If a function has a limit from the right but not from the left, is it still continuous?
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1$\begingroup$ No, because that limit does not exist. $\endgroup$– JnxFSep 5, 2016 at 3:32
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7$\begingroup$ You need to consider the domain over which the function is defined. The function $f:[0,\infty)\rightarrow \mathbb{R}$ given by $f(x) = \sqrt{x}$ is continuous. $\endgroup$– MichaelSep 5, 2016 at 3:34
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$\begingroup$ Related: math.stackexchange.com/questions/637280/… $\endgroup$– Hans LundmarkDec 27, 2019 at 17:10
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3 Answers
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It is continuous at $0$.
By construction, the domain of the square-root function is $\mathbb R_+=[0,\infty)$. Now, for any sequence $(x_n)_{n\in\mathbb N}$ in the domain (that is, $x_n\geq 0$ for all $n\in\mathbb N$) that converges to $0$, one has that the corresponding function values $\sqrt{x_n}$ also converge to $\sqrt{0}=0$.
And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.
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2$\begingroup$ Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts. $\endgroup$– LubinJan 23, 2019 at 16:32
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$\begingroup$ @Lubin in what sense 1/x is discontinuous is a false statement? Usually when saying this, textbooks assume the so called infinity type of discontinuity, which apply precisely to points where a function is not defined and tends to infinity. I do understand 1/x is continuous on (0,infty) if you mean that, but I wouldn’t say it is false to say that as a function on R it has an infinity type discontinuity at zero. $\endgroup$– xyzApr 26, 2021 at 0:56
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1$\begingroup$ Golly @GGG , what is “infinity”? Is the reciprocal function defined at $0$ or not? If it is defined, then I guess you have to specify what the value is. That one value would have to be some “$\infty$” creature, and then you’d need to specify what its neighborhoods were. Once you do all of that, then I’ll talk to you about the function being defined but discontinuous at zero. $\endgroup$– LubinApr 26, 2021 at 18:49
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$\begingroup$ @Lubin Pardon but isn t that exactly what the extended real line is for? $\endgroup$– xyzApr 27, 2021 at 16:33
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$\begingroup$ Well, @GGG , which extended real line do you choose? One where there is an infinity at each end, or one with a single infinity? If the former, you have to decide which infinity you choose for a value at zero, in which case the reciprocal function certainly is discontinuous. If a single infinity, you’re speaking, essentially, of the projective line, and the reciprocal function is continuous. Too many choices, for my money. $\endgroup$– LubinApr 27, 2021 at 19:42
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One can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we can't talk about limits for values $x < 0$.
$\sqrt x $ is continuous everywhere it exists.
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$\begingroup$ Well, $0$ is in the domain of $\sqrt x$, so...? It is absolutely valid to say that a function $f:[0,\infty) \to \mathbb R$ is continuous (or not) at $0$. $\endgroup$– Pedro ♦Sep 5, 2016 at 4:10
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1$\begingroup$ Lots of downvotes for an obvious typo. How rude! $\endgroup$ Sep 5, 2016 at 6:54
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2$\begingroup$ @MathematicsStudent1122: The problem wasn't $\le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected. $\endgroup$ Sep 7, 2016 at 19:47
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By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous. In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.
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1$\begingroup$ So, @user361424, restricting it to real numbers would make it discontinuous at x = 0? $\endgroup$ Sep 5, 2016 at 3:33
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3$\begingroup$ Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x \le 0 it simply doesn't make sense to talk of sqrt being continuous. $\endgroup$ Sep 5, 2016 at 3:37
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$\begingroup$ Yes, restricting it to real numbers would make it discontinuous at 0. $\endgroup$ Sep 5, 2016 at 3:41
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