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Math Help - some limits

  1. #1
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    some limits

    Find \displaystyle \lim_{x \rightarrow - \infty} \left(\sqrt{x^2 + 2x} - \sqrt{x^2 + 1} \right).
    Last edited by mr fantastic; December 22nd 2010 at 01:14 AM. Reason: Fixed and improved typesetting.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by danial View Post
    Find \displaystyle \lim_{x \rightarrow - \infty} \left(\sqrt{x^2 + 2x} - \sqrt{x^2 + 1} \right).
    Consider that \displaystyle \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}}
    Last edited by mr fantastic; December 22nd 2010 at 01:20 AM. Reason: Originally a response to an incompetent post. Now directed to OP.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by danial View Post
    lim \sqrt{X^2 + 2x} - \sqrt{x^2 + 1}
    x \rightarrow - \infty
    \displaystyle\lim_{x \to \infty} \left[ \sqrt{x^2+2x}-\sqrt{x^2+1}\right]=1

    You can see this by expanding the square roots as power series after taking a factor of x out of each or by using Drexel28's method above.

    CB
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  4. #4
    Senior Member BAdhi's Avatar
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    f(x) = \sqrt{(x+1)^2 -1} - \sqrt{x^2 +1}
    f(x)= \frac{(x^2+2x) - (x^2+1)}{\sqrt{(x+1)^2 -1} + \sqrt{x^2 +1}}

    f(x)= \frac{2x-1}{\sqrt{(x+1)^2 -1} + \sqrt{x^2 +1}}

    f(x)= \frac{2-\frac{1}{x}}{\sqrt{(1+\frac{1}{x})^2 -\frac{1}{x^2}} + \sqrt{1+\frac{1}{x^2}}}}<br />
    Last edited by CaptainBlack; December 21st 2010 at 10:04 PM. Reason: to remove the last stage to leave something for the OP to do themselves
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  5. #5
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    I saw in a book that it had been solved like this:
    \displaystyle\lim_{x \to -\infty} \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}} =


    \frac{2x-1}{\sqrt{x^2(1+\frac{2}{x})}+ \sqrt{x^2(1+\frac{1}{x^2})}}=  \displaystyle\lim_{x \to -\infty} \left \frac{2x-1}{|x|+|x|}= \frac{2x-1}{-2x}=-1

    is it right?
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  6. #6
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    Quote Originally Posted by danial View Post
    I saw in a book that it had been solved like this:
    \displaystyle\lim_{x \to -\infty} \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}} =


    \frac{2x-1}{\sqrt{x^2(1+\frac{2}{x})}+ \sqrt{x^2(1+\frac{1}{x^2})}}=  \displaystyle\lim_{x \to -\infty} \left \frac{2x-1}{|x|+|x|}= \frac{2x-1}{-2x}=-1

    is it right?
    Dear danial,

    Yes it is correct. They have used the fact that, \sqrt{x^2}=\mid x\mid
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