Question

Find lim x→3 sqrt 2x+3-sqrt 3x/ x^2-3x. you must show your work or explain your work in words plsss I need help

Answer 1

I'm assuming the limit is supposed to be

$\displaystyle\lim_{x\to3}\frac{\sqrt{2x+3}-\sqrt{3x}}{x^2-3x}$

Multiply the numerator by its conjugate, and do the same with the denominator:

$\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)$

so that in the limit, we have

$\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}$

Factorize the first term in the denominator as

$x^2-3x=x(x-3)$

The $x-3$ terms cancel, leaving you with

$\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}$

and the limand is continuous at $x=3$, so we can substitute it to find the limit has a value of -1/18.