(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

来源：学生作业帮助网编辑：作业帮时间：2024/05/04 13:56:50

(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
从第二步到第三步是不对的.即
lim [cos^2x-(sin^2x/x^2)]/cos^2xsin^2x
=lim (cos^2x-1)/sin^2x
是不对的.如果这样做,就默认了
lim [cos^2x-(sin^2x)/x^2]/sin^2x
=[lim cos^2x-lim (sin^2x)/x^2]/lim(sin^2x)
但如果这样,因为 lim(sin^2x)=0,此时分母为0,分式无意义,也就是说此时极限的四则运算不成立,因此不能用.
不知道你学过洛必达法则没有,用洛必达法则是比较简单的.
原极限
=lim [cos^2x-(sin^2x/x^2)]/sin^2x
=lim (1/tan^2x - 1/x^2)
=lim (x^2-tan^2x)/(x^2tan^2x) (lim tan^2x/x^2=1)
=lim (x^2-tan^2x)/x^4
=lim (x+tanx)/x * lim (x-tanx)/x^3
=2lim (x-tanx)/x^3 (洛必达法则)
=2lim (1-sec^2x)/(3x^2)
=2lim (-tan^2x)/(3x^2)
=-2/3
即原式 = -2/3.

lim (x^2cos^2x-sin^2x)/x^2cos^2xsin^2x
=lim [cos^2x-(sin^2x/x^2)]/cos^2xsin^2x
mistake....

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