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Math Help - simple limit...

  1. #1
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    simple limit...

    simple limit...-limit-dqrt.jpg

    Hi guys,

    I have this limit, I fail to understand why the first stage is correct (the transformation of the original function to the state of it after the first stage)

    any explanation will be appreciated...thanks !
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  2. #2
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    They multiplied top and bottom by the bottom's conjugate...
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  3. #3
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    You have \lim_{x\to 3}\frac{x- 3}{\sqrt{x+1}- 2} and are asking how that becomes
    2\lim_{x\to 3}\sqrt{x+1}

    Easy answer- only by accident! The limits are the same but in general
    \frac{x- 3}{\sqrt{x+1}- 2} is NOT equal to 2\sqrt{x+ 1}.
    For example, if x= 8, the first is equal to \frac{8- 3}{\sqrt{8+1}- 2}= 5 while the second is 2\sqrt{8+ 1}= 6

    What is true is that you can "rationalize the denominator to get
    \frac{x- 3}{\sqrt{x+1}- 2}\fract{\sqrt{x+ 1}+ 2}{\sqrt{x+ 1}+ 2}= \frac{(x- 3)(\sqrt{x+ 1}+ 2)}{x- 3}= \sqrt{x+1}+ 2

    Now, it happens that both [itex]2\sqrt{x+1}[/itex] and [itex]\sqrt{x+1}+ 2[/itex] are equal to 4 when x= 3, so the limits are equal, but that is the only value of x for which they are the same
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  4. #4
    MHF Contributor chisigma's Avatar
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    In 'normal' circustances it shold be...

    \displaystyle \frac{x-3}{\sqrt{x+1}-2}= 2+\sqrt{x+1}

    In the jpg file however is...

    \displaystyle \frac{x-3}{\sqrt{x+1}-2}= 2\ \sqrt{x+1}

    May be that the '+' between 2 and the square root has been 'forgotten'...

    Kind regards

    \chi \sigma
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