当x→0时,有ln[(1-ax^2)/(1+ax^2)]~(sinx)^2~x^2,
∴ln[(1-ax^2)/(1+ax^2)]/x^2
→{(1+ax^2)/(1-ax^2)*[-4ax/(1+ax^2)^2]}/(2x)(罗比达法则)
=-2a/[(1-ax^2)(1+ax^2)]
→-2a=1,
∴a=-1/2.
当x→0时,有ln[(1-ax^2)/(1+ax^2)]~(sinx)^2~x^2,
∴ln[(1-ax^2)/(1+ax^2)]/x^2
→{(1+ax^2)/(1-ax^2)*[-4ax/(1+ax^2)^2]}/(2x)(罗比达法则)
=-2a/[(1-ax^2)(1+ax^2)]
→-2a=1,
∴a=-1/2.