1.运用洛必达法则,lim(x →O)ln(1-X)/X=Ⅰim(X→o)(-1/1-X)/1=-1(即分子,分母求导求极限)
2…运用等价无穷小变换
当X→O时,ln(1-X)~-X(好比是sin x~X一样,近似等于它)
即:lim(X→0)1n(1-X)/X=lim(X→O)-X/X=-1(此时不用管X→O)
以上仅供参考,不足请指正
1.运用洛必达法则,lim(x →O)ln(1-X)/X=Ⅰim(X→o)(-1/1-X)/1=-1(即分子,分母求导求极限)
2…运用等价无穷小变换
当X→O时,ln(1-X)~-X(好比是sin x~X一样,近似等于它)
即:lim(X→0)1n(1-X)/X=lim(X→O)-X/X=-1(此时不用管X→O)
以上仅供参考,不足请指正
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