## anonymous one year ago A limit (challenge) problem for smart people. In the question n is an integer Condition:- Try to solve it in not more than 3-4 steps.

1. ganeshie8

Is $$n$$ an integer ?

2. anonymous

3. anonymous

Yes n is an integer. :)

4. dan815

0

5. anonymous

no

6. dan815

infinity

7. dan815

1

8. anonymous

no

9. dan815

pi

10. dan815

e,

11. anonymous

u need to show your method , don't just give out answers like a idiot

12. dan815

lulz :)

13. dan815

maybe complex numbers i dont like the root in there

14. anonymous

no real sollution :)

15. dan815

you can get real solution from that

16. dan815

is it indeterminate?

17. dan815

18. anonymous

19. dan815

okay then ill work on it : )

20. anonymous

It dosen't have stupid answers like it does not exist etc

21. TrojanPoem

No.name is right. It got answer.

22. anonymous

23. anonymous

?

24. TrojanPoem

Nearly , but more than 3 steps.

25. anonymous

26. ganeshie8

im getting pi but im still messing wid it

27. anonymous

@ganeshie8 you are correct !

28. dan815

psh i already guessed pi lol

29. anonymous

Ok , before i post my sollution , how did u get it @ganeshie8

30. ganeshie8

bit embarrassing i kno |dw:1434985625298:dw|

31. TrojanPoem

haha @ganeshie8 Like ++

32. anonymous

You made my day @ganeshie8 :)

33. ganeshie8

I'm glad you find it was entertaining :) please provide a hint as im kinda stuck..

34. anonymous

no problem , never mind Well the hint is itself the whole sollution , so let me post the sollution itself ok?

35. dan815

no odont

36. ganeshie8

^ give us some time

37. anonymous

Yeah don't give up so easily , u will get it!

38. anonymous

After giving hint it is easy , so i will not give any hint :)

39. anonymous

Ok a small hint

40. ganeshie8

nope wait almost there

41. anonymous

ok ok

42. anonymous

Don't even think of binomial expansions , you will get the answer no doubt , but that's a waste of time and space

43. ganeshie8

\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ \end{align}}

44. ganeshie8

next i want to use sinx/x limit but im gonna need more paper, one sec..

45. dan815

its so weird i cant believe the 1 is skewing it form 2pi to pi

46. dan815

for a big number sqrt(1+n^2) is pretty much sqrt(n^2)

47. dan815

so i said it was just n*sin(2pi*n) and thats inf * 0 so from lohopitals ud get 2pi

48. dan815

but that +1 in there changed it to pi O_O still working

49. ganeshie8

I just figured out a really neat way to solve this w/o using lhopital xD

50. dan815

are you squaring and sqrting the whole limit

51. dan815

or k^2=1+n^2!

52. ganeshie8

nvm its not working, was trying to interpret the given expression as a variation of area of regular polygon formula... but no success

53. ganeshie8

as $$n\to\infty$$, the regular polygon becomes an unit circle evaluating the limit to $$\pi$$ thought it would be that simple but looks tricky to conclude like that

54. dan815

ya not sure what im doing wrong, i keep getting 2pi again

55. dan815

look at my work, can u spot something wrong?

56. ganeshie8

okie

57. dan815

|dw:1434988291816:dw|

58. dan815

first is that step legal

59. dan815

can i just square and sqrt the lmit like that, now this lets me use lhopitals rule a couple times with no problems

60. dan815

because k has remained an integer

61. ganeshie8

that is okay because sqrt(x) is a continuous function $\lim \sqrt{f(x)} = \sqrt{\lim f(x)}$ But aren't you assuming $$\sqrt{1+n^2}$$ is an integer which is incorrect ? let me go read it agian..

62. dan815

oh true okay throw that out then

63. anonymous

Does this count? Step 1: Make a conjecture about the limit value ($$\pi$$). Step 2: Prove it.

64. ganeshie8

i smell epsilon delta but it smells good to me :)

65. anonymous

I was actually thinking monotone convergence, but showing either condition might be a bit tricky. As for an $$\epsilon$$ proof, not sure how much analysis would be required relative to monotone convergence. We have to find some $$N$$ for which $$n>N$$ yields $\left|n\sin\left(2\pi\sqrt{1+n^2}\right)-\pi\right|<\epsilon$

66. anonymous

(It should go without saying that $$\epsilon>0$$)

67. anonymous

My point of view is: When a function has the limit where n tends to infinity, then the result from sine will eventually tend to undefinition, meaning the function becomes undefined.

68. ganeshie8

\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ &=\lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} +\mathcal{O}(1/n^2)\right) ~~~\color{gray}{(\text{taylor})}\\~\\ &= \lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} \right) + 0 \\~\\ &=\pi \end{align}}

69. xapproachesinfinity

can you do it without taylor?

70. ganeshie8

yeah just use $$\sin x\approx x$$ ;p

71. ganeshie8

(jk)

72. ParthKohli

Amazing!

73. anonymous

Well shall i give my sollution?

74. ganeshie8

Yes waiting

75. anonymous

76. ganeshie8

wow thats exact same as my solution

77. anonymous

I cannot see your sollution , it is in latex form , so i am not able to understand it. nice if u solved it

78. ganeshie8

what do you mean it is in latex form

79. anonymous

Like the only the code is visible not the output

80. ganeshie8

Ahh ok some problem with rendering here it is |dw:1435050449486:dw|

81. anonymous

nice but what is taylor

82. ganeshie8
83. anonymous

thnx

84. ganeshie8

Here taylor series of most common functions |dw:1435050691842:dw|

85. anonymous

Oh i know these expansions