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!

Hint: $\sin 2x -\sin Ax = 2\sin {\frac{2-A}{2}x} \cos {\frac{2+A}{2}x}$
Solution: if this hint is used, $$\lim_{x\to 0}{ \frac{2\sin \big( {\frac{2-A}{2}x}\big) \cos \big( {\frac{2+A}{2} x}\big) } {x(1+x^2)}}$$
$$2 \frac{2-A}{2} \lim_{x\to 0} {\frac{\cos \big( {\frac{2+A}{2}x}\big) }{1+x^2}}=2-A$$
And this is equal to $A^2$. Then $A=1$ or $A=-2$.
