Evaluating Limits: How Do I Substitute Infinity Into the Equation?

[danago]

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.

 

[unique_pavadrin]

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)

 

[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.

 

[danago]

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

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

 

[HallsofIvy]

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

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.