I am really struggling with this concept and I feel like no math book clarifies it sufficiently.

First off: I assume that if a one-sided limit exists then it is a real value and not $\pm \infty$. And a two-sided limit exists if both one-sided limits exist (i.e. both limits are real values) and are also equal to each other in value.

My understanding of continuity falls into three cases / mental notes:

  1. A function $f$ is considered continuous (i.e. continuous everywhere) iff it is continuous at every point in its domain.

  2. For a real endpoint $a$ of a domain interval: The function is continuous at $a$ if the one-sided limit $L$ exists and $L = f(a)$ (if the endpoint is on the left side of the interval, we refer to the one-sided limit from the right, and if the endpoint is on the right side of the interval, we refer to the one-sided limit from the left). Moreover $a$ as well as the infinitely-close $x$ value to $a$ must also be in the domain.

  3. For a real point $a$ within a domain interval, i.e. not the endpoints. The function is continuous at $a$ if the two-sided limit $L$ exists and $L = f(a)$. Moreover $a$ as well as the infinitely-close $x$ values on both sides of $a$ must also be in the domain.

If this is true so far then:

Removable Singularities:

An undefined (not in the domain) or vertically-shifted (in the domain) single point $a$ where the two-sided limit at $x=a$ nevertheless exists. This would be considered a discontinuity in the function either way due to definition 3 above. The two-sided limit $L$ exists, but $L \neq f(a)$ (although I am unsure if it makes sense to use $\neq$ if we're talking about $f(a)$ when undefined at $a$).

Jump Discontinuities

This is any value of $x=a$ in the domain where the left-sided limit and right-sided limit exist, but they approach different real values ($f(a)$ itself may or may not be defined). Now here's where I start to get fuzzy. If $f(a)$ is defined, the function is discontinuous by definition 3 since the two-sided limit does not exist. However, if $f(a)$ were undefined, then this would weirdly enough be continuous since we move into definition 2 for two intervals. Since $a$ is not part of the domain, we've just split things into two separate intervals, both of which would be considered continuous on their own based on endpoint continuity definition. Is this right?

Infinite/Essential Discontinuities (are these also Nonremovable Singularities?)

This is where I am even fuzzier. I believe these exist only when $f(a)$ is undefined, i.e. not in the domain, when at least one of the one-sided limits on either side does not exist (either because it approaches $\pm \infty$ or because it's oscillating forever and never settles down anywhere).

I'm even less certain about functions that are only defined in one direction but still have an asymptote. For example if we defined $f(x) = 1/x$ for $x>0$ only, would the asymptote at $x=0$ be considered an infinite discontinuity?

But let's say we had $f(x) = 1/x$ where $x \neq 0$. Would we say there is an infinite discontinuity at $x=0$ since we approach infinity in at least one direction, and yet this would not affect the overall continuity of $f$ since $0$ is not part of the domain?

I'm really unclear on how all of this works.

  • $\begingroup$ What is the question? $\endgroup$
    – copper.hat
    Commented Feb 8, 2018 at 17:17
  • $\begingroup$ If I am understanding these definitions and their usages correctly $\endgroup$
    – user525966
    Commented Feb 8, 2018 at 17:19
  • $\begingroup$ It sounds like you have a pretty good understanding. $\endgroup$
    – Doug M
    Commented Feb 8, 2018 at 17:20
  • $\begingroup$ @copper.hat there are a few questions scattered about the text. $\endgroup$
    – John Doe
    Commented Feb 8, 2018 at 17:20
  • $\begingroup$ @DougM But am I wrong anywhere? I am really unsure about the jump and infinite discontinuities. $\endgroup$
    – user525966
    Commented Feb 8, 2018 at 17:42

1 Answer 1


The following aims to clarify some of the questions by focusing on essential aspects of continuity.

At first I'd like to state a proposition which characterises continuity.

  • Let the function $f$ be defined at $X$ and let $\xi\in X$ be an accumulation point from both $\{x\in X: x<\xi\}$ and $\{x\in X: x>\xi\}$. The function $f$ is continuous at $\xi$ if and only if the one-sided limits $f(\xi-)$ and $f(\xi+)$ exist and they are $=f(\xi)$.

This proposition is stated as Satz 39.2 in the book Lehrbuch der Analysis vol. 1 by Harro Heuser.

We observe that $f$ in order to be continuous at a point $\xi\in X$ has to be defined there and $\xi$ has to be an accumulation point. This should help to clarify a corresponding statement in the section Jump discontinuities.

The author continues to classify continuities as following:

  • According to proposition 39.2 the function $f:(a,b)\rightarrow \mathbb{R}$ is discontinuous at $\xi\in (a,b)$ if and only if the following holds: The limits $f(\xi-)$ and $f(\xi+)$

    • exist and coincide, but are $\ne f(\xi)$;

    • exist and differ;

    • at least one of them does not exist.

  • In the first two cases $\xi$ is called a discontinuity of the first kind also called jump discontinuity, in the last case $\xi$ is called discontinuity of the second kind.

This classification might help to determine continuity versus discontinuity not only here but also in similar situations.

  • $\begingroup$ So under this definition, endpoints of an interval are not continuous because they do not have one-sided limits from both sides? $\endgroup$
    – user525966
    Commented Feb 11, 2018 at 17:47
  • $\begingroup$ @user525966: No. In the situation of endpoints we have to adapt the proposition and consider one one-sided limit only. Everything else is deduced from this situation analogously as stated in the answer. $\endgroup$ Commented Feb 11, 2018 at 18:00

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