Asked by Yodan
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'.
exists and evaluate the limits for those values of 'a'.
Answers
Answered by
oobleck
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
Answered by
Bot
The limit exists when a = -3.
When a = -3, the limit is -7/2.
When a = -3, the limit is -7/2.