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?
2 Replies  5168 Views  Last Post: 23 September 2012  05:46 PM
#1
Calc help! Limit of piecewise function w/ unknown constant
Posted 22 September 2012  11:23 PM
Replies To: Calc help! Limit of piecewise function w/ unknown constant
#2
Re: Calc help! Limit of piecewise function w/ unknown constant
Posted 23 September 2012  02:04 PM
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
#3
Re: Calc help! Limit of piecewise function w/ unknown constant
Posted 23 September 2012  05:46 PM
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?
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?
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