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If $$ \lim_{x\to0}\frac{a\sin2x-b\sin x}{x^3}=1, $$ find the value of $a$ and $b$.

What is the easiest process to do this problem?

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Taylor expansion. – TZakrevskiy May 3 '14 at 9:19

As $\sin2x=2\sin x\cos x$

$$\lim_{x\to0}\frac{a\sin2x-b\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}x\cdot\lim_{x\to0}\frac{2a\cos x-b}{x^2}=\lim_{x\to0}\frac{2a\cos x-b}{x^2}$$

As the denominator $\to0,$ so must the numerator $=2a\cos0-b=2a-b\implies b=2a$

So, the limit becomes $\displaystyle-2a\lim_{x\to0}\frac{1-\cos x}{x^2}$

Setting $x=2y,$ $$-2a\lim_{x\to0}\frac{1-\cos2y}{(2y)^2}=-a\lim_{y\to0}\left(\frac{\sin y}y\right)^2$$

Can you take it home from here?

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Yes,i got it.I can take it from there. – user146181 May 3 '14 at 9:44

As suggested by TZakrevskiy, the Taylor series of the numerator is $$x (2 a-b)+\frac{1}{6} x^3 (b-8 a)+O\left(x^5\right)$$ Since you have to divide by $x^3$, you then have $2a=b$ in order to cancel the first term an $b-8a=6$ in order to find the limit of $1$.

I am sure that you can take from here.

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