One-sided limits and existence of the two-sided limit at x = 0

For $x < 0$, $f(x) = \cos(x)$, and as $x \to 0^-$, $\cos(x) \to \cos(0) = 1$. So $\lim_{x \to 0^-} f(x) = 1$. For $x \ge 0$, $f(x) = 1 - x^2$, and as $x \to 0^+$, $1 - x^2 \to 1 - 0 = 1$. So $\lim_{x \to 0^+} f(x) = 1$.

Since the left-hand limit equals the right-hand limit, both equal to $1$, the two-sided limit exists and $\lim_{x \to 0} f(x) = 1$. Note this matches $f(0) = 1$, though this equality is not required for the limit to exist.