高等数学求解，谢谢
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解法一：洛必达法则
lim (x-sinx)/(x+sinx)
x→0
=lim (1-cosx)/(1+cosx)
x→0
=(1-cos0)/(1+cos0)
=(1-1)/(1+1)
=0/2
=0
解法二：等价无穷小
lim (x-sinx)/(x+sinx)
x→0
=lim [x-x+(1/6)x³]/[x+x-(1/6)x³]
x→0
=lim (1/6)x³/[2x-(1/6)x³]
x→0
=lim (1/6)/[2/x² -(1/6)]
x→0
=0
lim (x-sinx)/(x+sinx)
x→0
=lim (1-cosx)/(1+cosx)
x→0
=(1-cos0)/(1+cos0)
=(1-1)/(1+1)
=0/2
=0
解法二：等价无穷小
lim (x-sinx)/(x+sinx)
x→0
=lim [x-x+(1/6)x³]/[x+x-(1/6)x³]
x→0
=lim (1/6)x³/[2x-(1/6)x³]
x→0
=lim (1/6)/[2/x² -(1/6)]
x→0
=0
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