# Deriving definition of continuity from from definition of limit of a function

Often, continuity is derived as a "special case" of the limit of a function, which is defined as: Let D be a subset of R. The function $$f:D \rightarrow R$$ is said to converge to $$y \in R$$ for $$x \rightarrow x0$$ (where $$x0$$ must be a limit point of D), i.e., $$\lim_{x \rightarrow x_0} f(x) = y$$ if and only if $$\forall \varepsilon >0 \quad\exists \delta>0 \quad\forall x \in D: \quad0<|x-x_0|<\delta: \quad|f(x)-y|<\varepsilon$$ I want to emphasize that $$x_0$$ need not be an element of the domain D of the function f (but it has to be a limit point thereof). As a result, $$x$$ must not be equal to $$x_0$$ which is expressed via $$\quad0<|x-x_0|$$. (I am aware that there is a newer definition of the limit of a function which allows that $$x=x_0$$, but I dont want to discuss this case here)

Now, continuity is often just defined by the requirement that $$y=f(x_0)$$ in the definition of the limit of a function, i.e., $$\lim_{x \rightarrow x_0} f(x) = f(x_0)$$ With the additional requirement that now $$x_0$$ need to be in D.

If I now just plug $$y=f(x_0)$$ into the above defintion of the limit of a function I will obtain the epsilon delta definition: $$\forall \varepsilon >0 \quad\exists \delta>0 \quad\forall x \in D: \quad 0<|x-x_0|<\delta: \quad|f(x)-f(x_0)|<\varepsilon$$

HOWEVER, in essentially all books, in the definition of continuity the requirement $$0<|x-x_0|$$ is ommitted, i.e., $$x=x_0$$ is allowed.

My question is: Why is this requirement $$0<|x-x_0|$$ ommitted in the definition of continuity?

What concerns me is just that the expression $$\lim_\limits{x \rightarrow x_0} f(x)$$ has a specific, fixed meaning which is the epsilon delta criterion of the limit of a function. And just by plugging in $$f(x_0)$$ for $$y$$ in that definition (and thereby obtaining the defintion of continuity) should not alter the definition. In other words, these two definitions seem not to be equivalent with continuity just being a special case where $$y=f(x_0)$$. Because, if I start out from the definition of continuity and want to recover the definition of the limit of a function and just substitute $$f(x_0)$$ by $$y$$, then I will come to a wrong definition of the limit of a function, missing the requirement that $$0<|x-x_0|$$.

• Because $f$ has to be defined at the point $x_0$ in order to be continuous at $x_0$. – Tab1e May 16 '20 at 7:13
• Yes, I know. But that doesnt mean that I have to allow that I also look at $x=x_0$ for the x in the $\delta$ neighbourhood of $x_0$. The definition would be just fine without allowing $x=x_0$ to my understandings – guest1 May 16 '20 at 7:50
• NO. If $x_0$ is an isolated point, it has to be the case $x=x_0$. – Tab1e May 16 '20 at 21:23
• Thats true. Great thanks! – guest1 May 18 '20 at 7:21

¿$$x=x_0$$ make sense? $$|x-x_0|$$ is usually omitted because we asume $$x_0\in B(x,\delta)$$ (or the equivalent of B in other topological space). Obviously $$x_0\in B(x_0,\delta)\;$$. Try to take a look of $$f:X\longrightarrow Y$$ is continue if $$\;\;\forall V$$ (open) in $$Y \; | \; f(x_0)\in Y$$ exist $$U\in X$$ (open) $$\; |\; x_0\in U$$ and $$f(U)\subset V$$