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    limit as n tends to infinity

    ((e^-n+n)/n+2))^2n

    I thought if I found the limit of the bracket to be 1 and 1^2n is always 1 therefore the limit is 1?
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    (Original post by CammieInfinity)
    limit as n tends to infinity

    ((e^-n+n)/n+2))^2n

    I thought if I found the limit of the bracket to be 1 and 1^2n is always 1 therefore the limit is 1?
    Try a large number for n on your calculator. It isn't 1 is it?
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    I can't remember much about limits but this can be turned into:

    \lim_{x\to \infty} ((1+\frac{1}{x})^x)^{-4} where n=2x

    Your notation is ambiguous by the way - you could do with a bracket around n + 2.
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    How did you get it to 1+1/x? The division yields (1+(e^(-n)-2/n+2))^2n
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    (Original post by CammieInfinity)
    How did you get it to 1+1/x? The division yields (1+(e^(-n)-2/n+2))^2n
    If you still can't be bothered to use unambiguous notation I can't be bothered to make the effort to try to read that. I used the fact that e^{-x} tends to zero as x tends to infinity.
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    I got it despite your short comings

    https://www.youtube.com/watch?v=wrZRrzfMF3o
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    (Original post by CammieInfinity)
    ...
    Don't start being unpleasant or your time here could be very short.
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    Oh no say it isn't so, whatever will I do without your guidance. Plus i have several e-mails and signing up takes two ticks - you're fighting a lost battle hun. Xoxo
 
 
 
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