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