Question

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.

Answer 1

Is \(n\) an integer ?

Answer 2

Yes n is an integer. :)

Answer 3

0

Answer 4

no

Answer 5

infinity

Answer 6

1

Answer 7

no

Answer 8

pi

Answer 9

e,

Answer 10

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

Answer 11

lulz :)

Answer 12

maybe complex numbers i dont like the root in there

Answer 13

no real sollution :)

Answer 14

you can get real solution from that

Answer 15

is it indeterminate?

Answer 16

is the answer even converging

Answer 17

no there is a answer

Answer 18

okay then ill work on it : )

Answer 19

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

Answer 20

No.name is right. It got answer.

Answer 21

You got the answer @TrojanPoem

Answer 22

?

Answer 23

Nearly , but more than 3 steps.

Answer 24

its ok tell your answer

Answer 25

im getting pi but im still messing wid it

Answer 26

@ganeshie8 you are correct !

Answer 27

psh i already guessed pi lol

Answer 28

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

Answer 29

bit embarrassing i kno

Answer 30

haha @ganeshie8 Like ++

Answer 31

You made my day @ganeshie8 :)

Answer 32

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

Answer 33

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

Answer 34

no odont

Answer 35

^
give us some time

Answer 36

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

Answer 37

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

Answer 38

Ok a small hint

Answer 39

nope wait almost there

Answer 40

ok ok

Answer 41

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

Answer 42

\[\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}}\]

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

are you squaring and sqrting the whole limit

Answer 50

or k^2=1+n^2!

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

look at my work, can u spot something wrong?

Answer 55

okie