- #1
Zargawee
[SOLVED] Small Limits Question
Hi There,
I have this Simple Question In Limits :
If limx->3 (f(x) - 2) / (x - 3) = 7
Then limx->3 (x2f(x) - 18) / (x-3) = ??
I solved the question this way :
Since the denominator equals Zero , and limit exists , then the numerator equals zero .
[4]f(x) - 2 = 0 ---> f(x) = 2
limx->3 (x2f(x) - 18) / (x-3) =
limx->3 (2x2 - 18) / (x-3) =
limx->3 2(x2- 9) / (x-3) =
limx->3 2(x-3) (x+3) / (x-3) =
limx->3 2(x+3) =
2 (3+3) = 12
But I also solved in this way :
limx->3 (x2f(x) - 18) / (x-3) =
limx->3 (x2f(x) - 2x2 + 2x2 - 18) / (x-3)
limx->3 (x2f(x) + (2x2) /(x-3) - 2x2 - 18) / (x-3)
limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
32 * 7 + limx->3 (x+3)
(9 * 7) + (3 + 3) = 63 + 12 = 75
What's the wrong with the first One ?
Please help.
Hi There,
I have this Simple Question In Limits :
If limx->3 (f(x) - 2) / (x - 3) = 7
Then limx->3 (x2f(x) - 18) / (x-3) = ??
I solved the question this way :
Since the denominator equals Zero , and limit exists , then the numerator equals zero .
[4]f(x) - 2 = 0 ---> f(x) = 2
limx->3 (x2f(x) - 18) / (x-3) =
limx->3 (2x2 - 18) / (x-3) =
limx->3 2(x2- 9) / (x-3) =
limx->3 2(x-3) (x+3) / (x-3) =
limx->3 2(x+3) =
2 (3+3) = 12
But I also solved in this way :
limx->3 (x2f(x) - 18) / (x-3) =
limx->3 (x2f(x) - 2x2 + 2x2 - 18) / (x-3)
limx->3 (x2f(x) + (2x2) /(x-3) - 2x2 - 18) / (x-3)
limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
32 * 7 + limx->3 (x+3)
(9 * 7) + (3 + 3) = 63 + 12 = 75
What's the wrong with the first One ?
Please help.