lim（1/x^2 - cot^2 x ) (x趋近于0) 用洛必达法定则
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解决时间 2021-03-06 00:04
- 提问者网友：孤凫
- 2021-03-05 03:56
lim（1/x^2 - cot^2 x ) (x趋近于0) 用洛必达法定则
最佳答案
- 五星知识达人网友：怙棘
- 2021-03-05 05:14
原式=lim(1/x^2-cos^2x/sin^2x)
=lim(sin^2x-x^2*cos^2x)/(x^2*sin^2x)
=lim(sin^2x-x^2*cos^2x)/(x^4)
=lim[cos^2x(tan^2x-x^2)]/x^4
=lim(tan^2x-x^2)/x^4
=lim(2tanx*sec^2x-2x)/4x^3
=lim(sec^4x+2sec^2x*tan^2x-1)/6x^2
=lim(sec^4x-1)/6x^2+lim(2tan^2x*sec^2x)/6^2x
=lim[(sec^2x+1)(sec^2x-1)]/6x^2+1/3
=lim(sec^2x+1)*tan^2x/6x^2+1/3
=lim(sec^2x+1)/6+1/3
=1/3+1/3
=2/3
=lim(sin^2x-x^2*cos^2x)/(x^2*sin^2x)
=lim(sin^2x-x^2*cos^2x)/(x^4)
=lim[cos^2x(tan^2x-x^2)]/x^4
=lim(tan^2x-x^2)/x^4
=lim(2tanx*sec^2x-2x)/4x^3
=lim(sec^4x+2sec^2x*tan^2x-1)/6x^2
=lim(sec^4x-1)/6x^2+lim(2tan^2x*sec^2x)/6^2x
=lim[(sec^2x+1)(sec^2x-1)]/6x^2+1/3
=lim(sec^2x+1)*tan^2x/6x^2+1/3
=lim(sec^2x+1)/6+1/3
=1/3+1/3
=2/3
全部回答
- 1楼网友：梦中风几里
- 2021-03-05 05:19
其实不用那么麻烦
原式=lim(1/x^2-cos^2x/sin^2x)
=lim(sin^2x-x^2cos^2x)/(x^2sin^2x) sin^2x~x^2(x->0)
=lim(sinx+xcosx)/x(sinx-xcosx)/x^3 lim(sinx+xcosx)/x=2
=2lim(sinx-xcosx)/x^3
=2lim(xsinx)/3x^2
=2/3
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