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$\lim\limits_{n\to\infty}\large\frac{n^p\sin^2(n!)}{n+1}$$\qquad0 < p < 1$ is equal to

$(a)\;0\qquad(b)\;\infty\qquad(c)\;1\qquad(d)\;None\;of\;these$

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1 Answer

$\lim\limits_{n\to \infty}\large\frac{sin^2(n!)}{n^{1-p}(1+\Large\frac{1}{n})}$
$\Rightarrow \large\frac{\text{some number between 0 and 1}}{\infty}$
$\Rightarrow 0$
Hence (a) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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