## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{u\to2}\frac{\sqrt{4u+1}-3}{u-2}=\frac{2}{3}$
$A=\lim\limits_{u\to2}\frac{\sqrt{4u+1}-3}{u-2}$ Multiply both numerator and denominator by $\sqrt{4u+1}+3$ We notice that $(\sqrt{4u+1}-3)(\sqrt{4u+1}+3)=(4u+1)-9=4u-8=4(u-2)$ because $(a-b)(a+b)=a^2-b^2$ Therefore, $A=\lim\limits_{u\to2}\frac{4(u-2)}{(u-2)(\sqrt{4u+1}+3)}$ (divide both numerator and denominator by $u-2$) $A=\lim\limits_{u\to2}\frac{4}{\sqrt{4u+1}+3}$ $A=\frac{4}{\sqrt{4\times2+1}+3}$ $A=\frac{2}{3}$