求极限,最好有解题思路
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/06 07:57:56
求极限,最好有解题思路
g.e.= lim(x→0)[(e^x)-sinx-1]/[-(1/2)(x^2)] (等价无穷小替换)
= -2*lim(x→0)[(e^x)-sinx-1]/(x^2) (0/0)
= -2*lim(x→0)[(e^x)-cosx]/(2x)
= -lim(x→0)[(e^x)-cosx]/x (0/0)
= -lim(x→0)[(e^x)+sinx]/1
= -1
再问:
再问:
再问: ����
再答: ����20. �������滻 t = sqrt(2x-1) �ɽ⡣ �������һ�������� �� f(x) = (1/2)[1/(x+1) - 1/(x+3)] ������= (1/2){1/[(x-1)+2] - 1/[(x-1)+4]} ������= (1/4)/[1+(x-1)/2] - (1/8)/[1+(x-1)/4]}�� ������֪չ��ʽ ������1/(1+x) =��{n>=0}(-x)^n��|x|
= -2*lim(x→0)[(e^x)-sinx-1]/(x^2) (0/0)
= -2*lim(x→0)[(e^x)-cosx]/(2x)
= -lim(x→0)[(e^x)-cosx]/x (0/0)
= -lim(x→0)[(e^x)+sinx]/1
= -1
再问:
再问:
再问: ����
再答: ����20. �������滻 t = sqrt(2x-1) �ɽ⡣ �������һ�������� �� f(x) = (1/2)[1/(x+1) - 1/(x+3)] ������= (1/2){1/[(x-1)+2] - 1/[(x-1)+4]} ������= (1/4)/[1+(x-1)/2] - (1/8)/[1+(x-1)/4]}�� ������֪չ��ʽ ������1/(1+x) =��{n>=0}(-x)^n��|x|