数学极限 需要详细步骤
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解决时间 2021-03-22 10:05
- 提问者网友：雾里闻花香
- 2021-03-21 18:54
数学极限 需要详细步骤
最佳答案
- 五星知识达人网友：有你哪都是故乡
- 2021-03-21 19:20
(1)
lim(x->∞) (x^2+2x+1)/(x^4-3x+1)
=lim(x->∞) (1/x^2+2/x^3+1/x^4)/(1-3/x^3+1/x^4)
=0
(2)
lim(x->1)[ 1/(1-x) -3/(1-x^3)]
=lim(x->1) { 1/(1-x) -3/[(1-x)(1+x+x^2)] }
=lim(x->1) (1+x+x^2 -3)/[(1-x)(1+x+x^2)]
=lim(x->1) (x^2+x -2)/[(1-x)(1+x+x^2]
=lim(x->1) (x-1)(x+2)/[(1-x)(1+x+x^2)]
=lim(x->1) -(x+2)/(1+x+x^2)
=-1
(3)
lim(x->∞) x.[√(x^2+1) -x ]
=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/(1+1)
=1/2
lim(x->∞) (x^2+2x+1)/(x^4-3x+1)
=lim(x->∞) (1/x^2+2/x^3+1/x^4)/(1-3/x^3+1/x^4)
=0
(2)
lim(x->1)[ 1/(1-x) -3/(1-x^3)]
=lim(x->1) { 1/(1-x) -3/[(1-x)(1+x+x^2)] }
=lim(x->1) (1+x+x^2 -3)/[(1-x)(1+x+x^2)]
=lim(x->1) (x^2+x -2)/[(1-x)(1+x+x^2]
=lim(x->1) (x-1)(x+2)/[(1-x)(1+x+x^2)]
=lim(x->1) -(x+2)/(1+x+x^2)
=-1
(3)
lim(x->∞) x.[√(x^2+1) -x ]
=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/(1+1)
=1/2
全部回答
- 1楼网友：夜风逐马
- 2021-03-21 20:54
解：
望采纳
- 2楼网友：廢物販賣機
- 2021-03-21 20:30
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