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.
 A: 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,
$$
\lim_{n\to\infty}nI_n = \frac{1}{2020}.
$$
A: 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
$$ \lim_{n\to\infty}nI_n=\frac1{2020}.$$
A: 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 $Cx^n$. 
A: We can write
\begin{align}
{nx^n \over x^2+2019} &= {nx^n \over 2019} \cdot {1\over 1+{x^2\over 2019}} \\
&= {nx^n \over 2019} \cdot \left(1-{x^2\over 2019}+\left[{x^2\over 2019}\right]^2-\left[{x^2\over 2019}\right]^3+\cdots\right) \\
&={nx^n\over 2019}\cdot\sum_{k=0}^{\infty}{x^{2k}(-1)^k\over (2019)^k} \\
&={n\over2019}\cdot \sum_{k=0}^\infty {x^{n+2k}(-1)^k\over (2019)^k}.
\end{align}
Therefore,
\begin{align}
\int_0^1 {nx^n\over x^2+2019}\,dx &=\int_0^1 {n\over2019}\cdot \sum_{k=0}^{\infty}{x^{n+2k}(-1)^k\over (2019)^k}dx \\
&= {n\over2019}\cdot\sum_{k=0}^{\infty}{(-1)^k\over (2019)^k}\int_0^1  {x^{n+2k}}\,dx \\
&= \sum_{k=0}^\infty {n\over n+2k+1}\cdot{(-1)^k\over (2019)^{k+1}}.
\end{align}
Finally,
\begin{align}
\lim_{n\to \infty}\int_0^1 {nx^n\over x^2+2019}\,dx &= \lim_{n\to \infty}\sum_{k=0}^\infty {n\over n+2k+1}\cdot{(-1)^k\over (2019)^{k+1}}\\
&= \sum_{k=0}^\infty {(-1)^k\over (2019)^{k+1}}\\
&= {1\over 2020}.
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}\pars{n\int_{0}^{1}{x^{n} \over
x^{2} + 2019}\,\dd x}}\,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,
\lim_{n \to \infty}\bracks{n\int_{0}^{1}{\pars{1 - x}^{n} \over
x^{2} -2x + 2020}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{n\int_{0}^{1}{\expo{n\ln\pars{1 - x}} \over
x^{2} -2x + 2020}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\pars{n\int_{0}^{\infty}{\expo{-nx} \over
0^{2} -2 \times 0 + 2020}\,\dd x}\qquad\pars{\text{Laplace's Method}}
\\[5mm] = & \bbx{1 \over 2020} \approx 4.9505 \times 10^{-4}
\end{align}

Laplace's Method.

