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

Question

Christopher Allers · Accepted Answer

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#

Emily Ammerman · Answer

undefined To find the limit of a rational function as x approaches infinity, we compare the degrees of the numerator and denominator. In this case, both the numerator and denominator have a degree of 3. To determine the limit, we divide each term in the numerator and denominator by the highest power of x, which is x^3. This simplifies the expression to (1 - 2/x + 3/x^2 - 4/x^3) / (4 - 3/x + 2/x^2 - 1/x^3). As x approaches infinity, the terms with 1/x^2 and 1/x^3 become negligible, leaving us with the limit of (1/1) / (4/1), which simplifies to 1/4. Therefore, the limit of the given rational function as x approaches infinity is 1/4.