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 $0<|x-x_0|$.

  • $\begingroup$ Because $f$ has to be defined at the point $x_0$ in order to be continuous at $x_0$. $\endgroup$
    – Tab1e
    Commented May 16, 2020 at 7:13
  • $\begingroup$ 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 $\endgroup$
    – guest1
    Commented May 16, 2020 at 7:50
  • $\begingroup$ NO. If $x_0$ is an isolated point, it has to be the case $x=x_0$. $\endgroup$
    – Tab1e
    Commented May 16, 2020 at 21:23

1 Answer 1


¿$x=x_0$ 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 $f(U)\subset V$


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