Question Number 72633 by Rio Michael last updated on 30/Oct/19
$${prove}\:{using}\:{th}\:{sandwich}\:{or}\:{Squeez}\:{theorem}\:{that} \\ $$$${for}\:{any}\:\:{a}\:>\:\mathrm{0} \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$