Question

高数极限，划圈的两道题求解

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Answer 1

第一题　先通分得（2-x-x^2）/(1-x^3)再约分得（2+x）/(1+x+x^2),最后求x->1时，上式的极限得1
第二题　先将括号内分子分母同乘以两数之和得　x*2/(sqrt(x^2+1)+sqrt(x^2-1)),由于x>0,分子分母同除以x得　2/(sqrt(1+1/(x^2))+sqrt(1-1/(x^2)),因为x->正无穷大，所以1/(x^2)->0,最后上式变为2/(sqrt(1)+sqrt(1))=2/2=1

Answer 2

Answer 3

lim(x->1)[ 3/(1-x^3) - 1/(1-x) ]
=lim(x->1) [3- (1+x+x^2) ]/(1-x^3)
=lim(x->1) -(x^2+x^2-2)/(1-x^3)
=lim(x->1) -(x+2)(x-1)/(1-x^3)
=lim(x->1) -(x+2)(x-1)/[(1-x)(1+x+x^2)]
=lim(x->1) (x+2)/(1+x+x^2)
=1

(2)
√(x^2+1) -√(x^2-1)
=[√(x^2+1) -√(x^2-1)] . [√(x^2+1) +√(x^2-1)]/[√(x^2+1) +√(x^2-1)]
=2/[√(x^2+1) +√(x^2-1)]

lim(x->∞) x.[√(x^2+1) -√(x^2-1)]
=lim(x->∞) 2x/[√(x^2+1) +√(x^2-1)]
=lim(x->∞) 2/[√(1+1/x^2) +√(1-1/x^2)]
=2/2
=1