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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!

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    Can you use a Taylor series expansion for sin(cx) about $x=0$? – 2017-01-26
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    Using only $$\frac{\sin(x)}x\to1$$ one sees that $$\frac{\sin(cx)}x\to c$$ for every $c$, from which, using $$\frac{x}{x+x^3}=\frac1{1+x^2}\to1$$ the identity $$2-A=A^2$$ follows, and you are done. – 2017-01-26

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$$\begin{align}\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}&=\lim_{x\to 0} \left[\frac{\sin(2x)-\sin(Ax)}{x+x^3}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right]\\&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{\frac{x+x^3}{x}}\\ &=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{1+x^2}\\ &=\frac{2(1)-A(1)}{1+0}\\ &=2-A.\end{align}$$ 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 $A=1.$

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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 $A=-2$

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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 $A=-2$.

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    http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – 2017-01-26
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    Yes, exactly used these formulas – 2017-01-26
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$A=2$ 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 $A=-2$.

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    For the record, the equivalent $$\sin(2x)-\sin(Ax)\sim(2-A)x$$ and its consequence $$\lim_{x\to0}\frac{\sin(2x)-\sin(Ax)}{x+x^3}=2-A$$ also hold when $$A=2$$ hence there is no need to first eliminate this case. – 2017-01-26
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    @ Did - That's true. Nevertheless, some people are reluctant to write $f(x)\sim0$ ... That's why I prefered to separate the case – 2017-01-26