Calculate limit of ratio of these definite integrals How do I evaluate the following limit?
$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}$
 A: This is just a matter of substitution. Note that
$$\int_0^2 (4v^2-2v-2)^n dv = \frac{1}{2}\int_0^1 (x^2-x-2)^n dx$$
by substituting $2v=x$.
Hence
$$\lim_n \frac{\int_0^1 (x^2-x-2)^n dx}{\int_0^2 (4v^2-2v-2)^n dv} = 
\lim_n \frac{\int_0^1 (x^2-x-2)^n dx}{\frac{1}{2}\int_0^1 (x^2-x-2)^n dx} =
\lim_n 2 = 2$$
A: $\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}=\lim_{n\to\infty}\frac{\int_{0}^1\left(2+x-x^2\right)^n dx}{\int_{0}^1\left(2+2x-4x^2\right)^n dx}$
Let $I_{n}=\int_{0}^1\left(2+x-x^2\right)^n dx$
$(2+x-x^2)^n\geq 2^n$ for $0\leq x\leq 1$.Hence,$I_{n}\geq 2^n$
Let $J_{n}=\int_{0}^1\left(2+2x-4x^2\right)^n dx$
Now,$J_{n}=\frac{1}{2}\int_{0}^2(2+x-x^2)^n dx=\frac{1}{2}\int_{0}^1(2+x-x^2)^n dx+\frac{1}{2}\int_{1}^2(2+x-x^2)^n dx$
Let $K_{n}=\int_{1}^2(2+x-x^2)^n dx$
Now,$x^2-2x+1\geq 0$.Therefore, $2+x-x^2\leq 3-x$.Thus,
$K_{n}=\int_{1}^2(2+x-x^2)^n dx\leq\int_{1}^2(3-x)^n dx=\frac{2^{n+1}-1}{n+1}$
$K_{n}=\int_{1}^2(2+x-x^2)^n dx\leq\frac{2^{n+1}-1}{n+1}$
$0\leq \frac{K_{n}}{I_{n}}\leq \frac{2^{n+1}-1}{2^n(n+1)}$ 
$\lim_{n\to \infty} \frac{K_{n}}{I_{n}}=0$
$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}=\lim_{n\to \infty}\frac{I_{n}}{J_{n}}=\lim_{n\to \infty}\frac{2I_{n}}{I_{n}+K_{n}}=\lim_{n\to \infty}\frac{2}{1+\frac{K_{n}}{I_N}}=2$
Is there any other straight forward way.
A: Maybe what follows may lead to a complete answer. Let$$
I(a)=\int_0^1 (a^2x^2 -ax -2)^n dx
= 
a^{2n}\int_0^1 ((x-\frac{1}{2a})^2 -\frac{9}{4a^2})^n dx
\\
= 
a^{2n}\int_{-\frac{1}{2a}}^{1-\frac{1}{2a}} (y^2 -\frac{9}{4a^2})^n dy
$$
Another change of variable
$$y=\frac{3}{2a}z
$$
later we get
$$
I=a^{2n}\frac{3}{2a}(\frac{3}{2a})^{2n}\int_{-\frac 1 3}^{\frac{2a-1}{3}}(z^2-1)^n dz
=\frac 1 a (\frac{3}{2})^{2n+1}J
$$
with
$$
J
=
\int_{-\frac 1 3}^{\frac{2a-1}{3}}(z^2-1)^n dz.
$$
Your problem amounts to evaluate (warning: shabby notation ahead)
$$
\frac{J(1)}{J(2)}=\frac{\int_{-1/3}^{1/3}}{\int_{-1/3}^{1}}
=
1-\frac{\int_{1/3}^{1}}{\int_{-1/3}^{1}}
.$$
Clearly the integrand of $J$ is the greatest around $z=0$, however I could not obtain a condition showing what this ratio tends to.
$$
%\int_{\pm 1/3}^1 (z-1)^n(z+1)^ndz = - \frac{n}{n+1}\int_{\pm 1/3}^1(z-1)^{n+1}%(z+1)^{n-1}dz
%+[\frac{1}{n+1}(z-1)(z^2-1)^{2n}]_{\pm 1/3}^1
%$$
