The limit exists when a = -3.
When a = -3, the limit is -7/2.
Find the value of constant 'a' so that lim(x-> -2) ((x^3)+(ax^2)+3x-1)/((x^2)-4)
exists and evaluate the limits for those values of 'a'.
2 answers
AAAaannndd the bot gets it wrong yet again!
x^3 + 3x^2 + 3x - 1 = (x-1)^3 - 2
at x = -2, (x^3 + 3x^2 + 3x - 1)/(x^2-4) does not exist.
You want the numerator to be a multiple of (x+2) so that at x=-2 you have a quotient of 0/0 which is a removable hole.
That is, (x^3 + ax^2 + 3x - 1) = (x-k)(x-m)(x+2)
So, x^3 + (2-k-m)x^2 + (km-2k-2m)x + 2km = x^3 + ax^2 + 3x - 1
Looks like 2km = -1, so
2-k-m = a
-1/2 -2k-2m = 3
so a = 15/4
x^3 + 15/4 x^2 + 3x - 1 = (x+2)^2 (x - 1/4)/(x^2-4)
lim(x → -2) (x^3 + 15/4 x^2 + 3x - 1)/(x^2-4) = 0
since for x ≠ -2, (x+2)(x - 1/4) = 0
x^3 + 3x^2 + 3x - 1 = (x-1)^3 - 2
at x = -2, (x^3 + 3x^2 + 3x - 1)/(x^2-4) does not exist.
You want the numerator to be a multiple of (x+2) so that at x=-2 you have a quotient of 0/0 which is a removable hole.
That is, (x^3 + ax^2 + 3x - 1) = (x-k)(x-m)(x+2)
So, x^3 + (2-k-m)x^2 + (km-2k-2m)x + 2km = x^3 + ax^2 + 3x - 1
Looks like 2km = -1, so
2-k-m = a
-1/2 -2k-2m = 3
so a = 15/4
x^3 + 15/4 x^2 + 3x - 1 = (x+2)^2 (x - 1/4)/(x^2-4)
lim(x → -2) (x^3 + 15/4 x^2 + 3x - 1)/(x^2-4) = 0
since for x ≠ -2, (x+2)(x - 1/4) = 0
1 answer