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amyna
 one year ago
Find limit of a sequence. Please help i don't know where to start!
(1(1/n))^(n)
amyna
 one year ago
Find limit of a sequence. Please help i don't know where to start! (1(1/n))^(n)

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amyna
 one year ago
Best ResponseYou've already chosen the best response.1and i haven't learned l'hopitals rule, so i can't use it in this problem

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[e^k=\lim_{n \rightarrow \infty}(1+\frac{k}{n})^n \text{ is a well known limit }\]

amyna
 one year ago
Best ResponseYou've already chosen the best response.1yes but how to i use it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1in place of k you have 1...

amyna
 one year ago
Best ResponseYou've already chosen the best response.1since its a minus instead of a plus

freckles
 one year ago
Best ResponseYou've already chosen the best response.11something is the same as 1+(something)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444272525064:dw

amyna
 one year ago
Best ResponseYou've already chosen the best response.1so the answer would just be e^1

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1This is true \(\forall k\)

amyna
 one year ago
Best ResponseYou've already chosen the best response.1ahhh okay i think i got it ! thanks guys! :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1You can demostrate this using Stirling's approximation for example.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1well, 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
 one year ago
Best ResponseYou've already chosen the best response.1just 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
 one year ago
Best ResponseYou've already chosen the best response.1that you can use to rewrite the limit and reduce it (to e^k, if you do it correctly)

amyna
 one year ago
Best ResponseYou've already chosen the best response.1the formula is much easier. Thanks everyone!

amyna
 one year ago
Best ResponseYou've already chosen the best response.1how about this question: (1 + x/n)^3n what would i do to the exponent 3n?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{3n}}\) \(\large\color{slate}{\displaystyle \left(\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}\right)^3}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you know the limit inside the ( ... )\(^3\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1amyna, you aren't replying.... what's the problem?

amyna
 one year ago
Best ResponseYou've already chosen the best response.1i don't get why the whole limit is cubed

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Well, \(\Large \left(A\right)^{bc}=\left(A^b\right)^c\)

amyna
 one year ago
Best ResponseYou've already chosen the best response.1right, okay. then whats the next step?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1well, what is: \(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}}\) can you tell me that?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yes \((e^x)^3\), or \(e^{3x}\)

amyna
 one year ago
Best ResponseYou've already chosen the best response.1ok so thats the answer! Thank you for your help! :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1You are welcome:)
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