes.

3.
${\lim}_{x \to \pm \infty} f \frac{x}{x} = {\lim}_{x \to \pm \infty} \frac{3 {x}^{2} + 2 x - 3}{{x}^{2} - x} =$
$= {\lim}_{x \to \pm \infty} \frac{3 + \frac{2}{x} - \frac{3}{x} ^ 2}{1 - \frac{1}{x}} = 3 = a$
${\lim}_{x \to \pm \infty} \left(f \left(x\right) - a x\right) = {\lim}_{x \to \pm \infty} \left(\frac{3 {x}^{2} + 2 x - 3}{x - 1} - 3 x\right) =$
$= {\lim}_{x \to \pm \infty} \frac{3 {x}^{2} + 2 x - 3 - 3 {x}^{2} + 3 x}{x - 1} =$
$= {\lim}_{x \to \pm \infty} \frac{5 x - 3}{x - 1} = {\lim}_{x \to \pm \infty} \frac{5 - \frac{3}{x}}{1 - \frac{1}{x}} = 5 = b$
which gives us an oblique asymptote $y = 3 x + 5$.