limit n-->infinty

**metallica007** · December 21st 2009, 07:16 PM

Hi my friends
can you help to find the limit of the following equation

Area = $\lim_{n \rightarrow \infty}$ $\frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})$

**Bruno J.** · December 21st 2009, 07:31 PM

Well think about it. $\left|\sin\left(\frac{\pi(n+1)}{2}\right)\right|$ is always less than or equal to $1$ , and $\frac{\pi}{n}$ gets as small as you wish...

**firebio** · December 21st 2009, 07:41 PM

Originally Posted by metallica007

Hi my friends
can you help to find the limit of the following equation

Area = $\lim$ n--> $\infty$ $\frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})$

Well u can use the squeeze theorem.

-1<= sinx<=1
so
lim n => inf -pi/n<= pi/n *sinx<=pi/n
From this, is pretty obvious what the answer is...

**metallica007** · December 21st 2009, 07:52 PM

Guys your killing me

I did not get it
i know this :
lim n--> infinity pi/n =0
lim n--> infinity $sin(\frac{\pi*n+\pi}{2})$ does not exist
so the answer is ... 0 !!!

please correct me if i am wrong

**firebio** · December 21st 2009, 07:55 PM

Originally Posted by metallica007

Guys your killing me

I did not get it
i know this :
lim n--> infinity pi/n =0
lim n--> infinity $sin(\frac{\pi*n+\pi}{2})$ does not exist
so the answer is ... 0 !!!

please correct me if i am wrong

Sin always exist between -1 and 1...
and answer is correct

**metallica007** · December 21st 2009, 08:13 PM

but the answer doesn't seems right to me
because i tried to find the area of sin(x) from 0 to pi using limit --> infinity and Riemann sums and i got 0
but when i tried to integrate it using integration rules and i get 2 !!

**metallica007** · December 21st 2009, 09:58 PM

i am , doing it like this
$\Delta x =$ $\frac{\pi}{n}$

$\lim n-->\infty \sum Sin(xi) \Delta x$

$\lim n-->\infty\\\\ Sin(\frac{\pi}{n} \frac{n(n+1)}{2}) \frac{\pi}{n}$

$\lim$ n--> $\infty$ $\frac{\pi}{n} *sin(\frac{\pi n+\pi}{2})$

I dont know if that was the right steps to find area using Riemann sum and limit

please correct me if i am wrong

**VonNemo19** · December 22nd 2009, 08:50 PM

0... The answer is 0...

**metallica007** · December 22nd 2009, 09:06 PM

Originally Posted by VonNemo19

0... The answer is 0...

when a =0 , b = pi

does $\int\sin(x)= 0$ !!!!

**VonNemo19** · December 22nd 2009, 09:08 PM

Originally Posted by metallica007

when a =0 , b = pi

does $\int\sin(x)= 0$ !!!!

No, actually its 1

**Drexel28** · December 22nd 2009, 09:33 PM

Originally Posted by VonNemo19

No, actually its 1

2

**metallica007** · December 23rd 2009, 02:52 PM

Guys , what about my question
why i get 0 instead of 2 ( or 1 ) !!!!

**skeeter** · December 23rd 2009, 03:15 PM

Originally Posted by metallica007

Guys , what about my question
why i get 0 instead of 2 ( or 1 ) !!!!

as stated before ...

$-1 \le \sin(anything) \le 1$

so ...

$\lim_{n \to \infty} \, \frac{\pi}{n}(a \, value \, stuck \, between \, -1 \, and \, 1) = 0 <br />$

now look at the graph

**Plato** · December 23rd 2009, 03:37 PM

Originally Posted by metallica007

can you help to find the limit of the following equation
Area = $\lim _{n \to \infty }\frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})$

It seems to me that this thread is very confused.
It seem that the OP assumes that the above limit is an approximating sum for $\int\limits_0^\pi {\sin (x)dx}$ .
But that is totally wrong. A correct approximating sum (righthand sum) is $\frac{\pi }{n}\sum\limits_{k = 1}^n {\sin \left( {\frac{\pi }{n}k} \right)}$ .
Thus we want this limit: $\lim _{n \to \infty } \frac{\pi }{n}\sum\limits_{k =1 }^n {\sin \left( {\frac{\pi }{n}k} \right)}$ .

**skeeter** · December 23rd 2009, 03:57 PM

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