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Answer:
Correct option is A)
n→∞
lim
n
m+11 m+2 m+3 m+....n mn→∞
limr=1∑n
m+1r m
=
n→∞lim
[ n1∑(nr) m ]∫ 01
x dx=[ m+1x m+1] 01=m+11
Ans: A
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