I've been trying to compute the following limit:$$\lim_{n\ \to\ \infty}\int_{0}^{2\pi}\left[1 + \frac{\sin\left(x\right)}{n}\right]^{n/x}\,{\rm d}x$$
- I am not completely sure, but when $x\in (0,2\pi)$ then $\operatorname{f}_{n}\left(x\right) \to \exp\left(\sin(x)/x\right)$.
- So the pointwise limit of the sequence in $\left(0,2\pi\right)$ is $\exp\left(\sin(x)/x\right)$.
- When $x > 0$ it's easy to prove that $\sin\left(x\right)/x \leq 1$ so I think it's true that for all $x \in \left(0,2\pi\right)$.$$\left\vert\operatorname{f}_{n}\left(x\right)\right\vert \leq {\rm e} \in L^{1}\left(0,2\pi\right).$$
- At this point I assume I can use the
Lebesgue's Dominated Convergence Theorem, so$$\lim_{n \to \infty}\int_{0}^{2\pi}\operatorname{f}_{n}\left(x\right)\,{\rm d}x =\int_{0}^{2\pi}{\rm e}^{\sin\left(x\right)/x}\,\,\,{\rm d}x.$$I get stuck at this point, because I can't quite find the antiderivative. I tried integration by parts but it doesn't simplify things.