怎么做啊??
1个回答
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(5)因为分母的最高次数比分子大,所以极限为0
(6)原式=lim(x->1) (x+1)(x-1)/(x-1)(x-2)
=lim(x->1) (x+1)/(x-2)
=-2
(7)分子有理化
原式=lim(x->+∞) x(x^2+1-x^2)/[√(x^2+1)+x]
=lim(x->+∞) x/[√(x^2+1)+x]
=lim(x->+∞) 1/[√(1+1/x^2)+1]
=1/2
(8)分子分母分别有理化
原式=lim(x->1) (5x-4-x)[√x+1]/(x-1)[√(5x-4)+x]
=lim(x->1) 4(√x+1)/[√(5x-4)+x]
=4
(6)原式=lim(x->1) (x+1)(x-1)/(x-1)(x-2)
=lim(x->1) (x+1)/(x-2)
=-2
(7)分子有理化
原式=lim(x->+∞) x(x^2+1-x^2)/[√(x^2+1)+x]
=lim(x->+∞) x/[√(x^2+1)+x]
=lim(x->+∞) 1/[√(1+1/x^2)+1]
=1/2
(8)分子分母分别有理化
原式=lim(x->1) (5x-4-x)[√x+1]/(x-1)[√(5x-4)+x]
=lim(x->1) 4(√x+1)/[√(5x-4)+x]
=4
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