1. ## Removable discontinuity..

OK, f(x)=(3x^2 + 13x + 12)/(x^2 + x - 6)

This is what I got from that:

factored form f(x) = (3x + 4)/(x-2)

y-intercept: y = -2

x-intercept(s): x = -4/3

horizontal asymptote: y = 3

vertical asymptote: x = 2

removable discontinuity at x = 2

Now i have to find the limit of f(x) as x approaches the removable discontinuity. If I use 2, I get undefined.

Am I doing something wrong??

Am I doing something wrong??
No you are not. What happens as x tends to 2? Does f approach infinity?

The removable discontinuity... What is f when x is 3?

3. ohh, so it'd be infinity.. my brain checked out today or something.

ohh, so it'd be infinity.. my brain checked out today or something.
No, im sorry, that is the value given to the nonremovable discontinuity.

You want to know what f would be at the removable discontinuity.

5. all i can come up with would be undefined...

all i can come up with would be undefined...
Hang on a sec. This will take a sec, but I'll expain it better. Gimmie a minute.

all i can come up with would be undefined...
OK. As it stands now, this rational function is discontinuous in two places.

1. at x=2
2. at x=-3

$\frac{3x^2+13x+12}{x^2+x-6}$.

But if we factor...

$\frac{(3x+4)(x+3)}{(x-2)(x+3)}$

Then cancel

$\frac{3x+4}{x-2}$

I removed one of the discontinuities.

So, it is now possible to know $\lim_{x\to{-3}}\frac{3x^2+13x+12}{x^2+x-6}$ because it is equal to the $\lim_{x\to{-3}}\frac{3x+4}{x-2}$
So, now by direct substitution
$\lim_{x\to{-3}}\frac{3x+4}{x-2}=\frac{3(-3)+4}{(-3)-2}=\frac{-5}{-5}=1$

Does this make sense?

8. OK yes I think I understand! So instead of my answer being x = 2 for the removable discontinuity, it would be x = -3, because I am able to cancel it?