The value of limn→∞ 8n2−n33+n is

limn→∞ 8n2−n33+n=limn→∞ 8n2−n33+n33=limn→∞ 8n2−n3+n38n2−n32/3+n32/3−8n2−n31/3n=limn→∞ 8n2n28n−12/3+n2−n28n−11/3=limn→∞ 88n−12/3+1−8n−11/3=83

The value of limn→∞ 8n2−n33+n is

limn→∞ 8n2−n33+n=limn→∞ 8n2−n33+n33=limn→∞ 8n2−n3+n38n2−n32/3+n32/3−8n2−n31/3n=limn→∞ 8n2n28n−12/3+n2−n28n−11/3=limn→∞ 88n−12/3+1−8n−11/3=83

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The value of $lim_{n \to \infty} (\sqrt[3]{8 n^{2} - n^{3}} + n)$ is

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$\frac{1}{3}$

$\frac{8}{3}$

$- \frac{1}{3}$

$- \frac{8}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} lim_{n \to \infty} (\sqrt[3]{8 n^{2} - n^{3}} + n) \\ = lim_{n \to \infty} (\sqrt[3]{8 n^{2} - n^{3}} + \sqrt[3]{n^{3}}) \\ = lim_{n \to \infty} \frac{8 n^{2} - n^{3} + n^{3}}{{(8 n^{2} - n^{3})}^{2 / 3} + {(n^{3})}^{2 / 3} - {(8 n^{2} - n^{3})}^{1 / 3} n} \\ = lim_{n \to \infty} \frac{8 n^{2}}{n^{2} {(\frac{8}{n} - 1)}^{2 / 3} + n^{2} - n^{2} {(\frac{8}{n} - 1)}^{1 / 3}} \\ = lim_{n \to \infty} \frac{8}{{(\frac{8}{n} - 1)}^{2 / 3} + 1 - {(\frac{8}{n} - 1)}^{1 / 3}} \\ = \frac{8}{3} \end{array}$