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填空题第二题求详解,谢谢了(^~^)
1个回答
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(2)
lim (1-³√x)/(1-x)
x→0
=lim (1-³√x)/[(1-³√x)(1+³√x+³√x²)]
x→0
=lim 1/(1+³√x+³√x²)
x→0
=1/(1+1+1)
=1/3
lim (1-√x)/(1-x)
x→0
=lim (1-√x)/[(1-√x)(1+√x)]
x→0
=lim 1/(1+√x)
x→0
=1/(1+1)
=1/2
lim 2(1-√x)/(1-x)
x→0
=lim 2(1-√x)/[(1-√x)(1+√x)]
x→0
=lim 2/(1+√x)
x→0
=2/(1+1)
=1
可见,1-x的等价无穷小是2(1-√x)
1-³√x、1-√x都是1-x的同阶但不等价无穷小
lim (1-³√x)/(1-x)
x→0
=lim (1-³√x)/[(1-³√x)(1+³√x+³√x²)]
x→0
=lim 1/(1+³√x+³√x²)
x→0
=1/(1+1+1)
=1/3
lim (1-√x)/(1-x)
x→0
=lim (1-√x)/[(1-√x)(1+√x)]
x→0
=lim 1/(1+√x)
x→0
=1/(1+1)
=1/2
lim 2(1-√x)/(1-x)
x→0
=lim 2(1-√x)/[(1-√x)(1+√x)]
x→0
=lim 2/(1+√x)
x→0
=2/(1+1)
=1
可见,1-x的等价无穷小是2(1-√x)
1-³√x、1-√x都是1-x的同阶但不等价无穷小
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