Compute $\lim \limits_{n\to \infty} \int_3^4 (-x^2+6x-8)^\frac{n}{2} dx$

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 $a=e^{\log(a)}$ $$\frac{ \Gamma \left(\frac{n+2}{2}\right)}{ \Gamma \left(\frac{n+3}{2}\right)}=\frac {\sqrt 2 } {n^{1/2}}-\frac{3}{2 \sqrt{2}}\frac 1 {n^{3/2}}+\cdots$$