- #1

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## 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

lim n-> infinity tan(pi/n)/(n*sin^2(2/n))

Any hints would be great

- Thread starter cateater2000
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- #1

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Hi i'm having trouble computing this limit

lim n-> infinity tan(pi/n)/(n*sin^2(2/n))

Any hints would be great

lim n-> infinity tan(pi/n)/(n*sin^2(2/n))

Any hints would be great

- #2

Tide

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- #3

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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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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

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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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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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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.cateater2000 said:

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

[tex]\lim_{x \rightarrow 0} \frac {x \tan \pi x}{\sin^2 \pi x}[/tex]

- #8

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

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