怎么用导数求sin² x的极限？
f(x)=sin²x，先求它在x=0处的各阶导数。因为f(ⁿ)(x)=2^(n-1)sin[2x+(n-1)π/2]。所以f(0)=0，f′(0)=0，f″(0)=2，f(³)(0)=0，f(4)(0)=-8，……，f(ⁿ)(0)=2^(n-1)sin[(n-1)π/2]
所以展开后为f(x)=2x²/2!-2³x^4/4!+2^5x^5/5!-2^7x^7/7!+……
+2^(n-1)sin[(n-1)π/2]xⁿ/n!+……
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