Question

Evalaute:

${lim}_{x\to 3}\left(\frac{x^2-4x+3}{\sqrt{2x+3}-3}\right)$

Answer 1

${lim}_{x\to 3}\left(\frac{x^2-4x+3}{\sqrt{2x+3}-3}\right)$

$Multiplybytheconjugateof\sqrt{2x+3}-3:$

$={lim}_{x\to 3}\frac{(x^2-4x+3)(\sqrt{2x+3}+3)}{2x-6}$

$={lim}_{x\to 3}\frac{(x-1)(x-3)(\sqrt{2x+3}+3)}{2(x-3)}$

$={lim}_{x\to 3}\left(\frac{1}{2}(x-1)(\sqrt{2x+3}+3)\right)$

$=\frac{1}{2}(3-1)(\sqrt{2\cdot 3+3}+3)$

$=\frac{1\cdot 2(3+3)}{2}$

$=6$