当x趋近于0时,(1+ax^2)^1/3与cosx-1是等价无穷小量,求常数a
1个回答
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两种方法
1.
lim (1+ax²)^(1/3)/(cosx-1)
x→0
用洛必达法则
=lim (1/3)*2ax*(1+ax²)^(-2/3)/(-sinx)
等价无穷小sinx~x
所以
=lim (1/3)*2ax*(1+ax²)^(-2/3)/(-x)
=lim -(1/3)*2a*(1+ax²)^(-2/3)
=-(1/3)*2a*1
=-2a/3
因为是等价无穷小
所以-2a/3=1
a=-3/2
2.
先l直接用等价无穷小
(1+ax²)^(1/3)~(ax²/3)
cosx-1~(-x²/2)
lim (1+ax²)^(1/3)/(cosx-1)
x→0
=lim (ax²/3)/(-x²/2)
=-2a/3
得a=-3/2
1.
lim (1+ax²)^(1/3)/(cosx-1)
x→0
用洛必达法则
=lim (1/3)*2ax*(1+ax²)^(-2/3)/(-sinx)
等价无穷小sinx~x
所以
=lim (1/3)*2ax*(1+ax²)^(-2/3)/(-x)
=lim -(1/3)*2a*(1+ax²)^(-2/3)
=-(1/3)*2a*1
=-2a/3
因为是等价无穷小
所以-2a/3=1
a=-3/2
2.
先l直接用等价无穷小
(1+ax²)^(1/3)~(ax²/3)
cosx-1~(-x²/2)
lim (1+ax²)^(1/3)/(cosx-1)
x→0
=lim (ax²/3)/(-x²/2)
=-2a/3
得a=-3/2
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