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 $\displaystyle \lim_{x\to1}\frac{\sqrt x -1}{\sqrt{2x+3} - \sqrt5}$, start by multiplying top and bottom of the fraction by $\displaystyle \sqrt{2x+3} + \sqrt5$, to get $\displaystyle \frac{(\sqrt x -1)(\sqrt{2x+3} + \sqrt5)}{2x-2}$. Then factorise the denominator (difference of two squares) as $\displaystyle 2(\sqrt x +1)(\sqrt x -1)$. Now you can cancel the $\displaystyle (\sqrt x -1)$ 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?
Actually I multiplied by $\displaystyle \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt {2x + 3} + \sqrt 5 } \right)}}{{\left( {\sqrt {2x + 3} + \sqrt 5 }\right)\left( {\sqrt x + 1} \right)}}$.
As to other ways, in calculus one might use derivatives.
But for pre-calculus this is the best.