# Rules for evaluating limits?

1. Feb 12, 2007

### danago

Hi. We havnt started limits in class yet, but ive come across one, which i need to evaluate.

$$\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n}$$

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:

$$\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r$$

Thanks for the help.
Dan.

2. Feb 12, 2007

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

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: Feb 12, 2007
3. Feb 12, 2007

### Gib Z

$$\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r$$

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

$$\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r$$

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.

4. Feb 12, 2007

### danago

Ahh thanks very much for that. That answer is exactly what im after, considering the context im using the equation in

5. Feb 12, 2007

### HallsofIvy

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