About the limit of a function

Question

We know that if $A\subseteq\mathbb{R}$ and $x_{0}\in \mathbb{R}$ a cluster point of $A$. Then the real number $L$ is called limit of $f:A\rightarrow \mathbb{R}$ at $x_{0}$ if for all $\epsilon>0$, there is a $\delta_{\epsilon}>0$ such that $\forall x\in A$ such that $0<|x-x_{0}|<\delta_{\epsilon} \Rightarrow |f(x)-f(x_{0})|<\epsilon$. My questions are (i) Why we take a cluster point here instead of any real number? (ii) $\delta$ depends on $\epsilon$ but in the example $f(x)=\frac{1}{x}$ for all $x\in A=(0,\infty)$, we have for all $\epsilon>0$, a $\delta$ is $\delta=\inf\{\frac{x_{0}}{2},\frac{x_{0}^{2}}{2}\}$ so that $f(x)\rightarrow \frac{1}{x_{0}}$ as $x\rightarrow x_{0}$. Clearly $\delta$ depends on $\epsilon$ and $x_{0}$. In continuity, we called uniform continuity. Can we called here the convergence uniform convergence if $\delta$ depends only on $\epsilon$.

Answer 1

(i) Because if $x_0$ is not a cluster point, then any real number $L$ is a limit of $f$ at $x_0$.

(ii) No. Uniform convergence is about sequences of functions. But it is true that, in general, $\delta$ depends upon $\varepsilon$ and $x_0$.