Wednesday

December 17, 2014

December 17, 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

**Answer this Question**

**Related Questions**

Calculus - Find a value of the constant k such that the limit exists. x^2-k^2 ...

Calculus - Find a value of the constant k such that the limit exists. lim x->...

calculus - (a) Find the number c such that the limit below exists. Limit as x ...

Calculus - Find the positive integers k for which lim ->0 sin(sin(x))/x^k ...

Math - (a) Find the number c such that the limit below exists. Limit as x goes ...

Calculus - Find the constant c so that lim [x^2 + x + c]/[x^2 - 5x + 6] exists. ...

Calculus 3 - Find the limit, if it exists, or show that the limit does not ...

Calculus 3 - Find the limit, if it exists, or show that the limit does not exist...

Check my CALCULUS work, please! :) - Question 1. lim h->0(sqrt 49+h-7)/h = ...

Calculus - Evaluate each limit. If it exists. a) Lim x^3 + 1 x->1 ------- x...