Question

求极限lim(x趋向于0) (x-tanx)/xsinx^2

Answer 1

用洛必达法则对分子分母上下求导
原式
=lim(1-sec²x)/(2xsinx+x^2*cosx)
=lim(-2sinx/cos^3x)/(2sinx+2xcosx+2xcosx-x^2sinx)
=lim(-2-4sin^2x/cos^4x)/(2cosx+4cosx-4xsinx-2xsinx-x^2cosx)
=-1/3
实在无语，相似度有那么高吗，不就是答案一样吗

Answer 2

lim(x-tanx)/xsinx^2
=lim(x-tanx)/(x
*
x^2
*
sinx^2
/
x^2)
等价无穷小量:
=lim(x-tanx)/(x
*
x^2)
=lim(1/x^2
-
sinx/x
*
1/[x^2
*
(cosx)]
=lim(1
-
secx)/x^2
罗必塔法则：
=lim(-tanxsecx)/2x
罗必塔法则：
=-1/2limsecx(tan^2
x
+
sec^2
x)
=-1/2