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Math Help - Limits

  1. #1
    Member srirahulan's Avatar
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    Arrow Limits

    \lim_{x\rightarrow1}\frac{x^2-1}{\sqrt{3x+1}-\sqrt{5x-1}} ..How can i Solve this using Theroms??
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  2. #2
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    Re: Limits

    I'd suggest multiplying the limit by the conjugate of the bottom (multiply top and bottom by sqrt(3x+1)+sqrt(5x-1)). Work from there and see if you can simplify the limit a bit.
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  3. #3
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    Re: Limits

    Hello, srirahulan!

    What "theorems" are you referring to?


    \lim_{x\to1}\frac{x^2-1}{\sqrt{3x+1}-\sqrt{5x-1}}

    Multiply top and bottom by the conjugate.

    . . \lim_{x\to1}\frac{x^2-1}{\sqrt{3x+1} - \sqrt{5x-1}} \cdot \frac{\sqrt{3x+1} + \sqrt{5x+1}}{\sqrt{3x+1} + \sqrt{5x-1}}

    . . . . =\;\lim_{x\to1}\frac{(x^2-1)\left(\sqrt{3x+1} + \sqrt{5x-1}\;\!\right)}{(3x+1) - (5x-1)} \;=\;\lim_{x\to1}\frac{(x^2-1)\left(\sqrt{3x+1} + \sqrt{5x-1}\right)}{-2x + 2}

    . . . . =\;\lim_{x\to1} \frac{(x-1)(x+1)\left(\sqrt{3x+1} + \sqrt{5x-1}\right)}{-2(x-1)} \;=\;\lim_{x\to1}\left[-\frac{(x+1)\left(\sqrt{3x+1} + \sqrt{5x-1}\right)}{2}\right]

    . . . . =\;-\frac{2(\sqrt{4} + \sqrt{4})}{2} \;=\;-(2+2) \;=\;\boxed{-4}
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  4. #4
    Member srirahulan's Avatar
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    Re: Limits

    Thank you for your explain.
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