If $I_n=\int_0^1 \frac{x^n}{x^2+2019}\,\mathrm dx$, evaluate $\lim\limits_{n\to \infty} nI_n$

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

For fixed $\delta\in(0,1)$, $$ \int_0^1 \frac{nx^n}{x^2+2019}dx=\int_0^\delta \frac{nx^n}{x^2+2019}dx+\int_\delta^1 \frac{nx^n}{x^2+2019}dx. $$ Note $$ 0\le\int_0^\delta \frac{nx^n}{x^2+2019}dx\le\int_0^\delta nx^ndx=\frac{n}{n+1}\delta^{n+1}. \tag{1}$$ and $$ \int_\delta^1 \frac{nx^n}{2020}dx\le\int_\delta^1 \frac{nx^n}{x^2+2019}dx\le\int_\delta^1 \frac{nx^n}{\delta^2+2019}dx. \tag{2}$$ But $$ \int_\delta^1 \frac{nx^n}{2020}dx=\frac{1}{2020}\frac{n}{n+1}(1-\delta^{n+1}), \int_\delta^1 \frac{nx^n}{\delta^2+2019}dx=\frac{1}{\delta^2+2019}\frac{n}{n+1}(1-\delta^{n+1}). \tag{3} $$ From (1),(2) and (3), one has $$ \frac{1}{2020}\frac{n}{n+1}(1-\delta^{n+1})\le nI_n\le \frac{n}{n+1}\delta^{n+1}+\frac{1}{\delta^2+2019}\frac{n}{n+1}(1-\delta^{n+1}).$$ Letting $n\to\infty$ gives $$ \frac{1}{2020}\le\liminf nI_n\le\limsup nI_n\le \frac{1}{\delta^2+2019}.$$ Letting $\delta\to1^-$ gives $$ \liminf nI_n=\limsup nI_n=\frac1{2020} $$ or $$ \lim_{n\to\infty}nI_n=\frac1{2020}.$$

Let $J_n=n I_n$, notice that when you use integration by parts, you would get $$J_n=\left[ \frac{x^{n+1}}{x^2+2019}\right]_0^1 -\int_0^1\frac{x^{n}(2019-x^2)}{(x^2+2019)^2}\ dx= \frac{1}{2020}-\int_0^1\frac{x^{n}(2019-x^2)}{(x^2+2019)^2}\ dx$$

Now, you only have to show that the limit of the rightmost integral is zero, notice that the integrand can be Bounded by $Cx^n$.