Trouble computing this limit

  • #1

Main Question or Discussion Point

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

Any hints would be great
 

Answers and Replies

  • #2
Tide
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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
"How about using the small angle approximations to the sine and tangent"

not sure about that.

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
Hurkyl
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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
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
Hurkyl
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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
Tide
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cateater2000 said:
"How about using the small angle approximations to the sine and tangent"

not sure about that.

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
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.

[tex]\lim_{x \rightarrow 0} \frac {x \tan \pi x}{\sin^2 \pi x}[/tex]
 
  • #8
thanks hurkyl and tide I was able to compute the limit both ways. You were of great help!
 

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