# Conceptual doubts based on Continuity

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:

$$f(f(f(x)))$$

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.
• @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 $x\mapsto 1/x$? – fwd May 29 at 11:26
• @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 $[0, \pi/4]$ – fwd May 29 at 11:46