Consider $$f(t)= \frac{1}{t} \int_{0}^t \sin(e^s) ds.$$ What is $$\mathrm{lim \ inf}_{t \rightarrow \infty} f(t)$$ and $$\mathrm{lim \ sup}_{t \rightarrow \infty} f(t)?$$
Using $u$-substitution, Taylor expanding, and integrating term by term, I arrived at $$\frac{1}{t} \sum_{n=0}^{\infty} \frac{(-1)^n(e^{t (2n+1)}-1)}{(2n+1)(2n+1)!}.$$ At this point, I am stuck. Any help would be appreciated. Thanks.
Intuition: the function $\sin(e^s)$ oscillates so rapidly between $-1$ and $1$ that its integral shouldn't have any tendency to become very positive or very negative; in other words, that integral should converge.
One way to implement this intuition is to integrate by parts: \begin{align*} \int_0^t \sin(e^s)\,ds &= \int_0^t e^{-s} \cdot e^s\sin(e^s)\,ds \\ &= e^{-s} ({-}\cos(e^s))|_0^t - \int_0^t ({-}e^{-s})({-}\cos(e^s))\,ds, \end{align*} so that $$ \bigg| \int_0^t \sin(e^s)\,ds \bigg| \le \cos1+e^{-t}|\cos(e^t)| + \int_0^t e^{-s}\,ds < \cos1+e^{-t}+1. $$ This is enough to show that the limit you're interested in equals $0$ (both the lim sup and lim inf).
A slightly more careful argument shows that in fact $$ \lim_{t\to\infty} \int_0^t \sin(e^s)\,ds = {-}\cos1 - \int_0^\infty e^{-t}\cos(e^s)\,ds. $$