## Precalculus (10th Edition)

$-1$
Recall: (1) $\lim _{x\rightarrow c}\left( f\left( x\right) \right) ^{n}=\left( \lim _{x\rightarrow c}f\left( x\right) \right) ^{n}$ (2) $\lim _{x\rightarrow c}f(x) = f(c)$ Use the rules above to have: $\lim _{x\rightarrow -1}\left( 2x+1\right) ^{\frac {5}{3}} \\=\left( \lim _{x\rightarrow -1}\left( 2x+1\right) \right) ^{\frac {5}{3}} \\=\left(2(-1)+1\right)^{5/3} \\=\left( -2+1\right) ^{\frac {5}{3}} \\=-1^{\frac {5}{3}} \\=-1$