I'm assuming the limit is supposed to be
[tex]\displaystyle\lim_{x\to3}\frac{\sqrt{2x+3}-\sqrt{3x}}{x^2-3x}[/tex]
Multiply the numerator by its conjugate, and do the same with the denominator:
[tex]\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)[/tex]
so that in the limit, we have
[tex]\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}[/tex]
Factorize the first term in the denominator as
[tex]x^2-3x=x(x-3)[/tex]
The [tex]x-3[/tex] terms cancel, leaving you with
[tex]\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}[/tex]
and the limand is continuous at [tex]x=3[/tex], so we can substitute it to find the limit has a value of -1/18.
