∵x→0时，sin2x～x2，1-cos4x～(4x)22＝8x2∴limx→0(1sin2x?cos2xx2)＝limx→0x2?sin2xcos2xx2sin2x=limx→0x2?14sin22xx4＝limx→02x?12sin4x4x3＝limx→01?cos4x6x2＝limx→012(4x)26x2＝43．
∵x→0时,sin2x～x2,1-cos4x～(4x)22=8x2∴limx→0(1sin2x−cos2xx2)=limx→0x2−sin2xcos2xx2sin2x=limx→0x2−14sin22xx4=limx→02x−12sin4x4x3=limx→01−cos4x6x2=limx→012(4x)26x2=43. 结果三 题目 求limx→0(1sin2x−cos2xx2). 答案 ∵x→0时,sin 2x～x 2...
简单计算一下即可，答案如图所示
2,用Limx趋于0 sinx/x=1 而cosx趋于1
元旦快乐！Happy New Year ！1、本题是无穷大减无穷大型不定式，但可以转化为无穷小/无穷小型不定式；2、转化为无穷小/无穷小型不定式后，就可以有很多种解答方法；3、下面的解答方法是先化简，然后连续三次使用罗毕达求导法则得出答案。具体解答如下，若看不清楚，请点击放大：
原式=lim(x→0) (2sinxcosx*cosx)/(cos2x*sinx)=lim(x→0) 2(cosx)^2/cos2x =2*1/1 =2 望采纳
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结果1 题目 x2-sin2 xcos2x【 lim_(x→0)(1/(sin^2x)-(cos^2x)/(x^2))=lim_(x→0)(x^2-sin^2xcos^2x)/(x^2sin^2x))= lim0x2sin2x lim_(x→0)(x^2-1/4sin^22x)/(x^4)=lim_(x→0)(2x-cos2xsin2x)/(4x^2) lim_(x→∞)(x-1/4sinx)/(2x^3)=lim...
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原式=lim(x→0) (2sinxcosx*cosx)/(cos2x*sinx)=lim(x→0) 2(cosx)^2/cos2x =2*1/1 =2 望采纳