∫<上x,下0>(t-sint)dt =(1/2t^2+cost)|<上x,下0>=1/2x^2+cosx-1
lim
(1/2x^2+cosx-1)/ [(e^x^4)-1]
=lim(x-sinx)/ (4x^3*e^x^4)
=lim(1-cosx)/ (12x^2*e^x^4+16x^6*e^x^4)
實(shí)在搞不懂 e^x^4 的結(jié)構(gòu)(e^x)^4,還是e^(x^4)
剛才由后者算的,累人呀,下面用前者試試
lim(1/2x^2+cosx-1)/ [(e^x)^4)-1]
=lim(x-sinx)/ (4(e^x)^4=0呃,恐怕錯(cuò)了。書上正確答案是24你得將式子輸入清楚回到前者吧lim(1/2x^2+cosx-1)/ [(e^x^4)-1]=lim(x-sinx)/ (4x^3*e^x^4)x趨于0, e^x^4 --->1原式=lim(x-sinx)/ (4x^3)=lim(1-cosx)/ (12x^2) =lim(sinx)/ (24x)=lim(sinx)/ (24x)=lim(cosx)/ 24 =1/24