# Prove that $f:X\rightarrow\textbf{R}$ converges to $L$ at $x_{0}$ iff it maps convergence sequences onto convergence sequences.

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 $$L$$.

MY ATTEMPT

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?

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).