+9490 We want to find a such that the numerator is also zero when x = 3
That is, we want to find a such that...
2(3)2 - 3a(3) + 3 - a - 1 = 0
18 - 9a + 3 - a - 1 = 0
20 - 10a = 0
20 = 10a
2 = a
So yes, when a = 2, the limit exists.
Check:
limx→32x2−6x+x−3x2−2x−3 = limx→32x(x−3)+1(x−3)(x−3)(x+1) = limx→32x+1x+1 =74
It's obvious that, no matter what the value of "a" is, when x=3 the denominator becomes zero.
+9490 We want to find a such that the numerator is also zero when x = 3
That is, we want to find a such that...
2(3)2 - 3a(3) + 3 - a - 1 = 0
18 - 9a + 3 - a - 1 = 0
20 - 10a = 0
20 = 10a
2 = a
So yes, when a = 2, the limit exists.
Check:
limx→32x2−6x+x−3x2−2x−3 = limx→32x(x−3)+1(x−3)(x−3)(x+1) = limx→32x+1x+1 =74
