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Suppose I have ln(x), the domain is given as x > 0, range is all reals. Now suppose I asked for the points of discontinuity of ln(x). How would one answer this question?

Is there an infinite discontinuity at x = 0? Or is there an infinite set of discontinuous points x ≤ 0? Or is there no discontinuities since I effectively removed them by stating the domain for which ln is defined?

Any thoughts welcome,

mugged99
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1 Answers1

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A point of discontinuity is by definition in the domain of the function.

Another way of thinking of your question is to ask whether, given a function defined on part of the real line, there exists a continuous extension of it to other parts of the real line. For $\log$, it's pretty clear that there does not exist a continuous extension to $x=0$. So any way of defining $\log (0)$ would force the function to be discontinuous there.

Eric Auld
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  • How can there be a discontinuity if those points aren't in the domain; conversely how can points be in a domain if they are undefined. Thats my problem here; im not sure which school of thought is correct. – mugged99 Mar 22 '15 at 20:15
  • $0$ is not in the domain. So it is not a point of discontinuity. – Eric Auld Mar 22 '15 at 20:50
  • Right ok. What happens if you don't have a domain provided? – mugged99 Mar 22 '15 at 21:12
  • @mugged99 If you have a function that you know something about, but you're not sure where it can be defined, it's kind of a case-by-case thing. You might be able to say, "I have shown that I can define this function continuously on such and such a subset of $\mathbb{R}$," for example. There's no reason that every expression or every function should necessarily be able to take every real number as an input. Functions such as the square root and the logarithm have problems for negative numbers, for instance. You may want to get more down to specifics. – Eric Auld Mar 23 '15 at 02:47