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Question: \(\lim _ { n \rightarrow \infty }\) \(\frac{n^{k}\sin^{2}(n!)}{n + 2}\), 0 \< k \< 1, is equal to –...

lim⁡n→∞\lim _ { n \rightarrow \infty } nksin⁡2(n!)n+2\frac{n^{k}\sin^{2}(n!)}{n + 2}, 0 < k < 1, is equal to –

A

∞

B

1

C

0

D

None of these

Answer

0

Explanation

Solution

lim⁡n→∞\lim _ { n \rightarrow \infty } = lim⁡n→∞\lim _ { n \rightarrow \infty } nksin⁡2(n!)n(1+2n)\frac { \mathrm { n } ^ { \mathrm { k } } \sin ^ { 2 } ( \mathrm { n } ! ) } { \mathrm { n } \left( 1 + \frac { 2 } { \mathrm { n } } \right) }

= lim⁡n→∞\lim _ { n \rightarrow \infty } = a finite quantity ∞\frac { \text { a finite quantity } } { \infty }

[Q sin2 (n!) always lies between 0 and 1. Also, since 1 – k > 0, ∴ n1–k → ∞ as n → ∞] = 0.

Hence (3) is the correct answer.