Finding this weird limit involving periodic functions with periods 5 and 10. If $f(x)$ and $g(x)$ are two periodic functions with periods 5 and 10 respectively, such that:
$$\lim_{x\to0}\frac{f(x)}x=\lim_{x\to0}\frac{g(x)}x=k;\quad k>0$$
then for $n\in\mathbb N$, the value of :
$$\lim_{n\to\infty}\frac{f(5(4+\sqrt{15})^n)}{g(10(4+\sqrt{14})^n)}$$.
Note: Everything should be done by hand, any claculating/plotting, etc. device/software is not allowed.
What I've thought is make up such a function:
$$f(x)\propto\sin\frac{2\pi x}{5}\qquad g(x)\propto\sin\frac{2\pi x}{10}$$
But the condition for the limit at zero cna not be satisfied in anyways from here.
Also:
$$5(4+\sqrt{15})^n=5\times4^n(1+\sqrt{15}/4)^n\sim 5\times4^n+5\times4^n\times n\sqrt{15}/4+...$$
wheer $5\times4^n\in\mathbb Z$
So, by periodicity:
$$f(5\alpha+\beta)=f(\beta)\qquad \alpha\in\mathbb Z,\beta\in\mathbb R_{|\beta|<5}$$
And then:
$$L=\lim_{n\to\infty}\frac{f(5\times4^nn\sqrt{15}/4+\cdots)}{g(10\times4^nn\sqrt{14}/4+\cdots)}$$
But this proceeds nowhere, as I was wshing to apply L'Hospital or retain only the $(\beta, |\beta|<5)$ part, but :D.
 A: The trick is that the conjugates [Galois conjugates, not complex conjugates] of $4+\sqrt{15}$ and $4+\sqrt{14}$ have modulus less than $1$, and if $a,b\in \mathbb{Z}$ then
$$(a+\sqrt{b})^n + (a-\sqrt{b})^n \in \mathbb{Z}$$
for all $n\in \mathbb{N}$.
So we have
\begin{align}
f\bigl(5(4+\sqrt{15})^n\bigr) &= f\Bigl(5\bigl((4+\sqrt{15})^n + (4-\sqrt{15})^n\bigr) - 5(4-\sqrt{15})^n\Bigr) = f\bigl(-5(4-\sqrt{15})^n\bigr),\\
g\bigl(10(4+\sqrt{14})^n\bigr) &= g\Bigl(10\bigl((4+\sqrt{14})^n + (4-\sqrt{14})^n\bigr) - 10(4-\sqrt{14})^n\Bigr) = g\bigl(-10(4-\sqrt{14})^n\bigr).
\end{align}
Now the arguments of $f$ resp. $g$ converge to $0$ for $n\to \infty$, and
\begin{align}
\frac{f\bigl(5(4+\sqrt{15})^n\bigr)}{g\bigl(10(4+\sqrt{14})^n\bigr)}
&= \frac{f\bigl(-5(4-\sqrt{15})^n\bigr)}{g\bigl(-10(4-\sqrt{14})^n\bigr)}\\
&= \underbrace{\frac{f\bigl(-5(4-\sqrt{15})^n\bigr)}{-5(4-\sqrt{15})^n}}_{\to k}\cdot \underbrace{\frac{-10(4-\sqrt{14})^n}{g\bigl(-10(4-\sqrt{14})^n\bigr)}}_{\to \frac{1}{k}}\cdot \underbrace{\frac{5(4-\sqrt{15})^n}{10(4-\sqrt{14})^n}}_{\to 0}\\
&\to 0
\end{align}
since $0 < 4-\sqrt{15} < 4-\sqrt{14}$. If there was a typo and the power in the argument should have been $(4+\sqrt{15})^n$ for both, $f$ and $g$, then the limit would be $\frac{1}{2}$.
