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    Here it is:

    Does n(1-(1/n))^2 - n tends to 0 or to -2 ?
    I find different reuslts each time

    Thanks!
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    I get -2... I'm not sure how you'd get 0. Factorise it and see what cancels.
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    -2

    how did you get 0?
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    well yeah I got -2 here (cause you can factorise and see it) but I think I confused myself when I had an arbitrary a>1 rather than 2, what I mean is:

    n(1-(1/n))^a - n

    Do you find that it tends to -a ? (and how)

    thanks.
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    (Original post by hitheuk)
    well yeah I got -2 here (cause you can factorise and see it) but I think I confused myself when I had an arbitrary a>1 rather than 2, what I mean is:

    n(1-(1/n))^a - n

    Do you find that it tends to -a ? (and how)

    thanks.
    Write n=-\dfrac{1}{h} and factorise it, then take the limit as h \to 0 (which is the same limit as n \to \infty)... what does this new limit look like?
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    (Original post by nuodai)
    Write n=-\dfrac{1}{h} and factorise it, then take the limit as h \to 0 (which is the same limit as n \to \infty)... what does this new limit look like?
    well theway I was doing is:

    (1-(1/n))^a tends to 1 when n tends to infinity.
    Then n( (1-(1/n))^a ) - n tends to n.1-n which tends to 0.

    Where is the mistake ?
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    (Original post by hitheuk)
    well theway I was doing is:

    (1-(1/n))^a tends to 1 when n tends to infinity.
    Then n( (1-(1/n))^a ) - n tends to n.1-n which tends to 0.

    Where is the mistake ?
    Your mistake is in taking the limit at two separate points. What you've done is take a limit (of the bracket), simplify (to give n-n=0) and then take another limit -- you can't do that. Alternatively, you might have taken a limit (of the bracket), then split the limits to give \lim n - \lim n, and tried to subtract infinity from itself... which is never a good idea.

    When you're taking limits, you can only use the rules \lim (a_nb_n) = \lim a_n \lim b_n and \lim (a_n + b_n) = \lim a_n + \lim b_n when a_n and b_n both converge. Here, this doesn't happen, since a_n=n doesn't converge as n \to \infty.
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    (Original post by nuodai)
    Your mistake is in taking the limit at two separate points. What you've done is take a limit (of the bracket), simplify (to give n-n=0) and then take another limit -- you can't do that. Alternatively, you might have taken a limit (of the bracket), then split the limits to give \lim n - \lim n, and tried to subtract infinity from itself... which is never a good idea.

    When you're taking limits, you can only use the rules \lim (a_nb_n) = \lim a_n \lim b_n and \lim (a_n + b_n) = \lim a_n + \lim b_n when a_n and b_n both converge. Here, this doesn't happen, since a_n=n doesn't converge as n \to \infty.
    Ok thanks.

    (Original post by nuodai)
    Write n=-\dfrac{1}{h} and factorise it, then take the limit as h \to 0 (which is the same limit as n \to \infty)... what does this new limit look like?

    So I get (1/h) [1-(1+h)^a] and then the first bit goes to infinity and the second bit to 0 when h tends to 0 ...
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    (Original post by hitheuk)
    Ok thanks.




    So I get (1/h) [1-(1+h)^a] and then the first bit goes to infinity and the second bit to 0 when h tends to 0 ...
    Would it look more familiar if you wrote it as:
    \displaystyle -\lim_{h \to 0} \dfrac{(1+h)^a - 1^a}{h} ?
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    (Original post by nuodai)
    Would it look more familiar if you wrote it as:
    \displaystyle -\lim_{h \to 0} \dfrac{(1+h)^a - 1^a}{h} ?
    Derivative of the function - (1+x)^a at zero ! So the answer is - a
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    (Original post by hitheuk)
    Derivative of the function - (1+x)^a at zero ! So the answer is - a
    Correct!

    Alternatively you could have used a Taylor expansion for (1-\frac{1}{n})^a, but this way's more fun.
 
 
 
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