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 展开
谢谢了急用 展开
2个回答
展开全部
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...
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询