amyna
  • amyna
Find limit of a sequence. Please help i don't know where to start! (1-(1/n))^(n)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amyna
  • amyna
|dw:1444271970378:dw|
amyna
  • amyna
and i haven't learned l'hopitals rule, so i can't use it in this problem
freckles
  • freckles
\[e^k=\lim_{n \rightarrow \infty}(1+\frac{k}{n})^n \text{ is a well known limit }\]

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amyna
  • amyna
yes but how to i use it?
freckles
  • freckles
in place of k you have -1...
amyna
  • amyna
since its a minus instead of a plus
freckles
  • freckles
1-something is the same as 1+(-something)
SolomonZelman
  • SolomonZelman
|dw:1444272525064:dw|
amyna
  • amyna
so the answer would just be e^1
SolomonZelman
  • SolomonZelman
This is true \(\forall k\)
freckles
  • freckles
you mean e^(-1)?
amyna
  • amyna
ahhh okay i think i got it ! thanks guys! :)
SolomonZelman
  • SolomonZelman
You can demostrate this using Stirling's approximation for example.
SolomonZelman
  • SolomonZelman
well, stirling's approximation of n! I am talking about. (And consequentially it would work for only integers, but this is in fact true for all k)
SolomonZelman
  • SolomonZelman
just that for integers we got a demonstration. I won't post that, but x! ~x^x(√[2πx]) / e^x is the approximation of Stirling.
SolomonZelman
  • SolomonZelman
that you can use to re-write the limit and reduce it (to e^k, if you do it correctly)
SolomonZelman
  • SolomonZelman
bye
amyna
  • amyna
the formula is much easier. Thanks everyone!
amyna
  • amyna
how about this question: (1 + x/n)^3n what would i do to the exponent 3n?
amyna
  • amyna
|dw:1444273022800:dw|
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{3n}}\) \(\large\color{slate}{\displaystyle \left(\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}\right)^3}\)
SolomonZelman
  • SolomonZelman
you know the limit inside the ( ... )\(^3\)
SolomonZelman
  • SolomonZelman
amyna, you aren't replying.... what's the problem?
amyna
  • amyna
i don't get why the whole limit is cubed
SolomonZelman
  • SolomonZelman
Well, \(\Large \left(A\right)^{bc}=\left(A^b\right)^c\)
amyna
  • amyna
right, okay. then whats the next step?
SolomonZelman
  • SolomonZelman
well, what is: \(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}}\) can you tell me that?
amyna
  • amyna
e^x
SolomonZelman
  • SolomonZelman
yes, and when the whole thing is cubed, \(\large\color{red}{\displaystyle \left(\color{black}{\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}}\right)^3}\) then, what do you get?
amyna
  • amyna
e^x^3 ?
SolomonZelman
  • SolomonZelman
Yes \((e^x)^3\), or \(e^{3x}\)
amyna
  • amyna
ok so thats the answer! Thank you for your help! :)
SolomonZelman
  • SolomonZelman
You are welcome:)

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