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Math Help - fiNd the VaLue of

  1. #1
    Junior Member mathbuoy's Avatar
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    Wink fiNd the VaLue of

    Find the value of \lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2}).

    ^include working and explanation needed pls
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  2. #2
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    Dear mathbuoy,

    use this:
    n - 4 = (\sqrt 2)^2 - 2^2
    and
    a^2 - b^2 = (a+b)*(a-b)
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  3. #3
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    Hi

    Quote Originally Posted by mathbuoy View Post
    Find the value of \lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2}).

    ^include working and explanation needed pls
     lim_{n \to 4} \sqrt{n}-2 = 0
     lim_{n \to 4} n - 4 = 0

    L'Hospital:

     lim_{n \to 4} \frac{n-4}{\sqrt{n}-2} =
      lim_{n \to 4} \frac{1}{(\sqrt{n})' } =  lim_{n \to 4} \frac{1}{0.5 * n^{-0.5}}

     = lim_{n \to 4} n^{0.5}/0.5 =lim_{n \to 4}  n^{0.5}*2 = 4^{0.5} * 2 = 4
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by mathbuoy View Post
    Find the value of \lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2}).

    ^include working and explanation needed pls

    \lim_{n \to 4} \frac{n-4}{\sqrt{n} -2 }

    Multiply by conjugate

     \lim_{n \to 4} \bigg(\frac{n-4}{\sqrt{n} -2 }\bigg)\bigg(\frac{\sqrt{n} + 2 }{\sqrt{n} + 2}\bigg)

    \lim_{n \to 4} \frac{(n-4)(\sqrt{n} + 2)}{n-4}

    \lim_{n \to 4} \sqrt{n} +2

    4
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  5. #5
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    With Skalkaz hint (btw it has a typo in it)

     lim \frac{n-4}{\sqrt{n} - 2} = lim \frac{(\sqrt{n})^2 - 2^2}{\sqrt{n} - 2}

     = lim \frac{(\sqrt{n}-2)(\sqrt{n}+2)}{\sqrt{n}-2}

     =  lim_{n -> 4}\sqrt{n}+2 = \sqrt{4} + 2 = 4

    Regards Rapha
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