$$f_n(x) = x^n - x^{3n}, \ \ D = [0,1] \\ 1) x = 0 \Rightarrow lim_{n \to \infty} 0^n - 0^{3n} = 0 \\ 2) x = 1 \Rightarrow \lim_{n\to \infty} 1^n - 1^{3n} = 0 \\ 3) \ \ 0<x<1 \Rightarrow \lim_{n \to \infty} x^n-x^{3n} = 0 $$
So $\lim_{n \to \infty} f_n = 0.$ Vai $\forall \epsilon > 0 \ \ \exists n_0 \in N: \forall n \ge n_0 \Rightarrow |f_n-0| < \epsilon \ \ \text{ or }$ $$ \\ \lim_{n \to \infty} \sup_{x \in D} |f_n(x) - 0| = 0 \Rightarrow |x^n-x^{3n}| < \epsilon \Rightarrow |x||1-x^2| < \epsilon^{\frac{1}{n}} \\ \max_{x \in D}|1-x^2| = 1 \Rightarrow 0 < x < \epsilon^{\frac{1}{n}} $$
Or I could express $\epsilon$ the other way:
$$\max_{x \in D} |x| = 1 \Rightarrow |1-x^2| = 0 \Rightarrow 0 < \epsilon^{\frac{1}{n}}$$
I don't know whether I arrived at contradiction or my calculations is faulty? Also, I don't know how to interpret what I wrote.
You need to somehow find the $$ \sup_{x \in D} |f_n(x)| = \sup_{x\in D} x^n (1-x^{2n}) = [t = x^n] = \sup_{t\in D} t (1-t^{2}) $$ This can be found using derivative $$ \left( t - t^3 \right)' = 1 - 3t^2 = 0 \implies t =\frac{1}{\sqrt{3}} \implies x = x_n = \frac{1}{3^{\frac{1}{2n}}} $$
This gives us $$ \lim_{n \to \infty}\sup_{x\in D} x^n (1-x^{2n}) = \lim_{n\to \infty} f(x_n) = \frac{1}{\sqrt{3}}\left(1 - \frac{1}{3} \right) =\frac{2}{3\sqrt{3}}\neq 0. $$