lim(x→0)(x^2+cosx-2)/(x^3)*ln(1+x)怎么算
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lim(x→0)(x^2+cosx-2)/(x^3)*ln(1+x)怎么算
写错了,lim(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x).
写错了,lim(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x).
lim(x→0)(x^2+cosx-2)/(x^3)*ln(1+x)
=lim(x→0)(0+1-2)*(ln(1+x)/(x^3))
=lim(x→0) -(ln(1+x)/(x^3))
=im(x→0) -1/[(1+x)*(3x^2)]
=im(x→0) -1/(3x^2)
负无穷
再问: cos前面有个2。之前写错了,题目应该是是这样:lim(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x)
再答: im(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x) =lim(x→0)(x^2+2cosx-2)*(ln(1+x)/(x^3)) 把cosx,ln(1+x)展开 cosx=1-x^2/2+x^4/24…… ln(1+x)=x-x^2/2+x^3/2+…… lim(x→0)(x^2+2cosx-2)/[(x^3)*(ln(1+x)] =lim(x→0)(x^2+2-x^2+x^4/12-2)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)x =1/12
再问: 这一步:lim(x→0)(x^2+2cosx-2)*(ln(1+x)/(x^3))为什么(ln(1+x)到分子上了原式是在分母上的啊
再答: 那一步写错了,后面是对的 im(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x) =lim(x→0)(x^2+2cosx-2/[(x^3))(ln(1+x)] 把cosx,ln(1+x)展开 cosx=1-x^2/2+x^4/24…… ln(1+x)=x-x^2/2+x^3/2+…… lim(x→0)(x^2+2cosx-2)/[(x^3)*(ln(1+x)] =lim(x→0)(x^2+2-x^2+x^4/12-2)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)x =1/12
=lim(x→0)(0+1-2)*(ln(1+x)/(x^3))
=lim(x→0) -(ln(1+x)/(x^3))
=im(x→0) -1/[(1+x)*(3x^2)]
=im(x→0) -1/(3x^2)
负无穷
再问: cos前面有个2。之前写错了,题目应该是是这样:lim(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x)
再答: im(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x) =lim(x→0)(x^2+2cosx-2)*(ln(1+x)/(x^3)) 把cosx,ln(1+x)展开 cosx=1-x^2/2+x^4/24…… ln(1+x)=x-x^2/2+x^3/2+…… lim(x→0)(x^2+2cosx-2)/[(x^3)*(ln(1+x)] =lim(x→0)(x^2+2-x^2+x^4/12-2)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)x =1/12
再问: 这一步:lim(x→0)(x^2+2cosx-2)*(ln(1+x)/(x^3))为什么(ln(1+x)到分子上了原式是在分母上的啊
再答: 那一步写错了,后面是对的 im(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x) =lim(x→0)(x^2+2cosx-2/[(x^3))(ln(1+x)] 把cosx,ln(1+x)展开 cosx=1-x^2/2+x^4/24…… ln(1+x)=x-x^2/2+x^3/2+…… lim(x→0)(x^2+2cosx-2)/[(x^3)*(ln(1+x)] =lim(x→0)(x^2+2-x^2+x^4/12-2)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)[(x-x^2/2+x^3/2)] =lim(x→0)(x^4/12)/(x^3)x =1/12
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