# Determine whether the sequence is divergent or convergent. lim n→∞ -2n^3+sin^2(7n)/ n^3 +9?

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- Geeganage WLv 57 months ago
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.

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- ted sLv 77 months ago
it should be obvious that it converges to ' - 2 '

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- rotchmLv 77 months ago
Factor our n^3 from the top & bottom.

What do you get?

Evaluate the limit of that.

Done!

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- Φ² = Φ+1Lv 77 months ago
Convergent

Limit: -2

as n→∞ (-2n³)/n³ = -2

the sine function and the 9 are both negligible for very large values of n.

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