第1题求极限？
1个回答
展开全部
x->0
分子
(1-cosx)^(1/3) = (x^2/2 )^(1/3) + o(x^(2/3) )
e^(sinx ) = e^(x+o(x)) = 1+x +o(x)
e^(sinx) +(1-cosx)^(1/3) = 1+ x + (x^2/2 )^(1/3) + o(x^(2/3))
ln[e^(sinx) +(1-cosx)^(1/3)]
=x + (x^2/2 )^(1/3) +o(x^(2/3))
ln[e^(sinx) +(1-cosx)^(1/3)] -sinx =(x^2/2 )^(1/3) +o(x^(2/3))
分母
1-cosx = (1/2)x^2 +o(x^2)
4.(1-cosx)^(1/3) = 4(x^2/2 )^(1/3) + o(x^(2/3) )
arctan[4.(1-cosx)^(1/3) ] =4(x^2/2 )^(1/3) + o(x^(2/3) )
lim(x->0) { ln[e^(sinx) +(1-cosx)^(1/3)] -sinx }/ arctan[4.(1-cosx)^(1/3)]
=lim(x->0) (x^2/2 )^(1/3) / [4. (x^2/2 )^(1/3)]
=1/4
分子
(1-cosx)^(1/3) = (x^2/2 )^(1/3) + o(x^(2/3) )
e^(sinx ) = e^(x+o(x)) = 1+x +o(x)
e^(sinx) +(1-cosx)^(1/3) = 1+ x + (x^2/2 )^(1/3) + o(x^(2/3))
ln[e^(sinx) +(1-cosx)^(1/3)]
=x + (x^2/2 )^(1/3) +o(x^(2/3))
ln[e^(sinx) +(1-cosx)^(1/3)] -sinx =(x^2/2 )^(1/3) +o(x^(2/3))
分母
1-cosx = (1/2)x^2 +o(x^2)
4.(1-cosx)^(1/3) = 4(x^2/2 )^(1/3) + o(x^(2/3) )
arctan[4.(1-cosx)^(1/3) ] =4(x^2/2 )^(1/3) + o(x^(2/3) )
lim(x->0) { ln[e^(sinx) +(1-cosx)^(1/3)] -sinx }/ arctan[4.(1-cosx)^(1/3)]
=lim(x->0) (x^2/2 )^(1/3) / [4. (x^2/2 )^(1/3)]
=1/4
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