If \(f(x) = \begin{cases} x^2 - 1 & \text{if } x \geq 2 \\ x + 1 & \text{if } x < 2 \end{cases}\), then \(\lim\limits_{x \to 1} f(x) + \lim\limits_{x \to 2} f(x) = \)

For \(x \to 1\), since \(x < 2\), the function is \(f(x) = x + 1\).

\(\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (x + 1) = 1 + 1 = 2\)

For \(x \to 2\), we evaluate the left-hand and right-hand limits.

Left-hand limit at \(x = 2\):
\(\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (x + 1) = 2 + 1 = 3\)

Right-hand limit at \(x = 2\):
\(\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (x^2 - 1) = 2^2 - 1 = 3\)

Since the left-hand limit equals the right-hand limit, \(\lim\limits_{x \to 2} f(x) = 3\).

The required sum is \(\lim\limits_{x \to 1} f(x) + \lim\limits_{x \to 2} f(x) = 2 + 3 = 5\).

Answer: \(5\)