[JennaPeterson88]

I am asked to find all values of the constant k such that the limit as x approaches -2 of the function f(x) exists, given that f(x) is a piecewise function:
For x < -2, f(x) = (k^2)x
For x = -2, f(x) = k - 6
For x > -2, f(x) = 4k - x


I know that in order for a limit as x approaches a number to exist, the limit as x approaches the number from the left and right must also exist, and "all three" limits must be equal.

Lim x -> -2(-) = Lim x -> -2(+) = Lim x -> -2


From the piecewise definition of f(x) above,
Lim x -> -2(-) = -2k^2
Lim x -> -2(+) = 4k + 2
Lim x -> -2 = k - 6


Therefore -2k^2 = 4k + 2 = k - 6



-2k^2 = 4k + 2 -> 2k^2 + 4k + 2 = 0 -> k = -1
-2k^2 = k - 6 -> 2k^2 + k - 6 = 0 -> k = 3/2 or -2
4k + 2 = k - 6 -> 3k - 8 = 0 -> k = -8/3
2k^2 + 4k + 2 = k - 6 -> 2k^2 + 3k + 8 = 0 -> k has no solution
2k^2 + k - 6 = 4k + 2 -> 2k^2 - 3k - 8 = 0 -> k = -1.39 or 2.89 (roughly)


Either there is no value k for which the limit exists, or I'm completely missing a concept.
Which is it?

[blackcompe]

Why don't you ask this in a Math forum? We know a lot of Math around here, but usually the Math is related to Computer Science. This sounds like a Calc I homework problem.

This post has been edited by blackcompe: 23 September 2012 - 02:04 PM

[macosxnerd101]

Pick two functions, set them equal to each other and solve for k:
-2k^2 = 4k + 2
-2k^2 - 4k - 2 = 0
k^2 + 2k + 1 = 0

What are the roots for k? Does plugging k back into all three pieces satisfy the limit?