calculus - If $I_n=int_0^1 frac{x^n}{x^2+2019},mathrm dx$, evaluate $limlimits_{nto infty} nI_n$

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.

Answer

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,
$$\lim_{n\to\infty}nI_n = \frac{1}{2020}.$$