_{n}=n*sin(3π/n), which may be restated as

_{n}=sin(3π/n)/(1/n)=3π(sin(3π/n)/(3π/n))

Which of the following is the limit of the sequence with nth term an=n*sin(3pi/n)?

a) 1

b) pi

c) 2pi

d) 3pi

e) 4pi

The n for an is lower case. Do I supposed to take the limit as n approaches to infinity? Please show all your work.

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Michael F. | Mathematics TutorMathematics Tutor

Remember the limit as x→0 of (sinx)/x is 1

Your problem is a_{n}=n*sin(3π/n), which may be restated as

a_{n}=sin(3π/n)/(1/n)=3π(sin(3π/n)/(3π/n))

as n→∞ we have 3π/n→0 and so the limit is 3π×1=3π

“Which of the following is the limit of the sequence with nth term an=n*sin(3pi/n)?

a) 1

b) pi

c) 2pi

d) 3pi

e) 4pi"

==== Between the bars is wrong first answer.

Looks to me like an has no limit; in fact it goes to infinity as n goes to infinity.

The sine factor oscillates between -1 and +1, so does not reduce the growing n.

=====

This is right:

As n → ∞,

3pi/n → 0,

sin(3pi/n) → 3pi/n,

n*sin(3pi/n) → n*3pi/n = 3pi.

So the answer is D.

check: GeoGebra graph: http://www.wyzant.com/resources/files/267291/limit_of_x_sin_3_pi_x

a) 1

b) pi

c) 2pi

d) 3pi

e) 4pi"

==== Between the bars is wrong first answer.

Looks to me like an has no limit; in fact it goes to infinity as n goes to infinity.

The sine factor oscillates between -1 and +1, so does not reduce the growing n.

=====

This is right:

As n → ∞,

3pi/n → 0,

sin(3pi/n) → 3pi/n,

n*sin(3pi/n) → n*3pi/n = 3pi.

So the answer is D.

check: GeoGebra graph: http://www.wyzant.com/resources/files/267291/limit_of_x_sin_3_pi_x

While it's true that we can use the approximation sin(θ) ≈ θ radians for small θ, Michael's answer is the proof.

What's the answer then?

Sorry, Sun. Wasn't thinking hard enough first time.

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