构造函数g(x)=ln(1+x)
g'(x)=1/1+x b=x^2 ,a=sin^2 x
用拉格朗日中值定理：
ln(1+x^2)-ln(1+sin^2 x)=g(b)-g(a)=(b-a)g'(t) 其中t介于a,b之间
x趋向0时,t趋于0
lim(x趋向0)[ln(1+x^2)-ln(1+sin^2 x)]/xsin^3 x
=lim(x趋向0)(b-a)g'(t) /xsin^3 x
=lim(x趋向0)[x^2-sin^2 x]/x^4
=1/3(用sinx=x-1/6x^3+o(x^3))