lim(x->0)[cosx-e^(-x^2/2)]/x^2[x+ln(1-x)咋看分母是四阶的?
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lim(x->0)[cosx-e^(-x^2/2)]/x^2[x+ln(1-x)咋看分母是四阶的?
lim(x->0)[cosx-e^(-x^2/2)]/x^2[x+ln(1-x)咋看分母是四阶的?
lim(x->0)[cosx-e^(-x^2/2)]/x^2[x+ln(1-x)咋看分母是四阶的?
根据迈克劳林展开式
分母x^2[x+ln(1-x)]
=x^2[x+(-x)-(-x)^2/2+(-x)^3/3-...]
=x^2[-x^2/2-x^3/3-...]
=-x^4/2-x^5/3-...
因为最低次项为4,所以分母是四阶的无穷小量
原式=
lim{x->0}[1-x^2/2+x^4/24+o(x^4)-(1-x^2/2+x^4/8+o(x^4))]/[x^2(x-x+x^2/2+o(x^2)]
=lim{x->0}[-x^4/12+o(x^4)]/[x^4/2+o(x^4)]
=lim{x->0}[-/12+o(1)]/[1/2+o(1)]
=-1/6
因为分子是4阶无穷小,分母是高于3阶的无穷小,所以ln(1-x)至少要展开到平方项啦.
嗯,都是四阶
cosx=1-x^2/2+x^4/6+o(x^4)
e^(-x^2/2)=1-x^2/2+x^4/4+o(x^4)
ln(1-x)=-x+x^2/2-x^3/3+...
x^2[x+ln(1-x)]=x^4/2+o(x^4)
cosx-e^(-x^2/2)=-x^4/12+o(x^4)
所以:极限=-1/6
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