Let $f$ be a function continuous at $0$. Prove that $\lim_{n\rightarrow\infty}(1-\int_0^\frac{1}{n}f(x)dx)^n=e^{-f(0)}$.
Intuitively, the argument should be something along the lines of: for large enough $n$, $f(\frac{1}{n})\approx f(0)$. I don't think this is rigorous enough. I think I should provide an upper and lower bound of the left hand side and apply squeeze theorem, but I can't seem to find appropriate bounds.
Sketch. Fix $\varepsilon$. Since $f$ is continuous at $0$, there's an $n$ so that $|f(x) - f(0)| < \varepsilon$ for $|x|<1/n$. Then $$\left(1-\frac{f(0)+\varepsilon}{n}\right)^n \leq \left(1-\int_0^{1/n} f(x)dx\right)^n \leq \left(1-\frac{f(0)-\varepsilon}{n}\right)^n.$$
The limit of the leftmost and rightmost terms, as $n \rightarrow \infty$, is $e^{-f(0) - \varepsilon}$ and $e^{-f(0)+\varepsilon}$, respectively. Now use the fact that $\varepsilon$ was chosen arbitrarily.
There are details here that need filling in; I leave it to you to do so.
