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
http://i52.tinypic.com/2j2t2ti.gif
Can somebody help me? (:
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
http://i52.tinypic.com/2j2t2ti.gif
Can somebody help me? (:
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.
Hey! Thanks a lot for both anwers and lightening speed reply!
@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?