Lets $X$ be a subset of $\textbf{R}$, let $f:X\rightarrow\textbf{R}$ be a function, let $E$ be a subset of $X$, let $x_{0}$ be an adherent point of $E$, and let $L$ be a real number. Then the following two statements are logicall equivalent

(a) $f$ converges to $L$ at $x_{0}$ in $E$.

(b) For every sequence $(a_{n})_{n=m}^{\infty}$ which consists entirely of elements of $E$ and converges to $x_{0}$, the sequence $(f(a_{n}))_{n=m}^{\infty}$ converges to $L$.


I am mainly interested in the implication $(b)\Rightarrow (a)$.

Let us suppose the condition on $(b)$ holds and assume that $f$ does not converge to $L$ at $x_{0}$.

According to the corresponding definition, there is an $\varepsilon > 0$ such that for every $\delta > 0$ there is an $x\in E$ which satisfies \begin{align*} |x - x_{0}| \leq \delta\quad\wedge\quad |f(x) - L| > \varepsilon \end{align*}

If we choose $\delta = 1/n$, there corresponds a $x_{n}\in E$ such that $|x_{n} - x_{0}| \leq 1/n$ and $|f(x_{n}) - L| > \varepsilon$.

Hence we conclude (due to the squeeze theorem) that $\displaystyle \lim_{n\rightarrow\infty}x_{n} = x_{0}$.

On the other hand, $f(x_{n})$ does not necessarily converge, but if it does then $\displaystyle\lim_{n\rightarrow\infty}|f(x_{n}) - L| \geq \varepsilon > 0$, which contradicts our assumption. Therefore the proposed result holds.

Could someone verify if I am reasoning correctly?


The reasoning is correct, there is just a tiny detail in your last paragraph:

You have that $|f(x_n)-L|>\varepsilon$ for every $n\ge m$. That immediately implies that the sequence $\{f(x_n)\}_{n=m}^\infty$ does not converge to $L$. So you've already reached your contradiction, because the sequence $\{x_n\}_{n= m}^\infty$ did converge to $x_0$, and the two things put together contradict (b).

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