# Calculating a "0/0" limit

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• Jul 10th 2012, 08:04 AM
loui1410
Calculating a "0/0" limit
How can I calculate the following limit without using l'Hôpital's rule?

$\lim_{x\to 2}\frac {\sqrt{x+2}-\sqrt{3x-2}}{\sqrt{4x+1}-\sqrt{5x-1}}$
• Jul 10th 2012, 09:39 AM
Plato
Re: Calculating a "0/0" limit
Quote:

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 $\left( {\frac{{ - 2x + 4}}{{ - x + 2}}} \right)\left( {\frac{{\sqrt {4x + 1} + \sqrt {5x - 1} }}{{\sqrt {x + 2} + \sqrt {3x - 2} }}} \right).$
• Jul 10th 2012, 09:49 AM
loui1410
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.
• Jul 10th 2012, 09:59 AM
Plato
Re: Calculating a "0/0" limit
Quote:

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.

$\left( {\frac{{ - 2x + 4}}{{ - x + 2}}} \right)\left( {\frac{{\sqrt {4x + 1} + \sqrt {5x - 1} }}{{\sqrt {x + 2} + \sqrt {3x - 2} }}} \right)\to(2)\left( {\frac{6}{4} \right)$