Question

How to sove these question​

Answer 1

$\displaystyle\\\lim_{x\to 3}\left(\dfrac{\sqrt{2x+3}-\sqrt{4x-3}}{x^2-9}\right)=\\\\\lim_{x\to 3}\left(\dfrac{(\sqrt{2x+3}-\sqrt{4x-3})'}{(x^2-9)'}\right)=\\\\\lim_{x\to 3}\left(\dfrac{\dfrac{1}{2\sqrt{2x+3}}\cdot 2-\dfrac{1}{2\sqrt{4x-3}}\cdot 4}{2x}\right)=\\\\\lim_{x\to 3}\left(\dfrac{\dfrac{2}{2\sqrt{2x+3}}-\dfrac{4}{2\sqrt{4x-3}}}{2x}\right)=\\\\\dfrac{\dfrac{2}{2\sqrt{2\cdot 3+3}}-\dfrac{4}{2\sqrt{4\cdot 3-3}}}{2\cdot 3}=$

$\dfrac{\dfrac{2}{2\sqrt9}-\dfrac{4}{2\sqrt9}}{6}=\\\\\dfrac{-\dfrac{2}{6}}{6}=\\\\-\dfrac{1}{18}$