A) 0
B) \[\infty \]
C) 1
D) None of these
Correct Answer: A
Solution :
[a] Given limit can be written as \[\underset{n\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}(n!)}{{{n}^{1-p}}(1+1/n)}(0<p<1)\] \[=\frac{Some\text{ }real\text{ }number\text{ }in[0,1]}{\infty }=0\]\[(\because 1-p>0)\]
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