題目 3. (10%) Evaluate the limit. limx→0sin(2tan−1x)−tan(sin−12x)x3\lim_{x\to 0}\dfrac{\sin(2\tan^{-1}x)-\tan(\sin^{-1}2x)}{x^3}x→0limx3sin(2tan−1x)−tan(sin−12x) 解答 解法一 sin(2tan−1x)=2x1+x2=2x−2x3+更高次項,\sin(2\tan^{-1}x)=\frac{2x}{1+x^2}=2x-2x^3+\text{更高次項},sin(2tan−1x)=1+x22x=2x−2x3+更高次項, 而 tan(sin−12x)=2x1−4x2=2x+4x3+更高次項.\tan(\sin^{-1}2x)=\frac{2x}{\sqrt{1-4x^2}} =2x+4x^3+\text{更高次項}.tan(sin−12x)=1−4x22x=2x+4x3+更高次項. 因此所求極限為 −6\boxed{-6}−6。
