Question on number of solutions of $\sin (e^{x}) = e^{\sin x}$

Question

Let $S_n$ be the number of solutions of $\sin (e^{x}) = e^{\sin x}$ in the interval $[0,2n\pi]$

Then find the value of $\lim_{n \to \infty} \frac{ S_n}{e^{2n\pi}}$

Answer 1

HINT

We can show that

$$S_{n+1}-S_n\sim\frac{e^{2\pi n}(e^{2\pi}-1)}{\pi}$$

then note that by Stolz-Cesaro

$$\lim_{n \to \infty} \frac{ S_n}{e^{2n\pi}}=\lim_{n \to \infty} \frac{ S_{n+1}-S_n}{e^{2\pi(n+1)}-e^{2\pi n}}$$