这个数学极限怎么求？
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解决时间 2021-11-29 16:03
- 提问者网友：寂寞撕碎了回忆
- 2021-11-29 10:49
这个数学极限怎么求？
最佳答案
- 五星知识达人网友：痴妹与他
- 2021-11-29 12:29
全部回答
- 1楼网友：上分大魔王
- 2021-11-29 14:08
x->0
分子
tanx = x+(1/3)x^3 +o(x^3)
sinx = x-(1/6)x^3 +o(x^3)
tanx -sinx = (1/2)x^3 +o(x^3)
分母
e^tanx
= e^[x +(1/3)x^3]+o(x^3)
= 1 +[x +(1/3)x^3] + (1/2)[x +(1/3)x^3]^2 +(1/6)[x +(1/3)x^3]^3 +o(x^3)
=1 + x +(1/2)x^2 + [(1/3) + (1/6) ] x^3 +o(x^3)
=1 + x +(1/2)x^2 + (1/2)x^3 +o(x^3)
e^sinx
= e^[x -(1/6)x^3]+o(x^3)
= 1 +[x -(1/6)x^3] + (1/2)[x -(1/6)x^3]^2 +(1/6)[x -(1/6)x^3]^3 +o(x^3)
=1 + x +(1/2)x^2 + [-(1/6) + (1/6) ] x^3 +o(x^3)
=1 + x +(1/2)x^2 +o(x^3)
e^tanx -e^sinx = (1/2)x^3 +o(x^3)
/
lim(x->0) [√(4+tanx) - √(4+sinx) ]/[ e^tanx - e^sinx]
=lim(x->0) [(4+tanx) - (4+sinx) ]/{ [ e^tanx - e^sinx] . [√(4+tanx) + √(4+sinx) ] }
=(1/4)lim(x->0) (tanx -sinx)/ [ e^tanx - e^sinx]
=(1/4) lim(x->0) (1/2)x^3/ [(1/2)x^3]
=1/4
分子
tanx = x+(1/3)x^3 +o(x^3)
sinx = x-(1/6)x^3 +o(x^3)
tanx -sinx = (1/2)x^3 +o(x^3)
分母
e^tanx
= e^[x +(1/3)x^3]+o(x^3)
= 1 +[x +(1/3)x^3] + (1/2)[x +(1/3)x^3]^2 +(1/6)[x +(1/3)x^3]^3 +o(x^3)
=1 + x +(1/2)x^2 + [(1/3) + (1/6) ] x^3 +o(x^3)
=1 + x +(1/2)x^2 + (1/2)x^3 +o(x^3)
e^sinx
= e^[x -(1/6)x^3]+o(x^3)
= 1 +[x -(1/6)x^3] + (1/2)[x -(1/6)x^3]^2 +(1/6)[x -(1/6)x^3]^3 +o(x^3)
=1 + x +(1/2)x^2 + [-(1/6) + (1/6) ] x^3 +o(x^3)
=1 + x +(1/2)x^2 +o(x^3)
e^tanx -e^sinx = (1/2)x^3 +o(x^3)
/
lim(x->0) [√(4+tanx) - √(4+sinx) ]/[ e^tanx - e^sinx]
=lim(x->0) [(4+tanx) - (4+sinx) ]/{ [ e^tanx - e^sinx] . [√(4+tanx) + √(4+sinx) ] }
=(1/4)lim(x->0) (tanx -sinx)/ [ e^tanx - e^sinx]
=(1/4) lim(x->0) (1/2)x^3/ [(1/2)x^3]
=1/4
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