Q13/117) Find a such that limit of ((3x^2 + ax + a + 3) / (x^2 + x -2)) as x approaches -2.

(x^2+x-2) is zero at x = -2 so  of ((3x^2 + ax + a + 3) need to be zero else the ratio shall be infinite /zero form and it shall not exist

So f(x) = (3x^2 + ax + a + 3) should be zero at x = -2

f(x) = 12 -2a + a + 3 or a = 15

now ((3x^2 + ax + a + 3) = 3x^2 + 15x + 18 = 3(x+3)(x+2)

so ((3x^2 + ax + a + 3) / (x^2 + x -2)) = 3(x+3)(x+2)/((x-1)(x+2))

= 3(x+3)/(x-1) = 3 (1)/(-3) = - 1