一道导数连续题已知函数f(x)=(2x)/(x^2-1)-1/(x-1)条件x>1,f(x)=m条件x小于等于1.在x=1处连续,则实数m等于多少(f(x)是分段函数)

Question

一道导数连续题
已知函数f(x)=(2x)/(x^2 - 1) - 1/(x-1) 条件x>1,
f(x)=m 条件x小于等于1.在x=1处连续,则 实数m等于多少 (f(x)是分段函数)

Answer 1

∵lim(x->1+)f(x)=lim(x->1+)[2x/(x²-1)-1/(x-1)]
=lim(x->1+)[(x-1)/(x²-1)]
=lim(x->1+)[1/(x+1)]
=1/2
lim(x->1-)f(x)=m
又f(x)在x=1处连续
∴根据连续定义有lim(x->1+)f(x)=lim(x->1-)f(x)=f(1)
故 m=1/2.