## Precalculus (10th Edition)

$1$
Note that, if $f(x)\ge0$, then $\lim _{x\rightarrow c}\sqrt {f\left( x\right) }=\sqrt {\lim _{x\rightarrow c}f\left( x\right) }$ Thus, $\lim _{x\rightarrow 0}\sqrt {1-2x}=\sqrt {\lim _{x\rightarrow 0}\left( 1-2x\right) }$ Since $\lim _{x\rightarrow c}f(x) = f(c)$, then $\sqrt {\lim _{x\rightarrow 0}\left( 1-2x\right) } \\=\sqrt {1-2\times 0} =\sqrt {1} \\=1$