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Math Help - Finding the limit as x tends to infinity

  1. #1
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    Finding the limit as x tends to infinity

    The question:

    Find the limit as x tends to infinity \sqrt{x^2 + x} - x

    I'm not sure how to go about it. I'm thinking that I need to get rid of that square root, or at least, get it in a form which will make it easy to read off the limit.

    I've tried the following:

    \frac{\sqrt{x^2 + x} - x}{1} . \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}

    \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x}

    \frac{x}{\sqrt{x^2 + x} + x}

    I'm not sure how to factor out x in the denominator, I'm guessing that'd simplify it.

    If anyone could help, or offer a different method of solving this, I'd really appreciate it.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:

    Find the limit as x tends to infinity \sqrt{x^2 + x} - x

    I'm not sure how to go about it. I'm thinking that I need to get rid of that square root, or at least, get it in a form which will make it easy to read off the limit.

    I've tried the following:

    \frac{\sqrt{x^2 + x} - x}{1} . \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}

    \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x}

    \frac{x}{\sqrt{x^2 + x} + x}

    I'm not sure how to factor out x in the denominator, I'm guessing that'd simplify it.

    If anyone could help, or offer a different method of solving this, I'd really appreciate it.
    Divide both numerator and denominator by x to get \lim\limits_{x\to\infty}\dfrac{1}{\sqrt{1+\frac{1}  {x}}+1}

    Can you solve it now?
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  3. #3
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    Thanks.

    I know this is probably obvious (my brain is fried at the moment), but how did you divide the denominator? The square root is doing my head in. D:
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  4. #4
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    \dfrac{x}{\sqrt{x^2+x} + x} = \dfrac{x}{\sqrt{x^2\left(1+\frac{1}{x}\right)}+x} } = \dfrac{x}{x\sqrt{1+\frac{1}{x}}+x} }

    Then cancel x to give you what Chris gave.
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  5. #5
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    Or, (slightly) different way to do the same thing-
    \frac{\sqrt{x^2+ x}+ x}{x}= \sqrt{\frac{x^2+ x}{x^2}}+ 1= \sqrt{1+ \frac{1}{x}}+ 1

    You see how the "x" outside the square root became " x^2" inside the square root. As long as x is positive (which it certainly is here as x is going to \infty), x= \sqrt{x^2}.
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  6. #6
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    Ahh, thanks!
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  7. #7
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    Alternatively,

    \left(x^2+x\right)^n=\left(x^2\right)^n+\binom{n}{  1}\left(x^2\right)^{n-1}x+\binom{n}{2}\left(x^2\right)^{n-2}x^2+....

    For n=0.5, as x approaches infinity, all negative powers of x approach 0.
    Then the expansion reduces to

    \left(x^2\right)^{0.5}+0.5\left(x^2\right)^{-0.5}x=x+0.5

    \displaystyle\Rightarrow\ \lim_{x\rightarrow\infty}\left(\sqrt{x^2+x}-x\right)=x-x+0.5=0.5
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