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Rules for evaluating limits?

  • Thread starter danago
  • Start date
danago
Gold Member
1,122
4
Hi. We havnt started limits in class yet, but ive come across one, which i need to evaluate.

[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n}
[/tex]

For a limit like that, can i just substitute infinity into the equation? Or is there some rule against that? If so, would i be right in saying:

[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

Thanks for the help.
Dan.
 

Answers and Replies

Well sir dango

What your saying to me seams perfectly correct, however, limits are not my specialties. For the limit which you defined, that is the correct answer

unique_pavadrin

EDIT: dear sir dango what i had previously sated is incorrect, as n approaches n infinity, the function appraocd 2(pi)(r+R)
 
Last edited:
Gib Z
Homework Helper
3,344
4
[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

2pi r is a constant, take out of the limit. We have

[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.
 
danago
Gold Member
1,122
4
[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

2pi r is a constant, take out of the limit. We have

[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.
Ahh thanks very much for that. That answer is exactly what im after, considering the context im using the equation in :biggrin:
 
HallsofIvy
Science Advisor
Homework Helper
41,738
899
[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

2pi r is a constant, take out of the limit. We have

[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]
No, you have cancel the 2 [itex]2\pir[/itex] terms:
[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 0
[/tex][/quote]
Now what you say further makes sense.

Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.
 

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