This nice problem was in the analysis section of Putnam and Beyond: prove

\begin{align*} \lim_{n\to \infty} n^2 \int_0^{1/n} x^{x+1} \, dx = 1/2. \end{align*}

The solution is quite nice, and simply relies on the fact that $\lim_{n\to 0^+} x^x = 1$, hence for $n$ large enough, we can approximate the integral with $\int_0^{1/n} x\, dx$ instead. There’s an easy generalization of this problem: \begin{align*} \lim_{n\to \infty} n^{k+1} \int_0^{1/n} x^{x+k} \, dx = 1/(k + 1). \end{align*}

Generalizing this fact, we don’t even need the composite exponential as the proof just need a $f(x)$ to be a function such that $\lim_{x\to 0^+} f(x) = 1$ with an integral bound approaching $0$.