百度作业网大一数学题请高手帮忙做一下,我不会!
来源：学生作业帮 编辑：百度作业网作业帮 分类：数学作业 时间：2024/05/28 17:50:29
大一数学题请高手帮忙做一下,我不会!
http://hi.baidu.com/qwe352087404/album/item/e6f1bdc424b5b57be4dd3baa.html#
谢谢了急用
http://hi.baidu.com/qwe352087404/album/item/e6f1bdc424b5b57be4dd3baa.html#
谢谢了急用
1.
lim(x→-1)[(x^2-6x-7)/(2x^2+x-1)]=lim(x→-1)[(2x-6)/(4x+1)]=8/3
2.
lim(x→∞)[ln(1+x)/x]=lim(x→∞)[1/(x+1)]=0
3.[用两次]
lim(x→0)[ln(1+x^2)/(1-cosx)]=lim(x→0)[2x/((x^2+1)sinx)]=lim(x→0)[2/(x^2cosx+2xsinx+cosx)]=2
4.[用两次]
lim(x→0)(1/sinx-1/x)=lim(x→0)((x-sinx)/(xsinx))=lim(x→0)((1-cosx)/(xcosx+sinx))=lim(x→0)(sinx/(2cosx-xsinx))=0
老兄 基础啊基础...
lim(x→-1)[(x^2-6x-7)/(2x^2+x-1)]=lim(x→-1)[(2x-6)/(4x+1)]=8/3
2.
lim(x→∞)[ln(1+x)/x]=lim(x→∞)[1/(x+1)]=0
3.[用两次]
lim(x→0)[ln(1+x^2)/(1-cosx)]=lim(x→0)[2x/((x^2+1)sinx)]=lim(x→0)[2/(x^2cosx+2xsinx+cosx)]=2
4.[用两次]
lim(x→0)(1/sinx-1/x)=lim(x→0)((x-sinx)/(xsinx))=lim(x→0)((1-cosx)/(xcosx+sinx))=lim(x→0)(sinx/(2cosx-xsinx))=0
老兄 基础啊基础...