Calculate limit of ratio of these definite integrals How do I evaluate the following limit?
$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}$

$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}=\lim_{n\to\infty}\frac{\int_{0}^1\left(2+x-x^2\right)^n dx}{\int_{0}^1\left(2+2x-4x^2\right)^n dx}$
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.
