As a new year's resolution, I'm going through basic exercises and here is one that causes me some trouble. One needs to study the following limit using the dominated convergence theorem:
$$\lim_{n\rightarrow \infty}\ n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx$$
What I can say:
- By integration by parts, and solving a difference equation, one can certainly obtain the answer but that does not seem to be the goal of the exercise.
- The integral alone (without the $n^2$ factor) tends to 0: one can simply dominate by the constant function 1, and since the integrand converges to 0. Hence we are with an indetermined limit, something that goes to infinity times something that goes to 0.
- Doing the change of variable $y=nx$ leads to $$ n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx = n \int_0^n \left(1- \frac{y}{n}\right)^n sin\left(\pi \frac{y}{n}\right)\, dy$$ while $y=n^2 x$ leads to $$ n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx = \int_0^{n^2} \left(1- \frac{y}{n^2}\right)^n sin\left(\pi \frac{y}{n^2}\right)\, dy$$ In the first case the integral alone goes to 0 while n goes to infinity. The last formula seems promising, unfortunately I'm not able to dominate the integrand by an integrable function.
- I also thought about cutting the integral in two parts but at the moment it leads nowhere.
Ok, do this :$$n^2\int_0^1(1-x)^n\sin(\pi x)=\int_0^n\left(1-\frac xn\right)^n n\sin\left(\frac{\pi x}{n}\right).$$Of course here $$\left|n\sin\left(\frac{\pi x}{n}\right)\right|\le\pi x;$$a little calculus shows that $$\log(1-x/n)\le-x/n\quad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|\le g=\lim f_n=\pi xe^{-x}$.)