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

1. amyna

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2. amyna

and i haven't learned l'hopitals rule, so i can't use it in this problem

3. freckles

$e^k=\lim_{n \rightarrow \infty}(1+\frac{k}{n})^n \text{ is a well known limit }$

4. amyna

yes but how to i use it?

5. freckles

in place of k you have -1...

6. amyna

since its a minus instead of a plus

7. freckles

1-something is the same as 1+(-something)

8. SolomonZelman

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9. amyna

so the answer would just be e^1

10. SolomonZelman

This is true $$\forall k$$

11. freckles

you mean e^(-1)?

12. amyna

ahhh okay i think i got it ! thanks guys! :)

13. SolomonZelman

You can demostrate this using Stirling's approximation for example.

14. 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)

15. 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.

16. SolomonZelman

that you can use to re-write the limit and reduce it (to e^k, if you do it correctly)

17. SolomonZelman

bye

18. amyna

the formula is much easier. Thanks everyone!

19. amyna

20. amyna

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21. SolomonZelman

$$\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{3n}}$$ $$\large\color{slate}{\displaystyle \left(\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}\right)^3}$$

22. SolomonZelman

you know the limit inside the ( ... )$$^3$$

23. SolomonZelman

amyna, you aren't replying.... what's the problem?

24. amyna

i don't get why the whole limit is cubed

25. SolomonZelman

Well, $$\Large \left(A\right)^{bc}=\left(A^b\right)^c$$

26. amyna

right, okay. then whats the next step?

27. SolomonZelman

well, what is: $$\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(1+\frac{x}{n}\right)^{n}}$$ can you tell me that?

28. amyna

e^x

29. 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?

30. amyna

e^x^3 ?

31. SolomonZelman

Yes $$(e^x)^3$$, or $$e^{3x}$$

32. amyna