lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
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lim→0+[∫(上限x,下限0)ln(t+e^t)dt]/(1-cosx)lim→0+[∫(上限x,下限0)ln(t+e^t)dt]/(1-cosx)lim→0+[∫(上限x,下限0)ln(t+e^
lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
=lim→0+ [ln(x+e^x)] / (sinx)
=lim→0+ [1/(x+e^x)] *(1+e^x)/ (cosx)
=lim→0+ [(1+e^x)/(x+e^x)] / (cosx)
=2
用洛必达法则，原式=lim(x→0+)ln(x+e^x)/sinx=lim(x→0+)[(1+e^x)/(x+e^x)]/cosx=[(1+1)/(0+1)]/1=2
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