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? (:

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- May 1st 2011, 11:31 AMweberLimit 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 AMPlato
- May 1st 2011, 11:46 AMOpalg
Hello weber, and welcome to the Forum!

To find , start by multiplying top and bottom of the fraction by , to get . Then factorise the denominator (difference of two squares) as . Now you can cancel the factors and evaluate the result at x=1.

**Edit.**Sorry, I didn't see Plato's comment. - May 1st 2011, 01:19 PMweber
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 PMPlato