Prove $f:\mathbb{N} \rightarrow \mathbb{R}$ is continuous using the definition of sequential continuity The definition of sequential continuity is that $x_n \rightarrow x \implies f(x_n) \rightarrow f(x)$.
If the terms of the sequence $\{x_n\}$ are only natural numbers, I know that for all $\epsilon > 0$, we can find an $N \in \mathbb{N}$ such that for all $n \geq N$, $x_n = x$.
I'm not sure what to do from here

Any function $f : \mathbb N \to \mathbb R$ is continuous. To show this using sequential continuity, let $\{x_n\}$ be a sequence in $\mathbb N$ that converges to $x$. A convergent sequence in $\mathbb N$ is eventually constant. But then, $\{f(x_n)\}$ is also eventually constant. Say $f(x_n) = f(x)$ for $n > N$ for some $N$. It follows that $\left|f(x_n) - f(x)\right| = 0$ for $n > N$. Therefore, $f$ is continuous.
