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Question

limn((nn+1)α+sin1n)n when α ϵ Q is equal to

A
e1α
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B
-α
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C
e
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D
e1+α
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Solution

The correct option is A e1α
limn((nn+1)α+sin1n)n
Replace n with 1x,
Now as n,x0

Limit becomes
L=limx0((x+1)α+sinx)1x

Taking log on both sides, we get
lnL=limx0ln((x+1)α+sinx)x

It os forming the 00 form,
So, using L'Hospital's Rule, we get
lnL=limx0α(x+1)α1+cosx(x+1)α+sinx

lnL=1α
L=e1α

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