Follow up question to question involving $\lim_{n\to \infty}\frac{\int_0^1 \left(x^2-x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n \, dx}$ I tried to answer Q 4482921 by elementary calculus but got stuck :

The first step in my solution is to replace $x$ by $2u$ so that $dx=2du$. The integral becomes $$\lim_{n\to \infty}\frac{\int_0^1 \left(x^2-x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n \, dx}= \lim_{n\to \infty}\frac{\int_0^\frac{1}{2}\left(4u^2-2u-2\right)^n \ 2du}{\int_0^1 \left(4 x^2-2 x-2\right)^n \, dx}$$$$=2\left(1-\lim_{n\to \infty}\frac{\int_
\frac{1}{2}^1 \left(4x^2-2x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n \,
dx}\right)$$

Now I need to prove that $\displaystyle\lim_{n\to \infty}\frac{\int_
\frac{1}{2}^1 \left(4x^2-2x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n\, dx} =0.$ It certainly looks like it in the graph, and I also found that $$\int_{\frac 12}^1 \left(4 x^2-2 x-2\right)^n\, dx= \int_{-1}^{0} \left(4 x^2-2 x-2\right)^n\, dx.$$

$f(x)=x^2-x-2$ has a minimum at $x=1/2,$ decreases for $x<1/2$ and increases for $x>1/2.$
In particular, $f(x)<0$ on $(0,2).$
So for $1<2x<2,$ $0=f(2)>f(2x)>f(1)=-2.$ So:
$$\left|\int_{1/2}^1 f(2x)^n\,dx \right|\leq 2^{n-1}.$$
On the other hand, when $\frac16<x<\frac13,$ $f(2x)\leq f(1/3)=f(2/3)=-\frac{20}{9}.$ So:
$$\left|\int_{0}^1 f(2x)^n\,dx \right|\geq \frac16\left(\frac{20}9\right)^n.$$
So the absolute value of your quotient is less than: $$3\left(\frac{9}{10}\right)^n,$$ which converges to $0.$
We could have picked any symmetric interval around $\frac{1}{4}$ of length less that $\frac{1}{4},$ $(1/6,1/3)$ was just the simplest.

More generally, if $h(x)$ is a continuous non-negative function on $[a,b]$ which is not zero everywhere, and with all its maximum values taken in $[a,c)$ for some $c\in (a,b),$ then $$\frac{\int_c^b h(x)^n\,dx}{\int_a^b h(x)^n\,dx}\to 0.$$
Essentially, if $M=\sup_{x\in [a,b]} f(x)$ is the maximum, and $M_1=\sup_{x\in [c,b]} f(x)$ satisfies $M_1<M,$ then for some interval $(u,v)$ in $(a,c)$ we have the $h(x)>\frac{M_1+M}{2}$ for $x\in(u,v)$ and this:
$$\int_{c}^b f(x)^n\,dx \leq (b-c)M_1^n.$$
and:
$$\int_{a}^b f(x)^n\,dx \geq (v-u)\left(\frac{M_1+M}2\right)^n.$$
So you quotient is bounded above by $$\frac{b-c}{v-u}\left(\frac{2M_1}{M+M_1}\right)^n$$
We have $0<\frac{2M_1}{M+M_1}<1,$ so this goes to zero.
In your case, $h(x)=2+2x-4x^2,$ $a=0,b=1,c=1/2,$ $M=2\frac12$ at $x=1/4,$ and $M_1=h(1/2)=2.$
