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Thread: limit3

  1. #1
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    limit3

    2x-squareroot(4x^2+x+1)
    ----------------------------
    x-squareroot(x^2+2x)

    Find the limit when x goes to infinity.

    I got
    2x-2x
    ------=undefined
    x-x

    Answer must be 1/4
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  2. #2
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    Re: limit3

    $\begin{align*}\displaystyle \lim_{x \to \infty} \dfrac{2x-\sqrt{4x^2 + x + 1}}{x - \sqrt{x^2+2x}} \cdot \dfrac{x+\sqrt{x^2+2x}}{x+\sqrt{x^2+2x}} \cdot \dfrac{2x+\sqrt{4x^2+x+1}}{2x+\sqrt{4x^2+x+1}} & = \displaystyle \lim_{x \to \infty} \dfrac{\left[4x^2-(4x^2+x+1)\right]\left(x+\sqrt{x^2+2x}\right)}{\left[x^2-(x^2+2x)\right]\left(2x+\sqrt{4x^2+x+1}\right)} \\ & = \displaystyle \lim_{x \to \infty} \dfrac{(x+1)\left(x+\sqrt{x^2+2x}\right)}{2x\left( 2x+\sqrt{4x^2+x+1}\right)}\cdot \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}} \\ & = \displaystyle \lim_{x \to \infty} \dfrac{\left(1+\dfrac{1}{x}\right)\left(1+\sqrt{1+ \dfrac{2}{x}}\right)}{2\left(2+\sqrt{4+\dfrac{1}{x }+\dfrac{1}{x^2}}\right)} \\ & = \dfrac{(1+0)\left(1+\sqrt{1+0}\right)}{2(2+\sqrt{4 +0+0})} \\ & = \dfrac{1}{4}\end{align*}$

    I skipped a bunch of steps. Let me know if you have trouble following anything I did.
    Last edited by SlipEternal; May 31st 2017 at 01:56 PM.
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  3. #3
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    Re: limit3

    2x-2x
    ------=undefined
    x-x
    I wanted to solve by that way but it is impossible in this example. Your solution is the only way . Many Thanks.
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  4. #4
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    Re: limit3

    Quote Originally Posted by kastamonu View Post
    2x-2x
    ------=undefined
    x-x
    I wanted to solve by that way but it is impossible in this example. Your solution is the only way . Many Thanks.
    It's not the only way. Your fraction isn't that far off the idea, either.


    \sqrt{4x^2 + x + 1} can be estimated to the linear form of  2x + \tfrac{1}{4} by rewriting the radicand as (2x + \tfrac{1}{4})^2 + \tfrac{15}{16}

    by completing the square, for instance, or using the binomial theorem, for two of the ways.


    Likewise, \sqrt{x^2 + 2x} can be estimated to the linear form of  x + 1.


    Then the problem becomes:


    \displaystyle\lim_{x \to \infty} \dfrac{2x - (2x + \tfrac{1}{4})}{x - (x + 1)} =

    \displaystyle\lim_{x \to \infty} \dfrac{2x - 2x - \tfrac{1}{4}}{x - x - 1} =

     \dfrac{- (\tfrac{1}{4})}{ - 1} =

     \dfrac{1}{4}
    Last edited by greg1313; Jun 1st 2017 at 03:37 PM.
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