lim 1-cos(x)/sin(5x) x是趋向0的.分子是1-cos(x) 分母是sin(5x)过程写一下,
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lim(sin(x^2*cos(1/x)))/x怎么做?
用罗必达法则解 (见图)lim [(e^x)-1/sin(5x)]x->0lim [1-cos(5x)/1-cos(7x)]x->0lim x->0 [(8^x-7^x-1)/(x^2)-1]lim [1-(6/x)]^x x->∞ lim 7cos(3x)sec(9x)x->pi/2
lim 1-cos(x)/sin(5x) x是趋向0的.分子是1-cos(x) 分母是sin(5x)过程写一下,
(x^2cos^2x-sin^2x)/x^2cos^2xsin^2x (x->0)lim (x^2cos^2x-sin^2x)/x^2cos^2xsin^2x (x->0)=lim [cos^2x-(sin^2x/x^2)]/cos^2xsin^2x=lim (cos^2x-1)/sin^2x=lim -sin^2x/sin^2x=-1这样解错在哪啊?小弟感激不尽
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用洛必达法则求几个极限(有图)~用洛必达法则求以下极限(见图)lim [(e^x)-1/sin(5x)]x->0lim [1-cos(5x)/1-cos(7x)]x->0lim x->0 [(8^x-7^x-1)/(x^2)-1]lim [1-(6/x)]^x x->0 lim 7cos(3x)sec(9x)x->∞(搞错了~应该是>/
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