用洛必达法则做
Limit[In[sin x/2]/1 + cosx,x -> π])
=Limit[In[sin x/2]'/lim(1 + cosx)',x -> π])
=lim1/(2sinx/2)/lim(-sinx)
=-lim1/(2sinx/2*sinx)
=0
Limit[In[sin x/2]/1 + cosx,x -> π])求极限
3个回答
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