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.

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.
