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Q.

12limnx[(1+1n2)(1+22n2)(1+n2n2)]1/n=

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a

e

b

2e

c

2eπ22

d

eπ42

answer is D.

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Detailed Solution

y=limn((1+1n2)(1+22n2).......(1+n2n2))1/n

logy=limn1n

[log(1+1n2)+log(1+22n2)+.....+log(1+n2n2)]

logy=limn1nΣlog(1+r2n2)

=01log(1+x2)dx

=(log(1+x2)1dxddxlog(1+x2).1  dx)01

=(xlog(1+x2)2x21+x2dx)01=2eπ42

 

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12limn→x[(1+1n2)(1+22n2)−−−(1+n2n2)]1/n=