如题如题如题求极限?
创始人
2024-08-23 01:40:04
如题如题如题求极限?这个极限怎么求呢?
cos2x
= 1-(1/帆敬2)(2x)^2 +(1/24)(2x)^4 +o(x^4)
=1- 2x^2 + (2/3)x^4 +o(x^4)
1-cos2x =2x^2 - (2/3)x^4 +o(x^4)
(1+x^2).(1-cos2x)
=(1+x^2).[2x^2 - (2/3)x^4 +o(x^4)]
=[2x^2 - (2/3)x^4 +o(x^4)] +x^2.[2x^2 - (2/3)x^4 +o(x^4)]
=[2x^2 - (2/3)x^4 +o(x^4)] +[2x^4+o(x^4)]
=2x^2 + (4/3)x^4 +o(x^4)
(1+x^2).(1-cos2x) -2x^2 =(4/3)x^4 +o(x^4)
//
lim(x->0) [(1+x^2)(1-cos2x) -2x^2]/凯轿猛盯桥x^4
=lim(x->0) (4/3)x^4/x^4
=4/3
解神敬如下段瞎脊握渗图所示
cos2x
= 1-(1/帆敬2)(2x)^2 +(1/24)(2x)^4 +o(x^4)
=1- 2x^2 + (2/3)x^4 +o(x^4)
1-cos2x =2x^2 - (2/3)x^4 +o(x^4)
(1+x^2).(1-cos2x)
=(1+x^2).[2x^2 - (2/3)x^4 +o(x^4)]
=[2x^2 - (2/3)x^4 +o(x^4)] +x^2.[2x^2 - (2/3)x^4 +o(x^4)]
=[2x^2 - (2/3)x^4 +o(x^4)] +[2x^4+o(x^4)]
=2x^2 + (4/3)x^4 +o(x^4)
(1+x^2).(1-cos2x) -2x^2 =(4/3)x^4 +o(x^4)
//
lim(x->0) [(1+x^2)(1-cos2x) -2x^2]/凯轿猛盯桥x^4
=lim(x->0) (4/3)x^4/x^4
=4/3
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