Given the function $$f(x)=\begin{cases} x^2, & x<1 \\ 2x-1, & x\geq3 \end{cases}$$ May I say $f$ is continuous at $x=3$? My question is basically about the fact $$\lim_{x\to3^{-}}f(x)$$ doesn't exist.
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1Do I correctly understand, that your function is not defined in $[1,3)$? – zkutch Jun 20 '23 at 13:44
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@zkutch yes, that is it. – mvfs314 Jun 20 '23 at 13:44
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3Your map is continuous at $3$ as you can see using this definition of continuity when the domain is not the whole real line. By the way, it would be useful that you mention in your question what definition of continuity you use. Doing so, I imagine that you'll be able to answer your question by yourself! – mathcounterexamples.net Jun 20 '23 at 13:45
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2Then $\lim\limits_{x\to3^{-}}$ simply have no sense. You can speak about continuity only from right. – zkutch Jun 20 '23 at 13:45
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1Does this answer your question? What is the limit of a function which is not defined everywhere?. Yours is potentially a duplicate in abstract. – Jam Jun 20 '23 at 19:35
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Yes, the function is continuous, the limit does not need to exist for the funtion to be continuous. What continuity gives is that, if the right and left hand limit exist, then they are equal to the value of the function at that point.
The basic definition of continuity (at least which I learnt first) is the sequential definition, not the one using limits:
Let $f:A\rightarrow B$ be a function. It is said to be continuous at $c\in A$, if for every sequence $(x_n)$ in $A$ that converges to $c$, the sequence $(f(x_n))$ converges to $f(c)$.
Notice that it does not say that a sequence should exist that converges to $c$ (well, apart from the constant sequence).
Suppose we have a function $f:\mathbb{N}\rightarrow\mathbb{R}$, no matter how
we define $f$, it will be continuous. Does this answer your question?
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