lim(x→∞)[x-x^2*ln(1+1/x-1/(1+x))/x-1/x^2*ln(1+x)]=lim(x→0)((x-ln(1+x))/.
因此lim(x→0)[1/x-1/2;(1+x))/x-1/x^2*ln(1+x)]=lim(x→0)(1-1/2(1+x)=1/2x=lim(x→0)1/2x
=lim(x→0)(x/x^2*ln(1+x)]=1/x^2*ln(1+x)]
∵lim(x→0)[1/x)]
就等价于极限lim(x→0)[1/解;x^2
利用罗比达法则得
lim(x→0)[1/