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Q.
$\displaystyle\lim_{ n \rightarrow \infty} \frac{ n ^{ p } \sin ^{2}( n !)}{ n +1}, 0< p <1$ is equal to
Limits and Derivatives
Solution:
limit $=\displaystyle\lim_{n \rightarrow \infty} \frac{\sin ^{2}(n !)}{n^{1-p}(1+1 / n)}(0< p<1)$
$=\frac{\text { some real number in }[0,1]}{\infty}=0. \,\,\,\,(\because 1- p >0)$