# Math Help - Calculating a "0/0" limit

1. ## Calculating a "0/0" limit

How can I calculate the following limit without using l'Hôpital's rule?

$\lim_{x\to 2}\frac {\sqrt{x+2}-\sqrt{3x-2}}{\sqrt{4x+1}-\sqrt{5x-1}}$

2. ## Re: Calculating a "0/0" limit

Originally Posted by loui1410
How can I calculate the following limit without using l'Hôpital's rule?
$\lim_{x\to 2}\frac {\sqrt{x+2}-\sqrt{3x-2}}{\sqrt{4x+1}-\sqrt{5x-1}}$
The original expression is equal $\left( {\frac{{ - 2x + 4}}{{ - x + 2}}} \right)\left( {\frac{{\sqrt {4x + 1} + \sqrt {5x - 1} }}{{\sqrt {x + 2} + \sqrt {3x - 2} }}} \right).$

3. ## Re: Calculating a "0/0" limit

I got to that and didn't know how to continue.. Only now did I notice that we can simply reduce the fraction Thanks.

4. ## Re: Calculating a "0/0" limit

Originally Posted by loui1410
I got to that and didn't know how to continue.. Only now did I notice that we can simply reduce the fraction Thanks.
$\left( {\frac{{ - 2x + 4}}{{ - x + 2}}} \right)\left( {\frac{{\sqrt {4x + 1} + \sqrt {5x - 1} }}{{\sqrt {x + 2} + \sqrt {3x - 2} }}} \right)\to(2)\left( {\frac{6}{4} \right)$