\(\text{Is there a number $a$ such that the limit}\\ lim_{x\to 3}\frac{2x^2-3ax+x-a-1}{x^2-2x-3}\,\text{exists?}\)
+9466 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:
\(\lim\limits_{x\to3}\frac{2x^2-6x+x-3}{x^2-2x-3}\ =\ \lim\limits_{x\to3}\frac{2x(x-3)+1(x-3)}{(x-3)(x+1)}\ =\ \lim\limits_{x\to3}\frac{2x+1}{x+1}\ =\frac74\)
It's obvious that, no matter what the value of "a" is, when x=3 the denominator becomes zero.
+9466 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:
\(\lim\limits_{x\to3}\frac{2x^2-6x+x-3}{x^2-2x-3}\ =\ \lim\limits_{x\to3}\frac{2x(x-3)+1(x-3)}{(x-3)(x+1)}\ =\ \lim\limits_{x\to3}\frac{2x+1}{x+1}\ =\frac74\)
