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Math Help - Limit, derivation, help please

  1. #1
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    Limit, derivation, help please

    Hi! I need some help in the following problems.

    1. What is the limit if n tends to infinity:

    [(n+2) choose 2]/(n choose 2)]^n

    2. Derivation

    y=(4^x-x^4)/log(7)x; that is the seven-based log of x

    3. Derivation

    y=ln[e^x+e^(-x)]

    4. Limit if x tends to zero

    1/[e^(2x)-1]-1/2x

    Thanks a lot.
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  2. #2
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    Quote Originally Posted by csodacsiga View Post
    [(n+2) choose 2]/(n choose 2)]^n
    Hint: {a\choose 2} = \frac{a(a-1)}{2}

    2. Derivation

    y=(4^x-x^4)/log(7)x; that is the seven-based log of x
    This is the same as,
    y= \frac{e^{\ln 4 x} - x^4}{\left( \frac{\ln x}{\ln 7} \right)}
    Now apply the quotient rule.

    y=ln[e^x+e^(-x)]
    With chain rule we get,
    (e^x + e^{-x})' \cdot \frac{1}{e^x + e^{-x}} = \frac{e^x - e^{-x}}{e^x + e^{-x}}


    4. Limit if x tends to zero

    1/[e^(2x)-1]-1/2x

    Thanks a lot.
    This one is not clear.
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  3. #3
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    For 1., {n+2\choose2}\bigg/{n\choose2} = \frac{\frac12(n+2)(n+1)}{\frac12n(n-1)} = 1+\frac{4n+2}{n(n-1)}. So \left({n+2\choose2}\bigg/{n\choose2}\right)^{\!\!n} = \left(1+\frac{4n+2}{n(n-1)}\right)^{\!\!n}. This looks as

    though it should be close to \bigl(1+\tfrac4n\bigr)^n, which has the limit e^4 as n\to\infty.

    To justify that guess, you could take the logarithm of \left(1+\frac{4n+2}{n(n-1)}\right)^{\!\!n} and show that it tends to 4 as n\to\infty.
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  4. #4
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    Thank you both for the great helps.

    1/[e^(2x)-1]-1/2x is

    (one over (e to the power of two x) minus one) minus one over two x
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  5. #5
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    Quote Originally Posted by csodacsiga View Post
    Thank you both for the great helps.

    1/[e^(2x)-1]-1/2x is

    (one over (e to the power of two x) minus one) minus one over two x
    \frac1{e^{2x}-1} - \frac1{2x} = \frac{1+2x-e^{2x}}{2x(e^{2x}-1)}. Now apply l'H˘pital two or three times.
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  6. #6
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    I assume this is it:

    \lim_{x\to 0}\frac{1}{e^{2x}-1}-\frac{1}{2x}?.

    We can be lazy and use ol' L'Hopital.

    Rewrite:

    \frac{-1}{2}\displaystyle{\lim_{x\to 0}\frac{-2x+e^{2x}-1}{x(e^{2x}-1)}}

    L'Hopital:

    \displaystyle{-\lim_{x\to 0}\frac{e^{2x}-1}{2xe^{2x}+e^{2x}-1}}

    L'Hopital again:

    \displaystyle{-\lim_{x\to 0}\frac{1}{2x+2}}

    Now, we can see it?.

    Now, try it without L'Hopital. That is 'funner'.
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