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Math Help - Limit Help

  1. #1
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    Smile 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

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

    Can somebody help me? (:
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  2. #2
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    Quote Originally Posted by weber View Post
    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)}}
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  3. #3
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    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.
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  4. #4
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    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?
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  5. #5
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    Quote Originally Posted by weber View Post
    @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.
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