The second part of the proof $f(x) \not \rightarrow L \longrightarrow \{f(x_n)\}\not \rightarrow L$,
My attempt: $f(x) \not \rightarrow L$ implies given $\varepsilon >0$, there exists $\delta$ such that $|f(x)-L|\ge\varepsilon$ for $x \in S\setminus \{c\}$, and $|x-c|<\delta.$ We know that $\lim_{n \to \infty}x_n =c$. So, find $M$ such that for $n\ge M, |x_n -c| < 0$. Then, for $n \ge M, |\{f(x_n)\}-L|\ge 0$. Therefore, $\{f(x_n)\}$ does converge to L.
In my textbook, the author uses $\delta = 1/n$ for the second part of proof, while she does not specify $\delta$ for the first part of the proof. Should we necessarily specify $\delta$ for the second part of the proof or can we prove it, using my method?
Thank you in advance.
For $f(x)\rightarrow L$ is not valid, it is not equivalent to say given $\epsilon>0$, there exists $\delta>0$ such that $|f(x)-L|\geq\epsilon$ for $x\in S-\{c\}$ and $|x-c|<\delta$, this is simply too strong to be equivalent.
Rather, if we formalize how we say $f(x)\rightarrow L$, it is somehow like \begin{align*} \forall\epsilon>0\exists\delta>0\forall x(0<|x-c|<\delta\rightarrow|f(x)-L|<\epsilon). \end{align*} To get the negation of this statement, it is like \begin{align*} &\neg(\forall\epsilon>0\exists\delta>0\forall x(0<|x-c|<\delta\rightarrow|f(x)-L|<\epsilon))\\ &=\exists\epsilon>0\neg(\exists\delta>0\forall x(0<|x-c|<\delta\rightarrow|f(x)-L|<\epsilon))\\ &=\exists\epsilon>0\forall\delta>0\neg(\forall x(0<|x-c|<\delta\rightarrow|f(x)-L|<\epsilon))\\ &=\exists\epsilon>0\forall\delta>0\exists x\neg(0<|x-c|<\delta\rightarrow|f(x)-L|<\epsilon)\\ &=\exists\epsilon>0\forall\delta>0\exists x(0<|x-c|<\delta\wedge\neg(|f(x)-L|<\epsilon))\\ &=\exists\epsilon>0\forall\delta>0\exists x(0<|x-c|<\delta\wedge|f(x)-L|\geq\epsilon), \end{align*} which reads as there exists an $\epsilon>0$ such that for every $\delta>0$, there exists an $x$ for which $0<|x-c|<\delta$ and $|f(x)-L|\geq\epsilon$.