Let $A\subset\mathbb{R}$ and consider a function $f:A\rightarrow\mathbb{R}$ and let $c\in\mathbb{R}$ be a cluster point of A.
Theorem: The function $f$ does not have a limit at $c$ if and only if there exists a sequence $(x_n)$ in A with $x_n\ne c$ for all $n\in \mathbb{N}$ such that the sequence $(x_n)$ converges to c but the sequence $(f(x_n))$ does not converge in $\mathbb{R}$.
My attempt: Assume $\lim(x_n)=c$, but $ \lim (f(x_n))$ does not converge in $\mathbb{R}$. Since there exists a sequence that does not coverge , then by sequential criterion, the limit does not exists at $c$.
Conversely, let f does not have a limit at c.
This implies there exists a neighbourhood $V$ of $L$ such that for every neighbourhood $W$ of $c$ there exists at least one element $x_w\in[W-{c}]\cap A$ for which $f(x_w)$ does not belong to $V$.
Let $W_1=N(c,1)$. Then there exists an element $x_1\in N'(c,1)\cap A$ such that $f(x_1)\notin V$.
Let Let $W_2=N(c,\frac 12)$. Then there exists an element $x_2\in N'(c,\frac 12)\cap A$ such that $f(x_2)\notin V$.
Proceeding in this manner we obtain a sequence $t_n=(x_1, x_2, x_3, ...)$ in A such that $limx_n=c$, sincr $x_n\in W_n=N(c,\frac 1n)$ for all $n\in N$. Now $t_n$ is unbounded in R. This implies that $f((x_n))$ is also unbounded and so divergent in $R$.
Is this proof correct?
