I know, from calculus the definition of continuity is given below: $$\lim_{x \to a} f(x) = f(a).$$ So most of cases, if $f(x)$ continuous at some point, the limit also exists. However, I have a 'little silly' thought and observation like this.
Actually the definition of continuity at $x=a$ is: $$\forall \epsilon > 0, \exists \delta > 0, 0<|x-a|<\delta \Rightarrow |f(x)-f(a)| < \epsilon.$$
Take a look about this function: $$f: \mathbb{N} \rightarrow \mathbb{R}, f(n) = 1.$$ This is a discrete function and continu at $n=1$.
We see, for every $\epsilon > 0$, we can choose $\delta = 1$ such that $$|n-1| < \delta \Rightarrow |f(n)-1| = 0 < \epsilon.$$ The implication is true, since $|n-1| < \delta$ is false. (Vacuous truth)
But the function above doesn't have limit at $n=1$ because there is no $\delta >0$ such that $$(n-\delta, n+\delta) \cup \mathbb{N} \not = \emptyset.$$ So $n=1$ is not accumulation point. Hence $$\lim_{n \to 1} f(n)$$ is undefined.
Is my reason correct? Or there is a flaw in reasoning (?) Thanks in advance.
