<s>[INST] 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
 [/INST] 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 $-2$.
$$\begin{align}
\lim_{x\to0}\frac{2\cos(x)-2}{x^2}&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\\
&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\frac{\cos(x)+1}{\cos(x)+1}\\
&=2\lim_{x\to0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)}\\
&=2\lim_{x\to0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}\\
&=-2\lim_{x\to0}\frac{\sin^2(x)}{x^2}\frac{1}{\cos(x)+1}\\
%&=-2\lim_{x\to0}\frac{1}{\cos(x)+1}\\
&=\ldots
\end{align}$$
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