## Calculus 10th Edition

$\lim\limits_{x\to-5}\sqrt[3]{x-3}=\sqrt[3]{-8}=-2.$
Using both Theorem $1.4: \lim\limits_{x\to c}\sqrt[n]{x}=\sqrt[n]{c}$ and Theorem $1.5: \lim\limits_{x\to c}(f(g(x)))=f(\lim\limits_{x\to c}g(x)).$ $\lim\limits_{x\to-5}\sqrt[3]{x-3}=\sqrt[3]{\lim\limits_{x\to-5}(x-3)}=\sqrt[3]{-5-3}=\sqrt[3]{-8}=-2.$