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) = $

If $\lim\limits_{x \to 3}\left(\dfrac{x^2 - ax - 3b}{x - 3}\right) = 5$, then $a + b = $

If $f(x) = \begin{cases} ax + 7 & \text{if } x < 1 \\ 3x - 1 & \text{if } x = 1 \\ \dfrac{b}{x + 3} & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then

$\lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$ is