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

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

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

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

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

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

    L'Hospital:

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

    $\displaystyle = lim_{n \to 4} n^{0.5}/0.5 =lim_{n \to 4} n^{0.5}*2 = 4^{0.5} * 2 = 4$
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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 $\displaystyle \lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2})$.

    ^include working and explanation needed pls

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

    Multiply by conjugate

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

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

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

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

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

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

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

    Regards Rapha
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