- #1

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I have no idea where to start. Can anyone who knows what to do give me a hint or tell me the first step? Thanks

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- Thread starter torasstripes
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- #1

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I have no idea where to start. Can anyone who knows what to do give me a hint or tell me the first step? Thanks

- #2

HallsofIvy

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That looks an awful lot like an exponential function to me. Take the logarithm of [itex](1+(x/n))^n[/itex] and you have a limit, as n goes to infinity, of the form [itex]0*\infty[/itex]. That can be rewritten as so that it is of the form "0/0" and, even though here n is an integer, you can use L'Hopital's rule.

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- #3

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That looks an awful lot like an exponential function to me. Take the logarithm of [itex]1+(x/n))^n and you have a limit, as n goes to infinity, of the form [itex]0*\infty[/itex]. That can be rewritten as so that it is of the form "0/0" and, even though here n is an integer, you can use L'Hopital's rule.

I'm pretty sure I can't pull the one out of the parenthesis like that. Like when you foil you always have a middle term? So with this we should have n+1 terms, but if we take the one out it chages the equation and we only have two terms if we expand it

- #4

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Maybe I'm wrong. If I can't come up with something better I will do that

- #5

HallsofIvy

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The logarithm of [itex](1+ (x/n))^n[/itex] is n log(1+ (x/n)) which, as I said, is of the form "[itex]0*\infty[/itex]". You can write that as log(1+ (x/n))/(1/n) so that it is now of the form "0/0". Apply L'Hopital's rule treating the n as a continuous variable.

- #6

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alright. thanks so much!

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