Finding the Limit of \(f(x)=\ \sqrt[3]{x^2+11}\)

Find \(\lim\limits_{x\rightarrow 4}\ \sqrt[3]{x^{2}+11}\).

Solution \begin{eqnarray*} \lim\limits_{x\rightarrow 4}\sqrt[3]{x^{2}+11}\underset{\underset{\color{#0066A7}{\hbox{Limit of a Root}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt[3]{\lim\limits_{x\rightarrow 4}(x^{2}+11)} \underset{\underset{\color{#0066A7}{\hbox{Limit of a Sum}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt[3]{\lim\limits_{x\rightarrow 4}x^{2}+\lim\limits_{x\rightarrow 4}11}\\ \underset{\underset{\color{#0066A7}{\hbox{\(\lim\limits_{x\rightarrow c}x^2=c^2\)}}}{\color{#0066A7}{\uparrow}}}{=} \sqrt[3]{4^{2}+11} =\sqrt[3]{27}=3 \\[-14pt] \end{eqnarray*}