第1个回答 2017-10-27
lim(x->∝) x^2ln(1+1/x) -x
=lim(x->∝) [xln(1+1/x)-1]/[(1/x)]
=lim(x->∝)[xln(1+1/x)-1]'/(1/x)'
=lim(x->∝)[ln(1+1/x)+x*(-1/x^2)/(1+1/x)]/(-1/x^2)
=lim(x->∝)-x^2ln(1+1/x)+x^2/(1+x)
=lim(x->∝)-x^2ln(1+1/x)+x -x/(1+x)
2lim(x->∝)x^2ln(1+1/x)-x =lim(x->∝)-x/(1+x)=lim(x->∝)-1/(1/x+1)=-1
lim(x->∝)x^2ln(1+1/x)-x=-1/2本回答被网友采纳
=lim(x->∝) [xln(1+1/x)-1]/[(1/x)]
=lim(x->∝)[xln(1+1/x)-1]'/(1/x)'
=lim(x->∝)[ln(1+1/x)+x*(-1/x^2)/(1+1/x)]/(-1/x^2)
=lim(x->∝)-x^2ln(1+1/x)+x^2/(1+x)
=lim(x->∝)-x^2ln(1+1/x)+x -x/(1+x)
2lim(x->∝)x^2ln(1+1/x)-x =lim(x->∝)-x/(1+x)=lim(x->∝)-1/(1/x+1)=-1
lim(x->∝)x^2ln(1+1/x)-x=-1/2本回答被网友采纳
第2个回答 2017-10-19
m（x）[0，pi]＝cosy，（pi，无穷大)-1追问
能详细一点吗
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