# Limit Help

• May 1st 2011, 11:31 AM
weber
Limit Help
Hello! It's my first thread here, so if i make something wrong, please correct me! =)

Well... i'm crushing my head and i couldn't find a solution for this limit (Headbang)

http://i52.tinypic.com/2j2t2ti.gif

Can somebody help me? (:
• May 1st 2011, 11:41 AM
Plato
Quote:

Originally Posted by weber
Hello! It's my first thread here, so if i make something wrong, please correct me! =)
Can somebody help me? (:

$\frac{{\sqrt x - 1}}{{\sqrt {2x + 3} - \sqrt 5 }} = \frac{{(x - 1)\left( {\sqrt {2x + 3} +\sqrt 5 } \right)}}{{(2x - 2)\left( {\sqrt x + 1} \right)}}$
• May 1st 2011, 11:46 AM
Opalg
Hello weber, and welcome to the Forum!

To find $\lim_{x\to1}\frac{\sqrt x -1}{\sqrt{2x+3} - \sqrt5}$, start by multiplying top and bottom of the fraction by $\sqrt{2x+3} + \sqrt5$, to get $\frac{(\sqrt x -1)(\sqrt{2x+3} + \sqrt5)}{2x-2}$. Then factorise the denominator (difference of two squares) as $2(\sqrt x +1)(\sqrt x -1)$. Now you can cancel the $(\sqrt x -1)$ factors and evaluate the result at x=1.

Edit. Sorry, I didn't see Plato's comment.
• May 1st 2011, 01:19 PM
weber
Hey! Thanks a lot for both anwers and lightening speed reply! :D
@Opalg Thanks for the tip! The key is in the factorise the denominator
@Plato Did you do the same as Opalg? (multiplying top and bottom of the fraction by \sqrt{2x+3} + \sqrt5) ?

Also, there is more options to solve this limit?
• May 1st 2011, 01:29 PM
Plato
Quote:

Originally Posted by weber
@Plato Did you do the same as Opalg? (multiplying top and bottom of the fraction by \sqrt{2x+3} + \sqrt5) ?
Also, there is more options to solve this limit?

Actually I multiplied by $\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt {2x + 3} + \sqrt 5 } \right)}}{{\left( {\sqrt {2x + 3} + \sqrt 5 }\right)\left( {\sqrt x + 1} \right)}}$.

As to other ways, in calculus one might use derivatives.
But for pre-calculus this is the best.