Question

Is there a number "a" such that the equation below exists? If so what is the value of "a" and its limit.

`lim_(x->-2)(3x^2+ax+a+3)/(x^2+x-2)`

Answer 1

`a = 15`

`lim_(x->-2) (3x^2+15x+18)/(x^2+x-2) = -1`

Explanation:

`x^2+x-2 = (x+2)(x-1)`

So the denominator contains exactly one factor `(x+2)`

So in order that `(3x^2+ax+a+3)/(x^2+x-2)` has a limit as `x->-2`, the only requirement is that:

`3x^2+ax+(a+3)" "` is divisible by `(x+2)`

Let `f(x) = 3x^2+ax+(a+3)`

This is divisible by `(x+2)` if and only if `f(-2) = 0`

Substituting `x=-2` we have:

`f(-2) = 3(color(blue)(-2))^2+a(color(blue)(-2))+a+3`

`color(white)(f(-2)) = 12-2a+a+3`

`color(white)(f(-2)) = 15 - a`

So we require `a=15`

With this value of `a`:

`f(x) = 3x^2+15x+18 = 3(x^2+5x+6) = 3(x+2)(x+3)`

`(3x^2+15x+18)/(x^2+x-2) = (3(color(red)(cancel(color(black)(x+2))))(x+3))/((color(red)(cancel(color(black)(x+2))))(x-1)) = (3(x+3))/(x-1)`

So:

`lim_(x->-2) (3x^2+15x+18)/(x^2+x-2) = lim_(x->-2) (3(x+3))/(x-1) = (3(color(blue)(-2)+3))/(color(blue)(-2)-1) = 3/(-3) = -1`