# Uniform convergence of $\frac{1}{x^3}\cos\left(\frac{x}{n}\right)$

Determine if $\frac{1}{x^3}\cos\left(\frac{x}{n}\right)$ uniformly converges on $E_1$ and $E_2$, where $E_1 = (0, 1)$ and $E_2 = (1, +\infty)$.

The pointwise convergence is trivial, since $\lim \limits_{n \to \infty} \frac{1}{x^3}\cos(\frac{x}{n}) = \frac{1}{x^3}$ for both $E_1$ and $E_2$.

I could also prove that it does not uniformly converge on $E_1$ by picking $x_n = \frac{1}{n^2}$ and plugging it into the original function.

However, I don't know how to prove uniform convergence on $E_2$, because I don't know how to evaluate $\sup f_n(x)$.

What are the methods to prove or disprove uniform convergence on $E_2$?

As regards $E_2=(1, +\infty)$ , note that by letting $t=x/n$, $$\sup_{x\in(1, +\infty)} \left|\frac{1}{x^3}-\frac{1}{x^3}\cos\left(\frac{x}{n}\right)\right|=\sup_{t\in(1/n, +\infty)} \frac{1-\cos(t)}{n^3t^3}\leq \frac{1}{2n^3}\sup_{t\in(1/n, +\infty)} \frac{1}{t}=\frac{n}{2n^3}=\frac{1}{2n^2}$$ where we used the fact that for $\frac{1-\cos(t)}{t^2}\leq \frac{1}{2}$ for any $t>0$ (it suffices to show that this function is bounded).