limx→π√1−cosx−√2sin2x
=limx→π√1−cosx−√2(1−cos2x)×√1−cosx+√2√1−cosx+√2
=limx→π1−cosx−2(1−cos2x)×√1−cosx+√2√1−cosx+√2
=limx→π−(1+cosx)(1−cos2x)×√1−cosx+√2√1−cosx+√2
=limx→π−1(1−cosx)(√1−cosx+√2)
=−1(1−(−1))(√1−(−1)+√2)
=−1(1+1)(√1+1+√2)
=−12(2√2)=−14√2