求第一题极限值为多少
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1 想办法有理化
上下各乘以(sqrt(n^2+7n+2)+sqrt(n^2+n+1))
原式=lim（n^2+7n+2-n^2-n-1）/(sqrt(n^2+7n+2)+sqrt(n^2+n+1))
=lim（6n+1）/(sqrt(n^2+7n+2)+sqrt(n^2+n+1))
=lim(6+1/n)/(sqrt(1+7/n+2/n^2)+sqrt(1+1/n+1/n^2))
=6/2
=3
上下各乘以(sqrt(n^2+7n+2)+sqrt(n^2+n+1))
原式=lim（n^2+7n+2-n^2-n-1）/(sqrt(n^2+7n+2)+sqrt(n^2+n+1))
=lim（6n+1）/(sqrt(n^2+7n+2)+sqrt(n^2+n+1))
=lim(6+1/n)/(sqrt(1+7/n+2/n^2)+sqrt(1+1/n+1/n^2))
=6/2
=3
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