# Trouble computing this limit

1. Sep 29, 2004

### cateater2000

Hi i'm having trouble computing this limit
lim n-> infinity tan(pi/n)/(n*sin^2(2/n))

Any hints would be great

2. Sep 29, 2004

### Tide

How about using the small angle approximations to the sine and tangent (Taylor series - if you're familiar with that)? Otherwise, l'Hopital might come in handy.

3. Sep 29, 2004

### cateater2000

"How about using the small angle approximations to the sine and tangent"

And wouldn't l'hopitals be a little nasty? I don't think it'll work out
(tan(pi/n)/n)/sin^2(2/n)
this has form 0/0 so I apply l'hopitals and get something real nasty that won't simplify. I'll try simplifying it sometime tonight though thanks for your help

4. Sep 29, 2004

### Hurkyl

Staff Emeritus
These problems can often be done by using known limits to "replace" messy things with simple ones.

For example, you know that the limit of tan(pi/n) / (pi/n) = 1 in this case (I hope!) For your more complicated limit, you could simplify it by pulling the tan(pi/n) off to the left and dividing it by (pi/n), then multiplying the other stuff by (pi/n).

5. Sep 29, 2004

### cateater2000

so I'd get (tan(pi/n)/(pi/n))*[(pi/n)/(n(sin^2(2/n))]

I don't really see how that helps.

[(pi/n)/(n(sin^2(2/n))]
Not sure how to computer the limit of that thing.

Any ideas?

6. Sep 29, 2004

### Hurkyl

Staff Emeritus
Would you agree that, at least, [(pi/n)/(n(sin^2(2/n))] looks simpler than [tan(pi/n)/(n*sin^2(2/n))]?

While this new expression can be simplified somewhat (for instance, pulling the 1/n in the numerator into an n on the denominator), the big thing to realize is you can keep applying the trick I mentioned...

7. Sep 30, 2004

### Tide

You might be able to see your way through l'Hopital if you replace 1/n with x and look at the limit as x -> 0.

$$\lim_{x \rightarrow 0} \frac {x \tan \pi x}{\sin^2 \pi x}$$

8. Sep 30, 2004

### cateater2000

thanks hurkyl and tide I was able to compute the limit both ways. You were of great help!