limx→0 2x/(xcosx/2+sinx/2)=limx→0 2/(3/2cosx/2-x/2sinx/2) 求助是怎么化过来的
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limx→0 2x/(xcosx/2+sinx/2)=limx→0 2/(3/2cosx/2-x/2sinx/2) 求助是怎么化过来的
limx→0 2x/(xcosx/2+sinx/2)=limx→0 2/(3/2cosx/2-x/2sinx/2) 求助是怎么化过来的
limx→0 2x/(xcosx/2+sinx/2)=limx→0 2/(3/2cosx/2-x/2sinx/2) 求助是怎么化过来的
lim(x->0) 2x/(xcos(x/2)+sin(x/2) ) (0/0 L's Hospital rule )
=lim(x->0) (2x)'/(xcos(x/2)+sin(x/2) )'
=lim(x->0) 2/[(cos(x/2) - (1/2)xsin(x/2) +1/2cos(x/2) )
=lim(x->0) 2/[(3/2)cos(x/2) - (1/2)xsin(x/2)]
错了吧,根据罗比达法则只能化为(2/(cosx-xsinx/2))
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