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Math Help - Find the Limit

  1. #1
    Member VitaX's Avatar
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    Find the Limit

    Find the limit as x goes to -3. (2 - Sqrt(x^2 - 5))/(x + 3)

    sorry for the sloppiness if I knew the codes or where to get the latex codes from I'd write this better. Hopefully someone can write this in the proper form so its easier to read. But I know the first step is to rationalize the limit by multiplying both the numerator and denominator by (2 + sqrt(x^2 - 5)). After this I get a little confused.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by VitaX View Post
    Find the limit as x goes to -3. (2 - Sqrt(x^2 - 5))/(x + 3)

    sorry for the sloppiness if I knew the codes or where to get the latex codes from I'd write this better. Hopefully someone can write this in the proper form so its easier to read. But I know the first step is to rationalize the limit by multiplying both the numerator and denominator by (2 + sqrt(x^2 - 5)). After this I get a little confused.
    \lim_{x \rightarrow -3} \frac{2-\sqrt{x^2-5}}{x+3}

    =\lim_{x \rightarrow -3} \frac{(2-\sqrt{x^2-5})(2+\sqrt{x^2-5})}{(x+3)(2+\sqrt{x^2-5)}}

    =\lim_{x \rightarrow -3} \frac{4-(x^2-5)}{(x+3)(2+\sqrt{x^2-5})}

    =\lim_{x \rightarrow -3} \frac{4-x^2+5}{(x+3)(2+\sqrt{x^2-5})}

    =\lim_{x \rightarrow -3} \frac{9-x^2}{(x+3)(2+\sqrt{x^2-5})}

    =\lim_{x \rightarrow -3} \frac{(3-x)(3+x)}{(x+3)(2+\sqrt{x^2-5})}

    =\lim_{x \rightarrow -3} \frac{3-x}{2+\sqrt{x^2-5}}

    =\frac{6}{4}

    =\frac{3}{2}
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  3. #3
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    \lim_{x \rightarrow -3} \frac{2-\sqrt{x^2-5}}{x+3}

    =\lim_{x \rightarrow -3} \frac{(2-\sqrt{x^2-5})(2+\sqrt{x^2-5})}{(x+3)(2+\sqrt{x^2-5)}}

    =\lim_{x \rightarrow -3} \frac{4-(x^2-5)}{(x+3)(2+\sqrt{x^2-5})}

    =\lim_{x \rightarrow -3} \frac{4-x^2+5}{(x+3)(2+\sqrt{x^2-5})}

    =\lim_{x \rightarrow -3} \frac{9-x^2}{(x+3)(2+\sqrt{x^2-5})}

    Oops sorry, forgot to think about the fact that the numerator factors.
    Last edited by seld; September 9th 2009 at 05:44 PM. Reason: bad solution
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  4. #4
    Member VitaX's Avatar
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    limit in the book is 3/2 by the way. Thanks for your help i understand where i went wrong mainly in the denominator.
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