lim x趋于0 [sin(a+x)sin(a+2x)-sin^2a]/x 求极限
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lim x趋于0 [sin(a+x)sin(a+2x)-sin^2a]/x 求极限
不用洛必达法则
不用洛必达法则
方法一:运用罗比达法则
原式=lim(x->0) [cos(a+x)sin(a+2x)+2sin(a+x)cos(a+2x)]
=cosasina+2sinacosa
=3sinacosa
=3/2*sin2a
方法二:运用积化和差和等价无穷小代换
原式=lim(x->0) [cosx-cos(2a+3x)+cos2a-1]/2x
=lim(x->0) [cosx-1+cos2a-cos2acos3x+sin2asin3x]/2x
=lim(x->0) (cosx-1)/2x + lim(x->0) cos2a(1-cos3x)/2x + lim(x->0) (sin2asin3x)/2x
=lim(x->0) -(1/2)x^2/2x + lim(x->0) cos2a*(1/2)(3x)^2/2x + lim(x->0) sin2a*3x/2x
=0+0+3/2*sin2a
=3/2*sin2a
原式=lim(x->0) [cos(a+x)sin(a+2x)+2sin(a+x)cos(a+2x)]
=cosasina+2sinacosa
=3sinacosa
=3/2*sin2a
方法二:运用积化和差和等价无穷小代换
原式=lim(x->0) [cosx-cos(2a+3x)+cos2a-1]/2x
=lim(x->0) [cosx-1+cos2a-cos2acos3x+sin2asin3x]/2x
=lim(x->0) (cosx-1)/2x + lim(x->0) cos2a(1-cos3x)/2x + lim(x->0) (sin2asin3x)/2x
=lim(x->0) -(1/2)x^2/2x + lim(x->0) cos2a*(1/2)(3x)^2/2x + lim(x->0) sin2a*3x/2x
=0+0+3/2*sin2a
=3/2*sin2a
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