Search 75,697 tutors
FIND TUTORS
Ask a question
1 0

Which of the following?

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.
Tutors, please sign in to answer this question.

2 Answers

Remember the limit as x→0 of (sinx)/x is 1
Your problem is an=n*sin(3π/n), which may be restated as
an=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

Comments

Comment

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

Comments

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

Comment