Limit of the function $\lim_{x\to0}\frac{\sin 2x + a\sin x}{x^3}$? Question

If for some real number $a$, $\lim_{x\to 0}\frac{\sin 2x + a\sin x}{x^3}$ exists, then the limit is equal to:

Here what i have done
since it is of $0/0$ form applying L' Hospital's rule$$\implies\lim_{x\to0}\frac{\sin 2x + a \sin x}{x^3} = \frac{2\cos 2x + a\cos x}{3x^2}$$ now what to do i am stuck here please help
Thanks
Akash
 A: Using no L'Hospital's Rule or Maclaurin series, but instead trig identities and that $\lim_{x\to0}\frac{\sin(x)}{x}=1$:
$$\begin{align}
\lim_{x\to0}\frac{\sin(2x)+a\sin(x)}{x^3}&=\lim_{x\to0}\frac{2\sin(x)\cos(x)+a\sin(x)}{x^3}\\
&=\lim_{x\to0}\frac{\sin{x}}{x}\frac{2\cos(x)+a}{x^2}\\
&=\lim_{x\to0}\frac{2\cos(x)+a}{x^2}\\
\end{align}$$
The denominator of this approaches $0$, so the only way the limit can exist is if the numerator also approaches $0$. So $a$ would have to be $-2$.
$$\begin{align}
\lim_{x\to0}\frac{2\cos(x)-2}{x^2}&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\\
&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\frac{\cos(x)+1}{\cos(x)+1}\\
&=2\lim_{x\to0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)}\\
&=2\lim_{x\to0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}\\
&=-2\lim_{x\to0}\frac{\sin^2(x)}{x^2}\frac{1}{\cos(x)+1}\\
%&=-2\lim_{x\to0}\frac{1}{\cos(x)+1}\\
&=\ldots
\end{align}$$
A: Since $3x^2\to 0,$ the numerator also has to go to $0$ when $x\to 0$. 
Hence, $2+a=0\iff a=-2.$
So, you'll have
$$\lim_{x\to 0}\frac{2\cos{2x}-2\cos x}{3x^2}.$$
You can use L' Hospital's rule twice.
A: You can use the Taylor series
$$ \frac{\sin2x + a\sin x}{x^3} = \frac{(2x-(2x)^3/3!+\dots)+a (x-x^3/3!+\dots)}{x^3}$$
$$ \sim_{x\to 0} \frac{(a+2)x-(2^3+1)x^3/3!}{x^3}\dots\,.  $$
Can you finish it?
A: This is not an answer as others have done a great job of it. I will try and explain how to think about the problem. In the end each of us approach a problem differently so how we think may be quite different. So, this is just my thinking.
The sine function has only odd powers since it is an odd function. So it will have $x$, $x^3$, $x^5$. Denominator has $x^3$, so if we can choose $a$ so that the $x$ term goes away, then we will be left with only $x^3$, $x^5$ etc. and after dividing by $x^3$, the $x^3$ term will be the limit and the rest will go to zero. 
The $x$ term of $\sin 2x$ is $2x$ and $x$ term of $a\sin x$ is $a x$. So to get rid of $x$ term, we need $a = -2$. 
The $x^3$ term of $\sin 2x$ is $-(2x)^3/3!$ and $x^3$ of $a\sin x$ is $-a x^3/3!$. So the limit is
$$
-8/3! -a /3! = -6/3! = -1$$
