The definition of the limit of a sequence is:
$L=\lim\limits_{n\to\infty}f(n)\Leftrightarrow\forall\epsilon>0:\exists N\in\mathbb{N}:\forall n\in\mathbb{N}:\left(n>N\Rightarrow d_Y\left(f(n),L\right)<\epsilon\right)\,,$
for $f:\mathbb{N}\to Y$, $L\in Y$, and $d_Y$ a metric on $Y$, and the definition of the limit in a metric space is
$L=\lim\limits_{x\to x_0}f(x)\Leftrightarrow\forall\epsilon>0:\exists \delta>0:\forall x\in X:\left(d_X(x,x_0)<\delta\Rightarrow d_Y\left(f(x),L\right)<\epsilon\right)\,,$
for $f:X\to Y$ and $d_X$ a metric on $X$.
Question: does a metric $d_X$ exist on $X=\mathbb{N}$ such that the first definition is a special case of the second definition?
My thoughts: Perhaps $d_\mathbb{N}(x,y)=\left|\frac{1}{x}-\frac{1}{y}\right|$ would do, but $\infty\notin\mathbb{N}$, so I would have to define the metric on $\mathbb{N}\cup\left\{\infty\right\}$, but then $f$ also needs to be defined for $\infty$, which is normally not the case.
One possibility is the following: you can define the metric in $\mathbb{N}\cup\infty$ as you suggest: $d(x,y)=|\frac{1}{x}-\frac{1}{y}|$, $d(x,\infty)=|\frac{1}{x}|$. Suppose you have a function $f:\mathbb{N}\longrightarrow Y$. Then
The function $f$ is extensible continuously to $\bar{f}:\mathbb{N}\cup\infty\longrightarrow Y$ if and only if the sequence related to $f$ has limit in the first sense, and in this case, $\bar{f}(\infty)=L$. Moreover, now the expression $\lim_{n\longrightarrow \infty}\bar{f}(n)=L$ is valid and is stated in the second sense (because we are using $\bar{f}$).