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July 30, 2014

July 30, 2014

Posted by **Blue** on Saturday, October 8, 2011 at 7:38pm.

lim [x^2 + x + c]/[x^2 - 5x + 6] exists.

x->3

For that value of c, determine the limit. (Hint: Find the value of c for which x - 3 is a factor of the numerator.)

- Calculus -
**Damon**, Saturday, October 8, 2011 at 7:55pmfactor the bottom

(x-3)(x-2)

factor the top with (x-3) a factor so we can cancel it

(x-3)(x+b) = x^2+x+c

x^2 - 3x +b x - 3 b = x^2 + x + c

so

b x -3 x = x so b = 4

then c = -12

so

(x^2 + x - 12)/(x^2 - 5 x + 6)

(x-3)(x+4) / [(x-3)(x-2)]

(x+4)/(x-2)

when x = 3

7/1 = 7

- Calculus -
**Steve**, Saturday, October 8, 2011 at 7:58pmhmmm. looks familiar. Anyway,

denominator is (x-3)(x-2), so we want the top to be

(x-3)(x-k) so that the fraction exists everywhere except at x=3, but has a finite limit there.

x^2 - (k+3)x + 3k = x^2 + x + c

k+3 = -1

k = -4

so**c = 3k = -12**

x^2 + x + c = x^2 + x -12 = (x-3)(x+4)

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