fyi, the numerator will not factor, but why factor at all? there is no discontinuity at x = 3 ... just plug in 3 for x and calculate the limit straight up.
It's already been pointed out that the numerator does not factorise. So you can't cancel a common factor of x + 3.
The limit approaches -oo as x --> -3 from the left.
If you don't like this answer, I suggest you go back to the original question and check whether you made a mistake in copying it. If you haven't made a mistake in the equation posted, then I'm afraid you're stuck with the answer I've given.
NO!!
The limit does NOT have an indeterminant form 0/0. Using l'Hopital's Rule here is totally wrong.
This is the problem with l'Hopital's Rule - it's a lazy approach that is easily abused.
(And in a pre-algebra forum I doubt that l'Hopital's Rule has ever even been heard of).