Quick question about limits involving sin. (1 Viewer)

[animeiswild]

Prove this:


Thanks =)

I can normally do this with n tending toward 0 but since n was tending toward infinity I got confused XD

 

[animeiswild]

Are you sure that's right?
http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+nsin(pi)/n

Sorry forgot some brackets XD it's supposed to be sin(pi/n) not sin(pi)/n

 

[LightXT]

Is this 2U?
http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+nsin%5B%28pi%29%2Fn%5D

 

[animeiswild]

Is this 2U?
http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+nsin%5B%28pi%29%2Fn%5D

It's from Yr 11 3u cambridge but could be 2u...

I can not understand the steps from that website... But that's the right equation.

 

[Shadowdude]

That's L'Hopital's rule for indeterminate forms of limits. That's first year uni stuff.

Are you trolling?

 

[animeiswild]

That's L'Hopital's rule for indeterminate forms of limits. That's first year uni stuff.

Are you trolling?

Sorry, are you referring to me? I am seriously asking a question here.

 

[LightXT]

Is it from the "Extension" part of Cambridge exercises? Because that shit is ridiculous.

 

[animeiswild]

Is it from the "Extension" part of Cambridge exercises? Because that shit is ridiculous.

No, it's not from the extension part, it's from the beginning of the development section.

 

[LightXT]

'sif. This is why I never used Cambridge. Because stuff like this never comes up in the 3U HSC.

 

[animeiswild]

If you don't believe me and still have your Cambridge book, I can tell you the page and question.

 

[Trebla]

lol this actually just uses a result from the syllabus that



By letting



the result follows noting that when



then



so we can rewrite the limit in terms of n rather than x after the substitution

Hence

 

[animeiswild]

lol this is actually just uses a result from the syllabus that



By letting , the result immediately follows

Sorry, I'm stupid. Please explain.

 

[study-freak]

That's L'Hopital's rule for indeterminate forms of limits. That's first year uni stuff.

Are you trolling?

My question: are YOU trolling?
It's doable with high school knowledge.

to OP: since nsin(pi/n)=sin(pi/n)/(1/n) since n=/=0

lim(n-->Inf) sin(pi/n)/(1/n)
=pi*lim(n-->Inf) sin(pi/n)/(pi/n)
=pi*1
=pi

using the limit: lim(m-->0) sin(m)/m=1
and lim(n-->Inf) (1/n)=0

 

[LightXT]

lol this is actually just uses a result from the syllabus that



By letting ,

the result immediately follows noting that



is equivalent to

This makes sense. Oh dear, my 3U is hazy.

 

[animeiswild]

. . . . . . I DOn't get it T____________T

Wait, gimme a moment to write it down....

 

[Trebla]

Sorry, I'm stupid. Please explain.

See the edited post.

 

[animeiswild]

See the edited post.

I saw. I could follow, then I lost it on the second last line. Where did the n at the beginning of sin go to?

Sorry... >.<

 

[Carrotsticks]

That's L'Hopital's rule for indeterminate forms of limits. That's first year uni stuff.

Are you trolling?

Although L'Hopital's rule is indeed a useful tool most of the time, it doesn't necessarily mean that we must use it the moment we see a limit problem

 

[animeiswild]

Wait, I understand. I get it now. Really, thank you for your help guys! =)