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

Question

Ezekiel Alexander · Accepted Answer

Factor and simplify the ratio.

Because #x=-1# makes the numerator #0#, we can be sure that #x-(-1) = x+1# is a factor of the numerator. Use division or trial and error to get #x^4+3x^3-x^2+x+4 = (x+1)(x^3+2x^2-3x+4)#. So, #lim_(xrarr-1)(x^4+3x^3-x^2+x+4 )/(x+1) = lim_(xrarr-1)((x+1)(x^3+2x^2-3x+4) )/(x+1) # # = lim_(xrarr-1)(x^3+2x^2-3x+4) = -1+2+3+4 = 8#

Samantha Adkison · Answer

Use L'Hôpital's rule
#lim_(xto-1)(x^3 + 3x^3 - x^2 + x + 4)/(x + 1) = 8# Because the expression becomes an indeterminate form, specifically #0/0# when evaluated at the limit, L'Hôpital's rule applies. Please understand, though the implied division can be performed without a remainder, this is not a valid way to determine the limit. The derivative of the denominator is 1, therefore, we do not need to write it into the resulting expression. The derivative of the numerator is: #4x^3 + 9x^2 - 2x + 1# #lim_(xto-1) 4x^3 + 9x^2 - 2x + 1 = -4 + 9 + 2 + 1= 8# L'Hôpital's rule stipulates that the limit of the original expression is the same. #lim_(xto-1)(x^3 + 3x^3 - x^2 + x + 4)/(x + 1) = 8#

Isaiah Allen · Answer

undefined To evaluate the limit of (x^4+3x^3-x^2+x+4)/(x+1) as x approaches -1, we can substitute -1 into the expression and simplify. Doing so, we get (-1)^4+3(-1)^3-(-1)^2+(-1)+4 / (-1+1), which simplifies to 1+(-3)-1+(-1)+4 / 0. Simplifying further, we have 0 / 0. This is an indeterminate form, so we can apply L'Hôpital's rule. Taking the derivative of the numerator and denominator separately, we get 4x^3+9x^2-2x+1 / 1. Substituting -1 into the derivative expression, we have 4(-1)^3+9(-1)^2-2(-1)+1 / 1, which simplifies to -4+9+2+1 / 1. Further simplifying, we get 8 / 1, which equals 8. Therefore, the limit of the given expression as x approaches -1 is 8.