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. #(x^3...etc.)/(4x^3...etc.)#) 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->oo)1/x=0# all the terms other than the #1/4# cancel to 0
Therefore #lim_(x->oo) (x^3−2x^2+3x−4)/(4x^3−3x^2+2x−1)=1/4#
