I have been studying a lot on continuity and differentiability the last 2 days and have had the following questions nagging me throughout that I think now is a good time to clear up. These question may be amateurish but please bare with me.

In the following, a is any Real number and assume domain and codomain of all functions are all real numbers.

Q1. If I have a function $f(x) = \frac1x$ and I must check the continuity at $x = 0$, we get the left hand limit $-\infty$ and right hand limit $+\infty$ and at f(0) the function is undefined. Here, is the type of discontinuity asymptotic (infinite) or missing point?

Q2. If only one of the limits in Q1 was either $+\infty$ or $-\infty$ what type of discontinuity would it be assuming the other limit is finite and the function remains undefined at that value?

Q3. If the LHL $=$ RHL $= +\infty$ at a given point '$a$' and $f(a)$ is undefined would this be removable by defining the function as $= +\infty$ at that point. (ie: would this be defined as a removable discontinuity considering LHL $=$ RHL $\ne f(a)$) or would this remain asymptotic owing to the fact that we cannot redefine a function at $x = a$ to be $+\infty$?

Q4. This is based on one of the exercises in my reference book. They have defined a function as follows:

$f(x) = \frac{x - 1}{x}$

Now, the question asked is what are the points and types of discontinuity in:


The way I had attempted this question was by noting that at $x = 0$ the function is not defined. Hence, when $f(x) = 0$ (ie: $x = 1$), $f(f(x))$ would be undefined and similarly when $f(f(x)) = 0$, (ie: $x = 0$), $f(f(f(x)))$ is undefined. So clearly, $x = \{0, 1\}$ are our points of discontinuity. However, when I saw that the book had defined them as 'missing point' discontinuities, that had me stumped considering that for $f(x)$ when $x = 0$, the LHL is $-\infty$ and RHL is $+\infty$ so (if the answer to Q1 is asymptotic as my intuition would predict) then at $x = 0$ and $x = 1$, we should have asymptotic discontinuities right?


To ask whether $f(x)$ is continuous at $a$, $a$ must be in the domain of $f$. So, for Q1 and Q2 we can't talk about continuity of $1/x$ at $0$. If one of the left and right hand limits were finite, it's still not a removable discontinuity. It might be helpful to adopt this classification of discontinuities: if $f$ is discontinuous at a point and both left and right hand limits exist (are finite), then $f$ has a simple discontinuity. Otherwise it has a discontinuity of the second kind. If $f$ has a simple discontinuity at a point in which the left and right limits agree, then $f$ is said to have a removable discontinuity at that point

For Q3, If $f$ is not allowed to take values in the extended real numbers $\mathbb{R} \cup \{-\infty, +\infty\}$, then we can't define $f(a) = \infty$ so there isn't a removable discontinuity.

For Q4, it is easy to check that $f(f(f(x))) = x$ although this is not defined at $0$ or $1$. But the left and right limits agree at both $1$ and $0$ for $x$ so the discontinuities are removable.

  • $\begingroup$ For your first statement, I had specifically said that let us assume that the domain and codomain of all functions are the set of Real numbers so I believe we can discuss the continuity of f(x) = 1/x when x = 0 as x = 0 is in the domain of the function. For example, tan x is discontinuous in the domain of Real numbers but is continuous in its own domain. Your classification of discontinuities is of great help! I'm assuming Q3 is a special case where although the limits agree, it is still not removable. For Q4 I had realised fofofx = x but didn't know we had to check for limits in that form! $\endgroup$ – Svee May 29 at 11:13
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    $\begingroup$ @Svee 0 is not in the domain of $1/x$ since it is not mapped. $f$ is a function, so every point in the domain of $f$ has an image. What would be the image of $0$ if it was indeed in the domain of $x\mapsto 1/x$? $\endgroup$ – fwd May 29 at 11:26
  • $\begingroup$ I understand this concept of 0 not being in the domain. I assumed we say that the function isn't defined at 0 as there are functions where the domain of the function includes some value where the function isn't defined. Hence, wherever such points occur, the function is discontinuous. For example, there is a question in my book where the function is (1 - tanx)/(4x - pi) where they have said the function is continuous in interval [0, pi/4] where clearly for pi/4, the function isn't defined so they have simply said if x is not equal to pi/4 what is the value of f(pi/4). $\endgroup$ – Svee May 29 at 11:33
  • $\begingroup$ For more information on the type of continuity where the function simply isn't defined at x = a, check point continuity here $\endgroup$ – Svee May 29 at 11:36
  • $\begingroup$ @Svee A precise definition of continuity of a function at a point requires that the function be defined at that point. In the example you provided, that function is continuous on $[0, \pi/4)$ but it can be extended to a function which is continuous on $[0, \pi/4]$ $\endgroup$ – fwd May 29 at 11:46

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