如果不想用等价无穷小与泰勒公式，可以充分利用四则运算法则
lim{x->0}sin(x)*[sin(x)-sin(sin(x))]/x^4
=lim{x->0}sin(x)/x*lim{x->0}[sin(x)-sin(sin(x))]/x^3
=lim{x->0}[cos(x)-cos(sin(x))*cos(x)]/(3*x^2) 注：lim{x->0}sin(x)/x=1
=lim{x->0}cos(x)*lim{x->0}[1-cos(sin(x))]/(3*x^2)
=lim{x->0}[sin(sin(x))*cos(x)]/(6*x)
=lim{x->0}cos(x)*lim{x->0}sin(sin(x))/(6*x)
=lim{x->0}cos(sin(x))*cos(x)/6
=1/6
sinx[sinx-sin(sinx)]里面第一个sinx是什么情况求解……是没有那个x还是是x和sinx-sin(sinx)的乘积……
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