lim[x-x^2ln(1+1/x)](X趋近于无穷大)算的过程是这样lim(x→+∞)[x-x²ln(1+1/x)]t=1/x,t→0=lim(t→0)[1/t-1/t²ln(1+t)]=lim(t→0)[t-ln(1+t)]/t²洛必达法则=lim(t→0)[1-1/(1+t)]/(2t)=lim(

Question

lim[x-x^2ln(1+1/x)] (X趋近于无穷大)
算的过程是这样
lim(x→+∞) [ x - x² ln(1+ 1/x ) ]
t = 1/x ,t→0
= lim(t→0) [1/t - 1/t² ln(1+t) ]
= lim(t→0) [ t - ln(1+t) ] / t²
洛必达法则
= lim(t→0) [ 1 - 1/(1+t) ] / (2t)
= lim(t→0) 1/ [ 2(1+t) ]
= 1/2
我想问问如果x不用t替换该怎么算?

Answer 1

一样的
分子和分母同时除以x^2化为
{1/x-ln(1+1/x)}/(1/x^2)
={-1/x^2+1/x-1/(x+1)}/[-2/(x^3)]
=(1/2)[1/(1+1/x)]
=1/2