# Composite exponential limit

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 $0$.

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