弯弯月儿挂林梢
幼苗
共回答了15个问题采纳率:93.3% 举报
lim(x->2) [√(2+x) -2]/[√(3x+3) -3]
=lim(x->2) [(2+x) -4][√(3x+3) +3]/[(3x+3) -9][√(2+x) +2]
=lim(x->2) [x-2][√(3x+3) +3]/[3x-6][√(2+x) +2]
=lim(x->2) [√(3x+3) +3]/3[√(2+x) +2]
=[3+3]/3[√4 +2]
= 1/2
lim(x->+∞) √(x^2+x+1) -√(x^2-x+1)
=lim(x->+∞) [(x^2+x+1) -(x^2-x+1)]/[√(x^2+x+1) +√(x^2-x+1)]
=lim(x->+∞) 2x/[√(x^2+x+1) +√(x^2-x+1)]
= 2*lim(x->+∞) 1/[√(1+1/x+1/x^2) +√(1-1/x+1/x^2)]
= 1
lim(x->-∞) √(x^2+x+1) -√(x^2-x+1)
=lim(x->-∞) [(x^2+x+1) -(x^2-x+1)]/[√(x^2+x+1) +√(x^2-x+1)]
=lim(x->-∞) 2x/[√(x^2+x+1) +√(x^2-x+1)]
= 2*lim(x->-∞) 1/[-√(1+1/x+1/x^2) -√(1-1/x+1/x^2)]
= -1
1年前
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