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November 27, 2014

November 27, 2014

Posted by **Ashley** on Monday, October 3, 2011 at 7:17pm.

- Calculus -
**Steve**, Tuesday, October 4, 2011 at 12:11amYou have a fraction with denominator=0 at x=2

So, you want the numerator to also be zero, or you have no hope of having a limit.

6x^2 + ax + a + 57 = 0

24 + 2a + a + 57 = 0

81 + 3a = 0

a = -27

so,

6x^2 - 27x + 30 / x^2 + x - 6

=(6x-15)(x-2) / (x+3)(x-2)

So, the limit = (6x-15)/(x+3) = -3/5

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