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 $$\int_{\frac 12}^1 \left(4 x^2-2 x-2\right)^n\, dx= \int_{-1}^{0} \left(4 x^2-2 x-2\right)^n\, dx.$$
 A: $f(x)=x^2-x-2$ 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 $M_1=h(1/2)=2.$
A: For limits as $n \to \infty$, we often don't need exact values of complicated terms, just a general comparison to the size of the result in terms of $n$.
First notice $4x^2 -2x-2 = 2(x-1)(2x+1)$ is negative on the region $0 \leq x < 1$, so $-4x^2+2x+2$ is positive (and both are zero when $x=1$).
To show convergence to zero, we can look at absolute values. When $n$ is odd, the integrand is never positive, and when $n$ is even, the integrand is never negative, so we don't have any "cancellation", and
$$ \left| \int_a^b (4x^2 -2x -2)^n\, dx \right| = \int_a^b |(4x^2-2x-2)^n| \, dx = \int_a^b (-4x^2+2x+2)^n\, dx $$
as long as $0 \leq a < b \leq 1$.
Also completing the square,
$$ -4x^2+2x+2 = -\left(2x-\frac{1}{2}\right)^2 + \frac{9}{4} $$
So if $0 \leq x \leq \frac{1}{2}$, then
$$ \left|2x-\frac{1}{2}\right| \leq \frac{1}{2} $$
$$ -4x^2+2x+2 \geq -\left(\frac{1}{2}\right)^2 + \frac{9}{4} = 2 $$
$$ \int_0^{1/2} (-4x^2+2x+2)^n \, dx \geq \int_0^{1/2} 2^n\, dx = 2^{n-1} $$
And since $(-4x^2+2x+2)^n$ is positive,
$$ \int_0^1 (-4x^2+2x+2)^n\, dx \geq \int_0^{1/2} (-4x^2+2x+2)^n\, dx \geq 2^{n-1} $$
The parabola is always less than or equal to the tangent line at $x=\frac{1}{2}$:
$$ -4x^2+2x+2 = -4\left(x-\frac{1}{2}\right)^2 -2x+3 \leq 3-2x $$
$$ \int_{1/2}^1 (-4x^2+2x+2)^n\, dx \leq \int_{1/2}^1 (3-2x)^n\, dx = \left. -2^n \frac{(\frac{3}{2}-x)^{n+1}}{n+1} \right|_{x=1/2}^{x=1} = -2^n \frac{2^{-n-1}-1}{n+1} $$
$$\int_{1/2}^1 (-4x^2+2x+2)^n\, dx \leq \frac{2^n - \frac{1}{2}}{n+1} < \frac{2^n}{n+1} $$
Finally,
$$ \left| \frac{\int_{1/2}^1 (4x^2-2x-2)^n\, dx}{\int_0^1 (4x^2-2x-2)^n\, dx} \right| = \frac{\int_{1/2}^1 (-4x^2+2x+2)^n\, dx}{\int_0^1 (-4x^2+2x+2)^n\, dx} < \frac{\frac{2^n}{n+1}}{2^{n-1}} = \frac{2}{n+1} $$
So the limit as $n \to \infty$ approaches zero.
