For a function to be continuous at an x-value of -3 you need to make sure two things line up:
First, we need to show that the limit from the left equals the limit from the right.
[tex]\lim_{x \to -3^{-}} f(x) = \lim_{x \to -3^{+}} f(x)[/tex]
Second, we need to show that this limit equals the functions value.
[tex]\lim_{x \to -3} f(x) = f(-3)[/tex]
The left hand limit involves the first piece, f(x) = x^2 - 9:
[tex]\begin{aligned} \lim_{x \to -3^{-}} f(x) &= \lim_{x \to -3^{-}} (x^2-9)\\[0.5em]&= (-3)^2-9\\[0.5em]&= 0\endaligned}[/tex]
The right hand limit invovles the second piece, f(x) = 0:
[tex]\begin{aligned} \lim_{x \to -3^{+}} f(x) &= \lim_{x \to -3^{+}} (0)\\[0.5em]&= 0\endaligned}[/tex]
Since the two one-sided limits do match, we can just say:
[tex]\lim_{x \to -3} f(x) = 0[/tex]
(no one-sided pieces needed now)
So that was step #1, to make sure the limit exists. Next we need to make sure the limit is headed to the same place where the functions. Since we're using x=-3, we'll use the top piece of the function because x=-3 fits with that piece ( x ≤ -3 ).
[tex]f(-3) = (-3)^2-9 = 0[/tex]
From this, we know that [tex]\lim_{x \to -3} f(x) = f(-3)[/tex], so the function is continuous at -3. (We also know parabolas and lines are continuous in general, so we only needed to check where the two pieces came together at x = -3.)
