Not understanding the concept well, I am trying to determint the pointwise and uniform convergence of the following sequence of function:
$$f_n(x) = \frac{\sin{nx}}{n^3}, x \in \mathbb{R}$$
The only part I understand so far is that I need $\lim_{x\to\infty}{f_n(x)}$ in which I have determined that (if I am correct):
$$\lim_{x\to\infty}{f_n(x)} = \lim_{x\to\infty}{\frac{\sin{nx}}{n^3}} = 0$$
Where do i go from here?
With what you wrote I think that you want to study the sequence $\left(f_n\right)_{n \in \mathbb{N}}$. With what you wrote, you have proved that the sequence $\left(f_n\right)_{n \in \mathbb{N}}$ converge pointwisely on $\mathbb{R}$ to the function $ \ f: x \mapsto 0$. Now you need to prove the following statement for proving that it converges uniformly : $$ \left\|f_n-f\right\|_{\infty,\mathbb{R}} \underset{n \rightarrow +\infty}{\rightarrow}0 $$ Here $f=0$ and for all $x \in \mathbb{R}$ $$ \left|f_n\left(x\right)-f\left(x\right)\right|=\left|\frac{\sin\left(nx\right)}{n^3} \right|\leq \frac{1}{n^3} $$ It is true for all $x \in \mathbb{R}$ so in particular for the upper bound $$ \left\|f_n-f\right\|_{\infty, \mathbb{R}}\leq \frac{1}{n^3} $$ Hence
$$ \left\|f_n-f\right\|_{\infty,\mathbb{R}} \underset{n \rightarrow +\infty}{\rightarrow}0 $$ The convergence is hence uniform.
Roughly, the deciding factor in uniform convergence is whether the "rate" of convergence of $f_n(x)$ to its limit is the same for all $x$.
A counterexample: let $f_n(x)=x/n$. The functions $f_n$ converge to $0$ pointwise. However, the time you have to wait for $f_n(x)$ to get within $0.01$ of $0$ depends on $x$. For example, when $x=3$, then the smallest $n$ such that $|f_n(x)|<0.01$ is $n=300$, while for $x=30$, the smallest such $n$ is $n=3000$. As $x$ gets further from $0$, the required $n$ to get $f_n(x)$ within $0.01$ of $0$ gets arbitrarily large, which means $f_n$ does not converge uniformly.
Back to your problem. The function $f_n(x)=\sin(nx)/n^3$ is more complicated, so given $\epsilon$, we cannot find the exact smallest $n$ for which $|f_n(x)|<\epsilon$. However, all we need to do is note that $|\sin nx|\le 1$ always, so that $|f_n(x)|\le 1/n^3$. Note that this bound does not depend on $x$; it gives us a guarantee that after $n$ is large enough, $f_n(x)$ will be close to $0$ for all $x$. Therefore, $f_n$ does indeed converge uniformly.
