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If
$$I_n=int_0^1 fracx^nx^2+2019,mathrm dx,$$ find $lim_nto infty nI_n$. Can somebody help me, please?

I've only found that $$frac12020 le nI_n le frac12019$$ 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,
beginalign
fracmathrm dmathrm dxbigg(x^ncdot fracxx^2+2019bigg) &= nx^n-1cdotfracxx^2+2019 + x^ncdotfracmathrm dmathrm dxbigg(fracxx^2+2019bigg) \
&= fracnx^nx^2+2019 + x^ncdot frac-x^2+2019(x^2+2019)^2
endalign
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
beginalign
int_0^1fracmathrm dmathrm dxbigg(x^ncdot fracxx^2+2019bigg),mathrm dx &= nint_0^1fracx^nx^2+2019,mathrm dx + int_0^1 x^ncdot frac-x^2+2019(x^2+2019)^2,mathrm dx \
leadstoquadfrac12020 &= nI_n + int_0^1 x^ncdot frac-x^2+2019(x^2+2019)^2,mathrm dx.
endalign
Hence,
beginalign
nI_n = frac12020 - int_0^1 x^ncdot frac-x^2+2019(x^2+2019)^2,mathrm dx.
endalign
By the triangle inequality and the fundamental theorem of calculus again,
beginalign*
bigg|int_0^1 x^ncdot frac-x^2+2019(x^2+2019)^2,mathrm dxbigg| &leqslant int_0^1bigg|x^ncdot frac-x^2+2019(x^2+2019)^2bigg|,mathrm dx \
&leqslant int_0^1 x^n,mathrm dx \
&= frac1n+1 to 0,quadtextas $ntoinfty$.
endalign*
Thus,
$$
lim_ntoinftynI_n = frac12020.
$$

For fixed $deltain(0,1)$,
$$ int_0^1 fracnx^nx^2+2019dx=int_0^delta fracnx^nx^2+2019dx+int_delta^1 fracnx^nx^2+2019dx. $$
Note
$$ 0leint_0^delta fracnx^nx^2+2019dxleint_0^delta nx^ndx=fracnn+1delta^n+1. tag1$$
and
$$ int_delta^1 fracnx^n2020dxleint_delta^1 fracnx^nx^2+2019dxleint_delta^1 fracnx^ndelta^2+2019dx. tag2$$
But
$$ int_delta^1 fracnx^n2020dx=frac12020fracnn+1(1-delta^n+1), int_delta^1 fracnx^ndelta^2+2019dx=frac1delta^2+2019fracnn+1(1-delta^n+1). tag3 $$
From (1),(2) and (3), one has
$$ frac12020fracnn+1(1-delta^n+1)le nI_nle fracnn+1delta^n+1+frac1delta^2+2019fracnn+1(1-delta^n+1).$$
Letting $ntoinfty$ gives
$$ frac12020leliminf nI_nlelimsup nI_nle frac1delta^2+2019.$$
Letting $deltato1^-$ gives
$$ liminf nI_n=limsup nI_n=frac12020 $$
or
$$ lim_ntoinftynI_n=frac12020.$$

Let $J_n=n I_n$, notice that when you use integration by parts, you would get
$$J_n=left[ fracx^n+1x^2+2019right]_0^1 -int_0^1fracx^n(2019-x^2)(x^2+2019)^2 dx= frac12020-int_0^1fracx^n(2019-x^2)(x^2+2019)^2 dx$$

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

We can write
beginalign
nx^n over x^2+2019 &= nx^n over 2019 cdot 1over 1+x^2over 2019 \
&= nx^n over 2019 cdot left(1-x^2over 2019+left[x^2over 2019right]^2-left[x^2over 2019right]^3+cdotsright) \
&=nx^nover 2019cdotsum_k=0^inftyx^2k(-1)^kover (2019)^k \
&=nover2019cdot sum_k=0^infty x^n+2k(-1)^kover (2019)^k.
endalign

Therefore,

beginalign
int_0^1 nx^nover x^2+2019,dx &=int_0^1 nover2019cdot sum_k=0^inftyx^n+2k(-1)^kover (2019)^kdx \
&= nover2019cdotsum_k=0^infty(-1)^kover (2019)^kint_0^1 x^n+2k,dx \
&= sum_k=0^infty nover n+2k+1cdot(-1)^kover (2019)^k+1.
endalign

Finally,

beginalign
lim_nto inftyint_0^1 nx^nover x^2+2019,dx &= lim_nto inftysum_k=0^infty nover n+2k+1cdot(-1)^kover (2019)^k+1\
&= sum_k=0^infty (-1)^kover (2019)^k+1\
&= 1over 2020.
endalign

$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
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newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
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newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]lim_n to inftyparsnint_0^1x^n over
x^2 + 2019,dd x,,,stackrelx mapsto 1 - x=,,,
lim_n to inftybracksnint_0^1pars1 - x^n over
x^2 -2x + 2020,dd x
\[5mm] = &
lim_n to inftybracksnint_0^1exponlnpars1 - x over
x^2 -2x + 2020,dd x
\[5mm] = &
lim_n to inftyparsnint_0^inftyexpo-nx over
0^2 -2 times 0 + 2020,dd xqquadparstextLaplace's Method
\[5mm] = & bbx1 over 2020 approx 4.9505 times 10^-4
endalign

Laplace's Method.