Explained: "Wn=3n5−2n3+n6n7+4n3−8, Yn=6n5+2n2−13n2(2n3+1), Zn=1−sin⁡(2nπ5)"

Janiyah Hays

Answered question

2022-03-30

W_{n} = \frac{3 n^{\sqrt{5}} - 2 n^{3} + n}{6 n^{\sqrt{7}} + 4 n^{3} - 8}

Y_{n} = \frac{6 n^{5} + 2 n^{2} - 1}{3 n^{2} (2 n^{3} + 1)}

Z_{n} = 1 - \sin (\frac{2 n π}{5})

Answer & Explanation

Adan Berry

Beginner2022-03-31Added 12 answers

Given
$W_{n} = \frac{3 n^{\sqrt{5}} - 2 n^{3} + n}{6 n^{\sqrt{7}} + 4 n^{3} - 8}$
$Y_{n} = \frac{6 n^{5} + 2 n^{2} - 1}{3 n^{2} (2 n^{3} + 1)}$
$Z_{n} = 1 - \sin (\frac{2 n π}{5})$
Now,
1) $lim_{n \to \infty} W_{n} = lim_{n \to \infty} \frac{3 n^{\sqrt{5}} - 2 n^{3} + n}{6 n^{\sqrt{7}} + 4 n^{3} - 8}$
$lim_{n \to \infty} W_{n} = lim_{n \to \infty} \frac{- 2 n^{3} + n + 3 n^{\sqrt{5}}}{4 n^{3} - 8 + 6 n^{\sqrt{7}}}$
$\lim_{n \to \infty} W_{n} = \lim_{n \to \infty} \frac{n^{3} (- 2 + \frac{1}{n^{2}} + \frac{3 n^{\sqrt{5}}}{n^{3}})}{n^{3} (4 - \frac{8}{n^{3}} + \frac{6 n^{\sqrt{7}})}{n^{3}})}$
$lim_{n \to \infty} W_{n} = lim_{n \to \infty} \frac{- 2 + \frac{1}{n^{2}} + \frac{3 n^{\sqrt{5}}}{n^{3}}}{4 - \frac{8}{n^{3}} + \frac{6 n^{\sqrt{7}}}{n^{3}}}$
$lim_{n \to \infty} W_{n} = \frac{- 2 + 0 + 0}{4 - 0 + 0}$
$lim_{n \to \infty} W_{n} = - \frac{1}{2}$
2) $lim_{n \to \infty} Y_{n} = lim_{n \to \infty} \frac{6 n^{5} + 2 n^{2} - 1}{3 n^{2} (2 n^{3} + 1)}$
$lim_{n \to \infty} Y_{n} = lim_{n \to \infty} \frac{n^{5} (6 + \frac{2}{n^{3}} - \frac{1}{n^{5}})}{3 n^{2} n^{3} (2 + \frac{1}{n^{3}})}$
$lim_{n \to \infty} Y_{n} = lim_{n \to \infty} \frac{n^{5} (6 + \frac{2}{n^{3}} - \frac{1}{n^{5}})}{3 n^{5} (2 + \frac{1}{n^{3}})}$
$lim_{n \to \infty} Y_{n} = lim_{n \to \infty} \frac{(6 + \frac{2}{n^{3}} - \frac{1}{n^{5}})}{3 (2 + \frac{1}{n^{3}})}$
$lim_{n \to \infty} Y_{n} = \frac{6 + 0 - 0}{3 (2 + 0)}$
$lim_{n \to \infty} Y_{n} = 1$
3) $lim_{n \to \infty} Z_{n} = lim_{n \to \infty} [1 - \sin (\frac{2 n π}{5})]$
As we don't know the value of the second term $\sin (\frac{2 n π}{5})$ when $n \to \infty$ $Z_{n}$ diverges.

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