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Question

Evaluate limn(1n2+1+1n2+2+1n2+3+...+nn2+n)

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Solution

limn(1n2+1+1n2+2+1n2+3+......+nn2+n)

Now dividing both numerator and denominator by n2
limn1n21+1n2+1n21+2n2+1n21+3n2+......+1n21+1n
=01+01+01........01
=0

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