Let $X \neq \emptyset$ and $f, f_n:X \to \mathbb R$ such that $f_n \to f$ pointwise, i.e., $f_n (x) \to f(x)$ for all $x \in X$.
I would like to ask if any of below statements is correct. The meaning is that the "deviation" of the "tail" of $f_n$ from the limit $f$ would be bounded by some $\varepsilon>0$ uniformly across all $x \in X$.
There are $N \in \mathbb N$ and $\varepsilon>0$ such that $$ |f_n (x) - f (x)| \le \varepsilon \quad \forall n \ge N, \forall x \in X. $$
There are $N \in \mathbb N$ and $\varepsilon>0$ such that $$ |f_n (x)| \le |f (x)| + \varepsilon \quad \forall n \ge N, \forall x \in X. $$
I guess above statements are not true for pointwise convergence, but I could not come up with a counter-example. Could you elaborate on them?
Consider $X=\mathbb R$ and $f_n(x)=\frac xn$, $x\in\mathbb R$. Then $f(x)=0$. But any of $f_n$ is unbounded, so "$|f_n(x)|\le\epsilon$ for all $x\in X$" cannot hold for any $\epsilon>0$ and $n$.
Here is another example where $X=[0,1]$ is compact, $f_n:[0,1]\to\mathbb R$ and $f:[0,1]\to\mathbb R$ are continuous, and $f_n\to f$ pointwise. But both statements in OP fail. Let $$f_n(x)=\begin{cases} 2n^2 x,& 0\leq x\leq\frac1{2n},\\ 2n-2n^2x, & \frac1{2n}\leq x\leq\frac1n,\\ 0, & \frac1n\leq x\leq 1.\end{cases}$$ Then $f(x)=0$. In this case, both statements in OP are eqivalent to that there exists $N\in\mathbb N$ and $\epsilon>0$ such that $|f_n(x)|<\epsilon$ for all $n\geq N$ and all $x\in[0,1]$, which implies that $$\sup_{n>N}\sup_{x\in[0,1]}|f_n(x)|<\infty.$$ This cannot be true for our example. Because in our example, $\sup_{x\in[0,1]}|f_n(x)|=n$ for all $n\in\mathbb N$.