# Rules for evaluating limits?

• danago

#### danago

Gold Member
Hi. We havnt started limits in class yet, but I've 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.

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

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.

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

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.

Ahh thanks very much for that. That answer is exactly what I am after, considering the context I am using the equation in $$\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r$$

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

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

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.