Limit of $a\sin x-\sin 2x$

Question

If $$\lim_{x \to 0}\frac{a\sin x-\sin 2x}{\tan^3x}$$ is finite, then find $a$ and the limit. Using series expansion, I got $a=2$, and then continuing I got the limit also $2$, which is wrong. I don't know where am I going wrong.

Answer 1

$$a\sin x-\sin2x=a\sin x-2\sin x\cos x=\sin x(a-2\cos x)$$ $$\lim_{x \to 0}\frac{a\sin x-\sin 2x}{\tan^3x}=\frac{\sin x(a-2\cos x)}{\tan^3x}=\frac{(a-2\cos x)\cdot\cos^3 x}{\sin^2x}$$ for limit to exist numerator should go to zero as well , thus $a=2$ thus limit is equal to $$\lim_{x \to 0}\frac{(a-2\cos x)\cdot\cos^3 x}{\sin^2x}=\frac{(2-2\cos x)\cdot\cos^3 x}{\sin^2x}=2 \cdot \frac{2\sin^2\frac{x}{2}}{\sin^2x}=1$$

Answer 2

Using Series Exansion,

$$\frac{a\sin x-\sin2x}{\tan^3x}=\frac{a\left[x-\dfrac{x^3}{3!}+\cdots\right]-\left[2x-\dfrac{(2x)^3}{3!}+\cdots\right]}{\left(x+\dfrac{x^3}3+\cdots\right)^3}$$

$$=\frac{x(a-2)+\dfrac{x^3}{3!}(2^3-a)+\cdots}{x^3+\cdots}$$

Clearly, $a=2$

$$a=2\implies\frac{x(a-2)+\dfrac{x^3}{3!}(2^3-a)+\cdots}{x^3+\cdots}=\frac{\dfrac{x^3}{3!}(2^3-2)+\cdots}{x^3+\cdots}=?$$

Answer 3

Hint: $$ \begin{align} \frac{2\sin(x)-\sin(2x)}{\tan^3(x)} &=2\color{#00A000}{\frac{\sin(x)}{\tan(x)}}\frac{\color{#0000FF}{1-\cos(x)}}{\tan^2(x)}\\ &=2\frac{\color{#00A000}{\cos(x)}}{\color{#0000FF}{1+\cos(x)}}\frac{\color{#0000FF}{\sin^2(x)}}{\tan^2(x)}\\ &=2\frac{\cos^3(x)}{1+\cos(x)} \end{align} $$