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Math Help - How do I calculate the limit of the function below?

  1. #1
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    How do I calculate the limit of the function below?

    Hello, this is my first post. I hope there's someone who can help me calculate the limit of the attached function.

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails How do I calculate the limit of the function below?-limit_calc_problem.bmp  
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  2. #2
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    Quote Originally Posted by azarue View Post
    Hello, this is my first post. I hope there's someone who can help me calculate the limit of the attached function.

    Thanks in advance.
    Dear azarue,

    Multiply both the numerator and denominator by, \sqrt{4x^2+4x+1}+\sqrt{4x^2-x}

    If you have any questions please don't hesitate to ask.
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  3. #3
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    Here's an answer I got to my private mailbox from Soroban

    I can't send PM's so Thank you Soroban.

    One comment, I do think there's a small mistake.
    \frac{2x+1-   \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}}   \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}} . =\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}

    The final result should be \;\frac{5x+1}{2x+1 + \sqrt{4x^2-x}}

    Thus the limit is \lim_{x\to\infty}=\;\;\frac{5}{4}

    Thank you all the helpers for helping me solve this.

    Adiel.

    Welcome aboard!

    \lim_{x\to\infty}\left(\sqrt{4x^2 + 4x + 1} - \sqrt{4x^2 -  x}\right)
    We have: . \sqrt{(2x+1)^2}  - \sqrt{4x^2-x}\;=\;2x+1 - \sqrt{4x^2-x}


    Multiply "top and bottom" by the conjugate:

    . . \frac{2x+1-  \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}}  \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}} . =\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}


    \text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{3x}{x} +  \dfrac{1}{x}}{\dfrac{2x}{x} + \dfrac{1}{x} + \dfrac{\sqrt{4x^2-x}}{x}}  \;\;=\;\;\frac{3+\dfrac{1}{x}}{2 + \dfrac{1}{x} +  \sqrt{\dfrac{4x^2-x}{x^2}}}

    . . . . . =\;\;\frac{3+\dfrac{1}{x}}{2 +  \dfrac{1}{x} + \sqrt{\dfrac{4x^2}{x^2} - \dfrac{1}{x^2}}}   \;\;=\;\;\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x} + \sqrt{4 -  \dfrac{1}{x^2}}}


    Therefore: . \lim_{x\to\infty}\left[\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x}  + \sqrt{4 - \dfrac{1}{x^2}}}\right]  \;\;=\;\;\frac{3+0}{2+0+\sqrt{4-0}}\;\;=\;\;\frac{3}{4}

    [/quote]
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by azarue View Post
    I can't send PM's so Thank you Soroban.

    One comment, I do think there's a small mistake.
    \frac{2x+1-   \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}}   \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}} . =\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}

    The final result should be \;\frac{5x+1}{2x+1 + \sqrt{4x^2-x}}

    Thus the limit is \lim_{x\to\infty}=\;\;\frac{5}{4}

    Thank you all the helpers for helping me solve this.

    Adiel.

    Welcome aboard!

    We have: . \sqrt{(2x+1)^2}  - \sqrt{4x^2-x}\;=\;2x+1 - \sqrt{4x^2-x}


    Multiply "top and bottom" by the conjugate:

    . . \frac{2x+1-  \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}}  \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}} . =\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}


    \text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{3x}{x} +  \dfrac{1}{x}}{\dfrac{2x}{x} + \dfrac{1}{x} + \dfrac{\sqrt{4x^2-x}}{x}}  \;\;=\;\;\frac{3+\dfrac{1}{x}}{2 + \dfrac{1}{x} +  \sqrt{\dfrac{4x^2-x}{x^2}}}

    . . . . . =\;\;\frac{3+\dfrac{1}{x}}{2 +  \dfrac{1}{x} + \sqrt{\dfrac{4x^2}{x^2} - \dfrac{1}{x^2}}}   \;\;=\;\;\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x} + \sqrt{4 -  \dfrac{1}{x^2}}}


    Therefore: . \lim_{x\to\infty}\left[\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x}  + \sqrt{4 - \dfrac{1}{x^2}}}\right]  \;\;=\;\;\frac{3+0}{2+0+\sqrt{4-0}}\;\;=\;\;\frac{3}{4}

    [/QUOTE]

    The error I see on this part :

    \frac{2x+1-   \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}}   \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}}

    is that the numerator should be:

    {(2x+1)^2 - (4x^2 - x)} = 4{x^2}+4x+1 - 4{x^2}+x = 5x+1

    this should take you to the desired conclusion
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  5. #5
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    I saw that issue. thank you for the comment

    There was an issue in Soroban's solution, but his way to the solution was good enough to understand the logic behind it.
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