limit question

**dandaman** · February 11th 2009, 04:05 PM

i know the lim as x approaches 1 of $\frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1}=-\frac{1}{2}$

however i don't know the steps to get there. Can someone help he? Thanks

**skeeter** · February 11th 2009, 04:47 PM

$\frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1} \cdot \frac{\sqrt{x+3}+\sqrt{3x+1}}{\sqrt{x+3}+\sqrt{3x+ 1}} =$

$\frac{(x+3) - (3x+1)}{(x-1)[\sqrt{x+3}+\sqrt{3x+1}]} =<br />$

$\frac{2(1 - x)}{(x-1)[\sqrt{x+3}+\sqrt{3x+1}]} =<br />$

$\frac{-2}{\sqrt{x+3}+\sqrt{3x+1}}$

now let $x \to 1$

**dandaman** · February 11th 2009, 05:55 PM

thanks man...u da man

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