Quick limit question... (1 Viewer)

[Js^-1]

If we say that:

p/q = 2 (n*sin(π/2n))²

What is the limiting value of p/q as n --> ∞

 

[conics2008]

p/q = 2 (n*sin(π/2n))2

n --> inf

we need to make it in the form of sinax/ax ok

from 2unit sinax/ax =1

so follow this.. i remember doing this, this is the 2007 exam paper it was straight forward and also the calculator is your friend..

 

[lolokay]

well lim n->∞ sin(pi/2n)/(pi/2n) = 1
(2/pi) n*sin(pi/2n) = 1
n*sin(pi/2n) = pi/2
2(n*sin(pi/2n))² = pi²/2
I think

 

[Timothy.Siu]

well lim n->∞ sin(pi/2n)/(pi/2n) = 1
(2/pi) n*sin(pi/2n) = 1
n*sin(pi/2n) = pi/2
2(n*sin(pi/2n))² = pi²/2
I think

seems correct

 

[Js^-1]

Ah yeh I remember the sinx/x thing.

So 2( sin(pi/2)/(pi/2) * (pi/2) )²
n-->inf = 2( (pi/2))²
= 2(pi)²/4
= (pi)²/2 ?

 

[Js^-1]

Awesome thanks heaps people.

 

[H4rdc0r3]

is this 2 unit?

 

[Timothy.Siu]

is this 2 unit?

can u read

 

[Js^-1]

It was the very last question of last years 2007 Extension 2 paper, but the method is essentially 2 unit I think.

 

[Timothy.Siu]

It was the very last question of last years 2007 Extension 2 paper, but the method is essentially 2 unit I think.

lol, not really 2unit i dont think, u dont learn sin x/x = 1 as x--> infinity
maybe 3unit

 

[conics2008]

its def 2unit..

its just basic limits question.. good on ya js^-1 but in the exax remember that the calculator is your firend.. xD

you can just sub in a really large number xD
and hope for the ebst on from there...

but keep in mind u need to use what ever u have..

 

[munch0r]

lol, not really 2unit i dont think, u dont learn sin x/x = 1 as x--> infinity
maybe 3unit

where exactly is sin x/x = 1 as x--> infinity in any textbook.
i can only find sin x/x = 0 as x --> infinity according to the "squeeze theorem"

 

[Js^-1]

Thanks conics...that question 8 was actually likeable - It didn't make me want to shoot myself just by looking at it.

 

[lolokay]

where exactly is sin x/x = 1 as x--> infinity in any textbook.
i can only find sin x/x = 0 as x --> infinity according to the "squeeze theorem"

as n->∞
pi/2n -> 0
so it's the same thing pretty much

(as x->∞, sinx/x ->0 since x keeps getting larger, but sinx doesn't exceed +-1)

 

[Js^-1]

where exactly is sin x/x = 1 as x--> infinity in any textbook.
i can only find sin x/x = 0 as x --> infinity according to the "squeeze theorem"

Maths in Focus: Two/Three unit Book two, page 182.

Its says: Lim_x-->0 sinx/x = 1

Substitute x for pi/2n

As n--> Infinity
x--> 0

 

[tacogym27101990]

i thought the rule was lim x>0 sinx/x =1
i cant remember learning that lim x>infinity sinx/x =1
???

 

[conics2008]

Thanks conics...that question 8 was actually likeable - It didn't make me want to shoot myself just by looking at it.

yeah believe it or not, 2007 paper is my favourite paper over all.. i actually did all of question 8 without encountring too much pain unlike some other past papers..

I hope this year paper is the same level..

try doing paper 1993 xD

 

[munch0r]

i thought the rule was lim x>0 sinx/x =1
i cant remember learning that lim x>infinity sinx/x =1
???

as x approaches infinity (sinx)/x = 0
but for the question

sin(pi/2n) / (pi/2n)
as n approaches infinity
pi/2n approaches 0

so think of pi/2n as x

(sinx)/x = 1 where x is approaching 0

 

[tacogym27101990]

as x approaches infinity (sinx)/x = 0
but for the question

sin(pi/2n) / (pi/2n)
as n approaches infinity
pi/2n approaches 0

so think of pi/2n as x

(sinx)/x = 1 where x is approaching 0

ooo i get it now
thanks

 

[Js^-1]

yeah believe it or not, 2007 paper is my favourite paper over all.. i actually did all of question 8 without encountring too much pain unlike some other past papers..

I hope this year paper is the same level..

try doing paper 1993 xD

Yeah same...maybe tomorow. My head hurts enough for one day.