# How do you find the limit of (x^3-2x^2+3x-4)/(4x^3-3x^2+2x-1) as x approaches infinity?

Apr 21, 2015

When youre taking the lim_(x->oo) (x^3−2x^2+3x−4)/(4x^3−3x^2+2x−1)

notice that the highest degree of x is the same in both the numerator and the denominator.
When this specific occasion is true of your f(x) (i.e. $\frac{{x}^{3.} . . e t c .}{4 {x}^{3.} . . e t c .}$) divide both the numerator and denomitator by the highest degree of x .

For our problem this is ${x}^{3}$

lim_(x->oo) ((x^3)/(x^3)−(2x^2)/(x^3)+(3x)/(x^3)−(4)/(x^3))/((4x^3)/(x^3)−(3x^2)/(x^3)+(2x)/(x^3)−(1)/(x^3))

simplifying we have

lim_(x->oo) (1−(2)/(x)+(3)/(x^2)−(4)/(x^3))/(4−(3)/(x)+(2)/(x^2)−(1)/(x^3))

Knowing that the ${\lim}_{x \to \infty} \frac{1}{x} = 0$ all the terms other than the $\frac{1}{4}$ cancel to 0

Therefore lim_(x->oo) (x^3−2x^2+3x−4)/(4x^3−3x^2+2x−1)=1/4