Answer
$\frac{2}{3}$
Work Step by Step
$\lim\limits_{u \to 2}\frac{\sqrt {4u+1}-3}{u-2}=\lim\limits_{u \to 2}\frac{\sqrt {4u+1}-3}{u-2}*\frac{\sqrt {4u+1}+3}{\sqrt {4u+1}+3}=\lim\limits_{u \to 2}\frac{4u+1-9}{(u-2)(\sqrt {4u+1}+3)}=\lim\limits_{u \to 2}\frac{4u-8}{(u-2)(\sqrt {4u+1}+3)}=\lim\limits_{u \to 2}\frac{4(u-2)}{(u-2)(\sqrt {4u+1}+3)}=\lim\limits_{u \to 2}\frac{4}{\sqrt {4u+1}+3}=\frac{4}{\sqrt {4(2)+1}+3}=\frac{4}{6}=\frac{2}{3}$
