Answer
$\lim\limits_{u \to -2}(\sqrt{u^4+3u+6})=4$
Work Step by Step
$\lim\limits_{u \to -2}(\sqrt{u^4+3u+6})$
$=\sqrt{\lim\limits_{u \to -2}(u^4+3u+6)}$ (root law) (we assume that $\lim\limits_{u \to -2}(u^4+3u+6)\geq0$)
$=\sqrt{\lim\limits_{u \to -2}u^4+\lim\limits_{u \to -2}3u+\lim\limits_{u \to -2}6}$ (addition law)
$=\sqrt{\lim\limits_{u \to -2}u^4+3\lim\limits_{u \to -2}u+\lim\limits_{u \to -2}6}$ (constant multiple law)
$=\sqrt{(-2)^4+3\times(-2)+6}$
$=4$
