计算lim(x-->0)[(tanx-sinx)/sin(x^3)]两种算法：算法一:x-->0时,tanx与x为等价无穷小,sinx与x为等价无穷小,sin(x^3)与x^3为等价无穷小则原式=lim(x-->0)[tanx/sin(x^3)]-lim(x-->0)[sinx/sin(x^3)]=lim(x-->0)(x/x^3)-lim(x-->0)(x/x^3)=0算法二:x-->0时,sinx与x为等价无穷小,1-cosx与(x^2)/2为等价无穷小,sin(x^3)与x^3为等价无穷小则原式=lim(x-->0)[(sinx/cosx-sinx)/sin(x^3)]=lim(x-->0)(sinx/cosx)(1-cosx)/sin(x^3)]=lim(x-->0)[(x/cosx)(x^2/2)/x^3]=lim(x-->0)[(x^3/2cosx)/x^3]=lim(x-->0)[1/2cosx]=1/2答案说第二种算法正确,那么第一种算法错在哪了?也就是说即使是两个相同的极限式，如

计算lim(x-->0)[(tanx-sinx)/sin(x^3)]
两种算法：
算法一:x-->0时,tanx与x为等价无穷小,sinx与x为等价无穷小,sin(x^3)与x^3为等价无穷小
则原式=lim(x-->0)[tanx/sin(x^3)]-lim(x-->0)[sinx/sin(x^3)]
=lim(x-->0)(x/x^3)-lim(x-->0)(x/x^3)=0
算法二:x-->0时,sinx与x为等价无穷小,1-cosx与(x^2)/2为等价无穷小,sin(x^3)与x^3为等价无穷小
则原式=lim(x-->0)[(sinx/cosx-sinx)/sin(x^3)]
=lim(x-->0)(sinx/cosx)(1-cosx)/sin(x^3)]
=lim(x-->0)[(x/cosx)(x^2/2)/x^3]
=lim(x-->0)[(x^3/2cosx)/x^3]
=lim(x-->0)[1/2cosx]=1/2
答案说第二种算法正确,那么第一种算法错在哪了?
也就是说即使是两个相同的极限式，如果都趋向于正无穷，那么他们相减也是没有意义的，也就是不能说=0。