Question

[ln(1+x+x^2)+ln(1-x+x^2)]/x^2 求x趋于零极限

Answer 1

x趋于0,x + x^2、-x+x^2亦趋于零

将分子在x + x^2处展开成带有皮亚诺余项的泰勒级数：

ln(1+x+x^2) = x+x^2 - [(x+x^2)^2]/2 + O(x^2)

ln(1-x+x^2) = -x+x^2 - [(-x+x^2)^2]/2 + O(x^2)

ln(1+x+x^2) + ln(1-x+x^2) = 2x^2 - (2x^4 + 2x^2)/2 + O(x^2) = x^2 + O(x^2)

所以

[ln(1+x+x^2)+ln(1-x+x^2)]/x^2 x趋于零

= [x^2 + O(x^2)]/x^2 x趋于零

= 1