(b)lim(√n2+ 1-n) = lim(n2+ 1)-n2√n2+ 1 +n= lim1√n2+ 1 +n= lim1nq1 +1n2+ 1=lim1nlimq1 +1n2+ 1=01= 0.68
(c)lim(√n2+n-n) = lim(n2+ 1)-n2√n2+n+n= limn√n2+n+n= lim1q1 +1n+ 1=1limq1 +1n+ 1=11 + 1=12.69
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- Fall '08
- Staff
- Limits, lim, Limit of a function, Limit of a sequence, Suppose sn
