Question

HELP !! Answer the questions shown

Answer 1

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.

     $\lim_{x \to -3^{-}} f(x) = \lim_{x \to -3^{+}} f(x)$

Second, we need to show that this limit equals the functions value.

    $\lim_{x \to -3} f(x) = f(-3)$

The left hand limit involves the first piece, f(x) = x^2 - 9:

    $\begin{aligned} \lim_{x \to -3^{-}} f(x) &= \lim_{x \to -3^{-}} (x^2-9)\\[0.5em]&= (-3)^2-9\\[0.5em]&= 0\endaligned}$

The right hand limit invovles the second piece, f(x) = 0:

    $\begin{aligned} \lim_{x \to -3^{+}} f(x) &= \lim_{x \to -3^{+}} (0)\\[0.5em]&= 0\endaligned}$

Since the two one-sided limits do match, we can just say:

    $\lim_{x \to -3} f(x) = 0$  

(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 ).

    $f(-3) = (-3)^2-9 = 0$

From this, we know that $\lim_{x \to -3} f(x) = f(-3)$, 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.)