老师x→∞时x-x²ln(1+1/x)的极限怎么求呢
发布网友
发布时间：2024-09-29 19:17
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热心网友
时间：2024-10-29 06:20
=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
