i know the lim as x approaches 1 of $\displaystyle \frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1}=-\frac{1}{2}$
however i don't know the steps to get there. Can someone help he? Thanks
$\displaystyle \frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1} \cdot \frac{\sqrt{x+3}+\sqrt{3x+1}}{\sqrt{x+3}+\sqrt{3x+ 1}} =$
$\displaystyle \frac{(x+3) - (3x+1)}{(x-1)[\sqrt{x+3}+\sqrt{3x+1}]} =
$
$\displaystyle \frac{2(1 - x)}{(x-1)[\sqrt{x+3}+\sqrt{3x+1}]} =
$
$\displaystyle \frac{-2}{\sqrt{x+3}+\sqrt{3x+1}}$
now let $\displaystyle x \to 1$