# About the limit of a function

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 $$\epsilon$$.

## 1 Answer

(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 $$x_0$$.