Yahoo Answers: Answers and Comments for Determine whether the sequence is divergent or convergent.
lim
n→∞ 2n^3+sin^2(7n)/ n^3 +9? [Mathematics]
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From FERNANDA
enGB
Tue, 22 Oct 2019 02:28:38 +0000
3
Yahoo Answers: Answers and Comments for Determine whether the sequence is divergent or convergent.
lim
n→∞ 2n^3+sin^2(7n)/ n^3 +9? [Mathematics]
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https://uk.answers.yahoo.com/question/index?qid=20191022022838AAw38VU
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From rotchm: Factor our n^3 from the top & bottom.
What...
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Tue, 22 Oct 2019 02:47:45 +0000
Factor our n^3 from the top & bottom.
What do you get?
Evaluate the limit of that.
Done!

From ted s: it should be obvious that it converges to '...
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Tue, 22 Oct 2019 04:57:04 +0000
it should be obvious that it converges to '  2 '

From Φ² = Φ+1: Convergent
Limit: 2
as n→∞ (2n³)/n³ = 2
th...
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Tue, 22 Oct 2019 02:35:35 +0000
Convergent
Limit: 2
as n→∞ (2n³)/n³ = 2
the sine function and the 9 are both negligible for very large values of n.

From Geeganage W: lim n→∞ [2n^3+sin^2(7n)]/ (n^3 +9) = lim n→...
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Tue, 22 Oct 2019 06:41:31 +0000
lim n→∞ [2n^3+sin^2(7n)]/ (n^3 +9) = lim n→∞ {2 + (1/n)[(sin7n)/(n)]^2}/[1+9/(n^3)] = (2+0)/1 = 2. It's convergent.
Here, sin(7n) = (7n)[(7n)^3]/3!+[(7n)^5]/5!…
(1/n)sin(7n) = 7  [(7^3*n^2)]/3!+[(7^5*n^4)]/5!…
n→∞, (1/n)[(1/n)sin(7n)]^2 =n→∞ (1/n){7  [(7^3*n^2)]/3!+[(7^5*n^4)]/5!…}^2 = 0*∞ =0.