Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, let $E$ be a subset of $X$, and let $f:X\to Y$ be a function. Let $x_{0}\in X$ be an adherent point of $E$ and $L\in Y$. Then the following statements are logically equivalent:
(a) $\displaystyle\lim_{x\rightarrow x_{0};x\in E}f(x) = L$
(b) For every sequence $(x_{n})_{n=1}^{\infty}$ in $E$ which converges to $x_{0}$ with respect to the metric $d_{X}$, the sequence $(f(x_{n}))_{n=1}^{\infty}$ converges to $L$ with respect to the metric $d_{Y}$.
MY ATTEMPT
Let us prove the implication $(a)\Rightarrow(b)$ first.
According to the definition of limit, for every $\varepsilon > 0$ there is a $\delta > 0$ such that for every $x\in E$ one has that \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),L) < \varepsilon \end{align*}
Consequently, if $x_{n}\to x_{0}$, for every $\delta > 0$, there is a natural number $N\geq 1$ such that \begin{align*} n\geq N \Rightarrow d_{X}(x_{n},x_{0}) < \delta \end{align*}
Given that $x_{n}\in E$ for every $n\geq N$, we can substitute $x$ by $x_{n}$, whence we conclude that for every $\varepsilon > 0$, there is a natural number $N\geq 1$ such that \begin{align*} n\geq N \Rightarrow d_{X}(x_{n},x) < \delta \Rightarrow d_{Y}(f(x_{n}),L) < \varepsilon \end{align*}
and $f(x_{n})\to L$.
Conversely, let us prove that $(b)\Rightarrow(a)$.
Let us suppose that $(b)$ holds, but $(a)$ does not.
Consequently, there is a $\varepsilon > 0$ such that for every $\delta > 0$ there is a $x\in E$ which satisfy \begin{align*} d_{X}(x,x_{0}) < \delta\quad\wedge\quad d_{Y}(f(x),L)\geq\varepsilon \end{align*} Thus for every $\delta = 1/n$ corresponds a $x_{n}\in E$ satisfying $d_{X}(x_{n},x_{0}) < 1/n$. Taking the limit, one concludes that $x_{n}\to x_{0}$.
On the other hand, $f(x_{n})\not\rightarrow L$, which contradicts our assumption.
Hence the proposed result holds.