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.

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- ganeshie8

Is \(n\) an integer ?

- anonymous

##### 1 Attachment

- anonymous

Yes n is an integer. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- dan815

0

- anonymous

no

- dan815

infinity

- dan815

1

- anonymous

no

- dan815

pi

- dan815

e,

- anonymous

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

- dan815

lulz :)

- dan815

maybe complex numbers i dont like the root in there

- anonymous

no real sollution :)

- dan815

you can get real solution from that

- dan815

is it indeterminate?

- dan815

is the answer even converging

- anonymous

no there is a answer

- dan815

okay then ill work on it : )

- anonymous

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

- TrojanPoem

No.name is right. It got answer.

- anonymous

You got the answer @TrojanPoem

- anonymous

?

- TrojanPoem

Nearly , but more than 3 steps.

- anonymous

its ok tell your answer

- ganeshie8

im getting pi but im still messing wid it

- anonymous

@ganeshie8 you are correct !

- dan815

psh i already guessed pi lol

- anonymous

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

- ganeshie8

bit embarrassing i kno
|dw:1434985625298:dw|

- TrojanPoem

haha @ganeshie8 Like ++

- anonymous

You made my day @ganeshie8 :)

- ganeshie8

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

- anonymous

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

- dan815

no odont

- ganeshie8

^
give us some time

- anonymous

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

- anonymous

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

- anonymous

Ok a small hint

- ganeshie8

nope wait almost there

- anonymous

ok ok

- anonymous

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

- ganeshie8

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

- ganeshie8

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

- dan815

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

- dan815

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

- dan815

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

- dan815

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

- ganeshie8

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

- dan815

are you squaring and sqrting the whole limit

- dan815

or k^2=1+n^2!

- ganeshie8

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

- 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

- dan815

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

- dan815

look at my work, can u spot something wrong?

- ganeshie8

okie

- dan815

|dw:1434988291816:dw|

- dan815

first is that step legal

- dan815

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

- dan815

because k has remained an integer

- 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..

- dan815

oh true okay throw that out then

- anonymous

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

- ganeshie8

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

- 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
\[\left|n\sin\left(2\pi\sqrt{1+n^2}\right)-\pi\right|<\epsilon\]

- anonymous

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

- 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.

- ganeshie8

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

- xapproachesinfinity

can you do it without taylor?

- ganeshie8

yeah just use \(\sin x\approx x\)
;p

- ganeshie8

(jk)

- ParthKohli

Amazing!

- anonymous

Well shall i give my sollution?

- ganeshie8

Yes waiting

- anonymous

##### 1 Attachment

- ganeshie8

wow thats exact same as my solution

- anonymous

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

- ganeshie8

what do you mean it is in latex form

- anonymous

Like the only the code is visible not the output

- ganeshie8

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

- anonymous

nice but what is taylor

- ganeshie8

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition

- anonymous

thnx

- ganeshie8

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

- anonymous

Oh i know these expansions

Looking for something else?

Not the answer you are looking for? Search for more explanations.