y=(1-x)arctg(1/1-x)在x=1处是没有定义的
而
lim(x趋向0+)(1-x)arctg(1/1-x)=(1-0)×π/2=π/2
lim(x趋向0-)(1-x)arctg(1/1-x)=(1-0)×(-π/2)=-π/2
即y(0+)≠y(0-)
所以间断点x=0是它的跳跃间断点
y=(1-x)arctg(1/1-x)在x=1处是没有定义的
而
lim(x趋向0+)(1-x)arctg(1/1-x)=(1-0)×π/2=π/2
lim(x趋向0-)(1-x)arctg(1/1-x)=(1-0)×(-π/2)=-π/2
即y(0+)≠y(0-)
所以间断点x=0是它的跳跃间断点