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

1 Answer
Jul 5, 2016

lim_{x to -1 ^ -} f(x) =- oo

lim_{x to -1 ^ +} f(x) =+ oo

the line y = 2x 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 < abs h "<<" 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 -oo

if h > 0, ie the right-sided limit, then the limit is +oo

so

lim_{x to -1 ^ -} f(x) =- oo

lim_{x to -1 ^ +} f(x) =+ oo

for horixontal and oblique asymptotes we look at x to pm oo

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 1/x term diminishes in the denominator

so the line y = 2x is an oblique asymptote