Differentiation Question 466
$ \underset{n\to \infty }{\mathop{\lim }},\frac{n^{p}{{\sin }^{2}}(n!)}{n+1},0<p<1 $ is equal to
Options:
0
B) $ \infty $
1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given limit can be written as $ \underset{n\to \infty}{\mathop{\lim }},\frac{{{\sin }^{2}}(n!)}{{n^{1-p}}(1+1/n)}(0<p<1) $ $ =\frac{Somerealnumberin[0,1]}{\infty }=0 $ $ (\because 1-p>0) $
BETA
