Solve $\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$ How do I solve for $A$? $$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$$
Since the denominator evaluates to $0$, I tried doing $$\lim_{x\to 0} [\sin(2x)-\sin(Ax)]=A^2 \cdot \lim_{x\to0}[x+x^3]$$ but it would go into $0=0$.
If I checked from the graph, then I believe that $A=1$.
Any advice on how to do this? I haven't learned l'Hopital's rule yet so I would rather avoid using it for now.
Thanks!

$$\begin{align}\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}&=\lim_{x\to 0} \left[\frac{\sin(2x)-\sin(Ax)}{x+x^3}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right]\\&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{\frac{x+x^3}{x}}\\
&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{1+x^2}\\
&=\frac{2(1)-A(1)}{1+0}\\
&=2-A.\end{align}$$ 
Since we want to have 
$$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}=A^2,$$ we get
$$2-A=A^2,$$
that is, $$A^2+A-2=0\iff (A+2)(A-1)=0.$$
Thus, $A=-2$ or $A=1.$
