Answer:
[tex]\boxed{c = 9}.[/tex]
Step-by-step explanation:
The limit [tex]\lim\limits_{x\to -3} f(x)[/tex] exists if, and only if both [tex]\lim\limits_{x\to -3^-} f(x)[/tex] and [tex]\lim\limits_{x\to -3^+} f(x)[/tex] exist and [tex]\lim\limits_{x\to -3^-} f(x) = \lim\limits_{x\to -3^+} f(x)[/tex].
Since [tex]f(x) = x^2 - 9[/tex] for [tex]x<-3[/tex], we have:
[tex]\lim\limits_{x\to-3^-}f(x) = \lim\limits_{x\to-3^-} (x^2 - 9) = (-3)^2 - 9 = 9-9 = 0.[/tex]
On the other hand, [tex]f(x) = -x^2+c[/tex] for [tex]x > -3[/tex], so we have:
[tex]\lim\limits_{x\to-3^+} f(x) = \lim\limits_{x\to-3^+} (c-x^2) = c-(-3)^2 = c-9.[/tex]
Assuming that [tex]c \in \mathbb{R}[/tex], both one-sided limits exist, which means they must be equal:
[tex]\lim\limits_{x\to -3^-} f(x) = \lim\limits_{x\to -3^+} f(x) \iff 0 = c-9 \iff c = 9.[/tex]
So we finally get:
[tex]\boxed{c = 9}.[/tex]
