# How do you find the asymptotes for f(x)=(2x²)/(x+1)?

Jul 5, 2016

${\lim}_{x \to - {1}^{-}} f \left(x\right) = - \infty$

${\lim}_{x \to - {1}^{+}} f \left(x\right) = + \infty$

the line $y = 2 x$ is an oblique asymptote

#### Explanation:

f(x)=(2x²)/(x+1)

for vertical asymptotes we look at where the demoninator is 0. here that means x = -1

f(-1)=(2(-1)²)/(-1+1) = 2/oo

if we explore $x = - 1 + h$ with $0 < \left\mid h \right\mid \text{<<} 1$ we have

f(h-1)=(2(h-1)²)/(h)

the numerator is always positive so

if $h < 0$, ie the left-sided limit, then the limit is $- \infty$

if $h > 0$, ie the right-sided limit, then the limit is $+ \infty$

so

${\lim}_{x \to - {1}^{-}} f \left(x\right) = - \infty$

${\lim}_{x \to - {1}^{+}} f \left(x\right) = + \infty$

for horixontal and oblique asymptotes we look at $x \to \pm \infty$

here, lim_{x to pm oo }(2x²)/(x+1) = pm infty as the quadratic power dominates the expression

but note also that

lim_{x to pm oo }(2x²)/(x+1) = lim_{x to pm oo }(2x)/(1+1/x) approx lim_{x to pm oo }(2x)/(1)  as the $\frac{1}{x}$ term diminishes in the denominator

so the line $y = 2 x$ is an oblique asymptote