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. Mar 28 | Sun from Los Angeles, CA | 2 Answers | 1 Vote Mark favorite Subscribe Comment
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 Mar 29 | Michael F. Comment Comments Thank you so much, Michael. Mar 29 | Sun from Los Angeles, CA 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. Mar 28 | Steve S. Comment Comments What's the answer then? Mar 28 | Sun from Los Angeles, CA Sorry, Sun. Wasn't thinking hard enough first time. Mar 29 | Steve S. Comment
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