当x→0时,3x-4sinx+sinxcosx与x^n为同阶无穷小量,则n=,由泰勒公式：sinx=x-(1/3!)x^3+(1/5!)x^5+

当x→0时,3x-4sinx+sinxcosx与x^n为同阶无穷小量,则n=,由泰勒公式：sinx=x-(1/3!)x^3+(1/5!)x^5+
o(x^5)
sinxcosx=(1/2)sin2x=1/2[2x-(1/6)(2x)^3+(1/120)(2x)^5+o(x^5)]
3x-4sinx+sinxcosx=(1/10)x^5+o(x^5)(x→0)
因此n=5
如上解法中,若将x的指数展开至3,或7或其他,结果就会改变,是否泰勒展开时指数默认展开至5?
实际运算中怎么确定展开的次数？