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Math Help - Find the Limit

  1. #1
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    Find the Limit

    Find the limit algebraically (without graphing)
    \lim_{x\to -\infty}\frac{x}{\sqrt{x^2+4}}

    lim as x approaches negative infinity

    I'm confused please help!
    Last edited by mr fantastic; September 19th 2009 at 02:19 AM. Reason: Inserted the latex command for infinity for better readability
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  2. #2
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    Krizalid's Avatar
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    \frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},

    |x|=-x since x<0.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},

    |x|=-x since x<0.
    Uh thanks but im still confused
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    \frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},

    |x|=-x since x<0.
    Actually when I looked it over and analyzed it more it makes sense except
    |x|=-x since x<0.

    how is absolute value of x = negative x?
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by yoman360 View Post
    Actually when I looked it over and analyzed it more it makes sense except
    |x|=-x since x<0.

    how is absolute value of x = negative x?
    I think what he meant to write is \frac{x}{|x|}=-1 since x<0.
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