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!
 A: $$\begin{align}\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}&=\lim_{x\to 0} \left[\frac{\sin(2x)-\sin(Ax)}{x+x^3}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right]\\&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{\frac{x+x^3}{x}}\\
&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{1+x^2}\\
&=\frac{2(1)-A(1)}{1+0}\\
&=2-A.\end{align}$$ 
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 $A=1.$
A: There are two solutions to be found.
For $x\approx0$ you can use a Taylor series expansion of a function about $x=0$.
The Taylor series of $sin(cx)$ is $cx+O(x^3)$. When $x$ is close to zero the $x^3$ term becomes negligible.
$\lim_{x\to 0} \frac{sin(2x)-sin(Ax)}{x+x^3} = A^2$
Using the Taylor series gives
$\lim_{x\to 0} \frac{2x-Ax+O(x^3)}{x+x^3} = A^2$
Simplify the fraction
$\lim_{x\to 0} \frac{2-A+O(x^2)}{1+x^2} = A^2$
Now it's easy to take the limit
$\frac{2-A}{1} = A^2$
This has solutions $A=1$ and $A=-2$
A: 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 $A=-2$.
A: $A=2$ is clearly not a solution of this equation.
Suppose $A\neq2$. Then we have, as $x\to0$ :
$$\sin(2x)-\sin(Ax)=\left(2x+o(x)\right)-\left(Ax+o(x)\right)$$
which reduces to :
$$\sin(2x)-\sin(Ax)\sim(2-A)x$$
Since $x+x^3\sim x$ as $x\to0$, we see that :
$$\lim_{x\to0}\frac{\sin(2x)-\sin(Ax)}{x+x^3}=2-A$$
Hence the equation is equivalent to $A^2+A-2=0$ and the solutions are $A=1$ and $A=-2$.
