Compute $\lim \limits_{n\to \infty} \int_3^4 (-x^2+6x-8)^\frac{n}{2} dx$ Compute $$\lim \limits_{n\to \infty} \int_3^4 (-x^2+6x-8)^\frac{n}{2}dx.$$
I am interested in a method to compute this as simply as possible. I know that by DCT this is $0$, but I am not allowed to use it. With the substitution $t=x-3$ I got that this is $\int\limits_0^1 (1-t^2)^\frac{n}{2}dt$ and by using that $e^x\ge x+1, \forall x\in \mathbb{R}$ I could show that the limit is $0$. This is anyway pretty complicated for the level of the exam where this was given, I would be interested in something even easier. Is it possible to write a recurrence relation for instance?
EDIT: Based on Ian's answer I came up with the following solution and I would like to know whether it works:
Let $\epsilon \in (0,1)$ and $I_n=\int_0^1 (1-t^2)^\frac{n}{2} dt$.
$$I_n=\int_0^{\epsilon}(1-t^2)^{\frac{n}{2}}dt+\int_{\epsilon}^{1}(1-t^2)^{\frac{n}{2}}dt\le \epsilon + (1-\epsilon^2)^\frac{n}{2}, \forall n\in \mathbb{N}$$ 
After we take the limit as $n\to \infty$ we get that $\lim\limits_{n\to\infty} I_n \le \epsilon, \forall \epsilon \in (0,1)$ and if we now let $\epsilon \searrow 0$ it follows that $\lim\limits_{n\to\infty} I_n \le0$ and since $I_n\ge 0$ we get that the limit is $0$.
I think this is basically what Ian did, but I would like to know whether it is correct to write it like this.
 A: May be too complex.
What you did is good. You end with
$$I_n=\int\limits_0^1 (1-t^2)^\frac{n}{2}\,dt$$ Now, make $t=\sin(u)$ to work with
$$I_n=\int_0^\frac \pi 2 \cos^{n+1}(u)\,du=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+2}{2}\right)}{ \Gamma
   \left(\frac{n+3}{2}\right)}$$ Now, take logarithms, use Stirling approximation to get
$$\log\left(\frac{ \Gamma \left(\frac{n+2}{2}\right)}{ \Gamma
   \left(\frac{n+3}{2}\right)}\right)=\frac{1}{2} \log \left(\frac{2}{n}\right)-\frac{3}{4
   n}+O\left(\frac{1}{n^2}\right)$$ Now, using $a=e^{\log(a)}$
$$\frac{ \Gamma \left(\frac{n+2}{2}\right)}{ \Gamma
   \left(\frac{n+3}{2}\right)}=\frac {\sqrt 2 } {n^{1/2}}-\frac{3}{2 \sqrt{2}}\frac 1 {n^{3/2}}+\cdots$$
A: For direct estimation (as opposed to evaluation in terms of special functions followed by estimation as in Claude's answer), your change of variable is convenient for computation. Introduce a parameter $\varepsilon \in (0,1)$ and write
$$\int_0^1 (1-t^2)^{n/2} dt = \int_0^{\varepsilon} (1-t^2)^{n/2} dt + \int_{\varepsilon}^1 (1-t^2)^{n/2} dt.$$
The first term is less than $\varepsilon$ because the integrand is bounded above by $1$ and the interval is of length $\varepsilon$; the second term is less than $(1-\varepsilon^2)^{n/2}$ for essentially the same reason. 
Now tune $\varepsilon(n)$ so that $\lim_{n \to \infty} \varepsilon(n)+(1-\varepsilon(n)^2)^{n/2} = 0$. In particular you find that you need $\varepsilon(n) \to 0$ and yet also $n \log(1-\varepsilon(n)^2) \to -\infty$. For the latter it suffices to have $n \varepsilon(n)^2 \to \infty$. So $\varepsilon(n)=n^{-1/3}$ will work.
Then the squeeze theorem gets you what you want.
You could also have done this without a change of variable, but then it would perhaps not be so obvious that $-x^2+6x-8$ is strictly decreasing starting from $1$ on $[3,4]$, which is really what we used here. Writing it as $1-(x-3)^2$ helps with seeing that.
A: Partial progress on another method : with the substitution $t=\sin u$ we get that $$\int_0^1 (1-t^2)^\frac{n}{2}dt=\int_0^\frac{\pi}{2}\cos^{n+1}(u)du$$
Let $I_n = \int\limits_0^\frac{\pi}{2}\cos^{n+1}(u)du$. By IBP we get that $I_n=\frac{n}{n+1}I_{n-2}$.
Now we may write that $I_{2n}=\frac{2n}{2n+1}I_{2n-2}$ and this gives us that $I_{2n}=\frac{2\cdot 4\cdot ...\cdot (2n)}{3 \cdot 3\cdot 5\cdot...\cdot (2n+1)},\forall n\in \mathbb{N}$ and I don't know how to compute this limit. I tried the ratio test, but it failed.
