Suppose that $f$ is a real-valued function defined on a set $S$ in $\mathbb{R}^{n}$ and that $c$ is a limit point of $S$. Prove that $\lim _{x \rightarrow c} f(x)=L$ if and only if, for every Cauchy sequence $\left(x_{k}\right)$ in $S$ that converges to $c, \lim _{k \rightarrow \infty} f\left(x_{k}\right)=L$.
It seems the crucial fact for this problem could be the convergence of the given Cauchy sequence, in which the "if" part becomes more manageable in that respect. However, the other direction becomes more complex and I could not come up with good approaches for that.
Suppose $\lim_{ x \rightarrow c}f(x)= L$ is given. Then for any $\epsilon>0~~~\exists~\delta>0$ such that \begin{equation} |f(x)-L|<\epsilon \end{equation} whenever $\|x-c\|<\delta$. For the above $\delta$ there exists $N\in \mathbb{N}$ such that \begin{equation} \|x_k-c\|<\delta \end{equation} for all $k >N$. Combining we get $|f(x_k)-L|<\epsilon$ for all $k>\mathbb{N}$. Thus $\lim_{k \rightarrow \infty}f(x_k)=L$.
For reverse part, suppose if possible $\lim_{ x \rightarrow c}f(x)\neq L$. Then there exists $\epsilon >0$ such that for all $\delta >0~~~\exists~x \in S$ for which $0<\|x-c\|<\delta$ but $|f(x)-L|>\epsilon$ ...(1)
Take a sequence $(\delta_n)$ in $(0,1)$ which converges to $0$. Now for each $\delta_i~~~\exists~x_i \in S$ for which \begin{equation} 0<\|x_i-c\|<\delta_i ~~~\textit{but}~~~ |f(x_i)-L|>\epsilon \end{equation}
This implies $\lim_{i \rightarrow \infty}x_i=c$ but $\lim_{i \rightarrow \infty}|f(x_i)-L|\geq \epsilon >0$. This is a contradiction So we must have $\lim_{x \rightarrow c}f(x)=L$.