If $I_n=\int_0^1 \frac{x^n}{x^2+2019}\,\mathrm dx$, evaluate $\lim\limits_{n\to \infty} nI_n$ If
$$I_n=\int_0^1 \frac{x^n}{x^2+2019}\,\mathrm dx,$$ find $\lim_{n\to \infty} nI_n$. Can somebody help me, please?
I've only found that $$\frac{1}{2020} \le nI_n \le \frac{1}{2019}$$ knowing that $x^2 \in [0,1]$, but it doesn't help to evaluate the limit.
I want a proof without the Arzelà-Ascoli theorem or dominated convergence.

The term $nx^n$ in the numerator of our integral is almost a derivative: $(x^n)' = nx^{n-1}$, so this suggests the following approach.
Notice that by the product rule,
\begin{align}
\frac{\mathrm d}{\mathrm dx}\bigg(x^n\cdot \frac{x}{x^2+2019}\bigg) &= nx^{n-1}\cdot\frac{x}{x^2+2019} + x^n\cdot\frac{\mathrm d}{\mathrm dx}\bigg(\frac{x}{x^2+2019}\bigg) \\
&= \frac{nx^n}{x^2+2019} + x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}
\end{align}
For $0 \leqslant x \leqslant 1$, the second term above is bounded above in absolute value by $|x|^n = x^n$. Integrating and applying the fundamental theorem of calculus gives
\begin{align}
\int_0^1\frac{\mathrm d}{\mathrm dx}\bigg(x^n\cdot \frac{x}{x^2+2019}\bigg)\,\mathrm dx &= n\int_0^1\frac{x^n}{x^2+2019}\,\mathrm dx + \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx \\
\leadsto\quad\frac{1}{2020} &= nI_n  + \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx.
\end{align}
Hence,
\begin{align}
nI_n = \frac{1}{2020} - \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx.
\end{align}
By the triangle inequality and the fundamental theorem of calculus again,
\begin{align*}
\bigg|\int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx\bigg| &\leqslant \int_0^1\bigg|x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\bigg|\,\mathrm dx \\
&\leqslant \int_0^1 x^n\,\mathrm dx \\
&= \frac{1}{n+1} \to 0,\quad\text{as $n\to\infty$.}
\end{align*}
Thus,
$$
\lim_{n\to\infty}nI_n = \frac{1}{2020}.
$$
