Question

Determine the value of a and b so that lim(x→0) (asin2x-bsinx/x³)=1

Answer 1

$$\lim\limits_{x\to 0}\dfrac{a\sin(2x)-b\sin x}{x^3}=[\dfrac{0}{0}]$$

$$=\lim\limits_{x\to 0}\dfrac{(a\sin(2x)-b\sin x)'}{(x^3)'}$$

$$=\lim\limits_{x\to 0}\dfrac{2a\cos(2x)-b\cos x}{3x^2}=[\dfrac{0}{0}]$$

$$2a\cos(2(0))-b\cos (0)=0=>2a=b$$

$$\lim\limits_{x\to 0}\dfrac{b\cos(2x)-b\cos x}{3x^2}=[\dfrac{0}{0}]$$

$$=\lim\limits_{x\to 0}\dfrac{(b\cos(2x)-b\cos x)'}{(3x^2)'}$$

$$=\lim\limits_{x\to 0}\dfrac{-2b\sin(2x)+b\sin x}{6x}=[\dfrac{0}{0}]$$

$$=\lim\limits_{x\to 0}\dfrac{(-2b\sin(2x)+b\sin x)'}{(6x)'}$$

$$=\lim\limits_{x\to 0}\dfrac{-4b\cos(2x)+b\cos x}{6}$$

$$=\dfrac{-4b\cos(2(0))+b\cos (0)}{6}$$

$$=-\dfrac{b}{2}=1=>b=-2, a=-1$$

$$\lim\limits_{x\to 0}\dfrac{-\sin(2x)+2\sin x}{x^3}=1$$

$$a=-1, b=-2$$