Shortlisted Problems for the Romanian NMO

Let us make the substitution $x=1-\frac{t}{n}$, so that as $x$ goes from $0$ to $1$, $t$ goes from $n(1-0)=n$ to $n(1-1)=0$ (i.e., $t$ goes from $n$ to $0$). The differential $dx=-\frac{dt}{n}$.

So, $${\int }_0^1{x}^{n}f(x)dx={\int }_{x=0}^{x=1}{x}^{n}f(x)dx={\int }_{t=n}^{t=0}{\left(1-\frac{t}{n}\right)}^{n}f\left(1-\frac{t}{n}\right)\left(-\frac{dt}{n}\right)$$ which is $$={\int }_{t=0}^{t=n}{\left(1-\frac{t}{n}\right)}^{n}f\left(1-\frac{t}{n}\right)\frac{dt}{n}$$ Now, $${\left(1-\frac{t}{n}\right)}^n=e^{n\ln (1-t/n)}\approx e^{-t}$$ as $n\to \infty$, by the standard limit ${\lim }_{n\to \infty }{\left(1-\frac{t}{n}\right)}^n=e^{-t}$.

Also, $f\left(1-\frac{t}{n}\right)\to f(1)$ as $n\to \infty$, but since $f(1)=0$, we need a more precise expansion. Since $f$ is differentiable at $1$ and $f(1)=0$, $$f\left(1-\frac{t}{n}\right)=f(1)-f^{\prime }(1)\frac{t}{n}+o\left(\frac{t}{n}\right)=-f^{\prime }(1)\frac{t}{n}+o\left(\frac{t}{n}\right)$$ So, $$n^2{\int }_0^1{x}^{n}f(x)dx\approx n^2{\int }_{t=0}^{t=n}{e}^{-t}\left(-f^{\prime }(1)\frac{t}{n}\right)\frac{dt}{n}$$ $$=n^2\cdot \left(-f^{\prime }(1)\right){\int }_{t=0}^{t=n}{e}^{-t}\frac{t}{n}\frac{dt}{n}$$ $$=-f^{\prime }(1)n^2{\int }_{t=0}^{t=n}{e}^{-t}\frac{t}{n^2}dt$$ $$=-f^{\prime }(1){\int }_{t=0}^{t=n}t{e}^{-t}dt$$ As $n\to \infty$, ${\int }_0^{n}t{e}^{-t}dt\to {\int }_0^{\infty }t{e}^{-t}dt=1$ (since ${\int }_0^{\infty }t{e}^{-t}dt=1$).

Therefore, $$\underset{n\to \infty }{\lim }{n}^2{\int }_0^1{x}^{n}f(x)dx=-f^{\prime }(1).$$