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Let's say I have a function that looks something like this. $$ f(x) = \begin{cases} x & x\leq 0 \\ 1 & x=1 \\ x & x\ge 2 \end{cases} $$ What would the values of $$\lim_{x\to1^-} f(x)$$ and $$\lim_{x\to1^+} f(x)$$ be? My guess is that it should be $0$ and $2$, respectively because we can ignore the regions $(0, 1)$ and $(1, 2)$ since the function isn't defined there.

Also, is the function continuous at $x = 1$?

What if I change the function to $$ f(x) = \begin{cases} 1 & x\leq 0 \\ 1 & x=1 \\ 1 & x\ge 2 \end{cases} $$ ? What would the limits be now? Would this new function be continuous?

Apologies if a similar question has been asked before. I didn't find anything relevant based on my searches.

My definition of limit is this: $$\forall\epsilon>0\ \exists\delta>0\ \forall x\in dom(f)\ (0 < |x - a| <\delta \implies |f(x) - l| < \epsilon)$$ So for the second case, if $l = 1$, for every $\epsilon\gt0$, we can choose $ = 2$, and we'll have $(0<|−|< ⟹ |()−|<)$. So limit at $1$ should be $1$. Is there a flaw in my understanding?

DevRS
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    What is your definition of limit? Can the limit exist (by definition) at an isolated point? – user Sep 27 '24 at 21:09
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    Both functions are continuous but limit at $x=1$ doesn't exist. – user Sep 27 '24 at 21:13
  • Refer to https://math.stackexchange.com/q/455296/505767 – user Sep 27 '24 at 21:13
  • @user my definition of limit is this: ∀>0 ∃>0 ∀ (0<|−|< ⟹ |()−|<) So for the second case, if = 1, for every >0, we can choose = 2, and we'll have (0<|−|< ⟹ |()−|<). So Iimit at 1 should be 1. Is there a flaw in my understanding? – DevRS Sep 27 '24 at 21:42
  • At an isolated point this is not true $\forall x$ with $0<|x-a|<\delta$, in other words we need that $x=a$ is an accumulation point. – user Sep 27 '24 at 22:33
  • @user I'm sorry but why is this not true? Just to clarify, when we say ∀x, we mean ∀x in $domain(f)$, right? – DevRS Sep 27 '24 at 22:41
  • Refer to what discussed here https://math.stackexchange.com/q/27429/505767 – user Sep 27 '24 at 22:54
  • The key point is that the definition works fine when $x=a$ is an accumulation point but for isolated points it doesn't work, you are right $L=1$ is the limit but also $L=2$ works by $\delta=1/2$. – user Sep 27 '24 at 22:58
  • You should add your first coment in your question. – user Sep 27 '24 at 22:59
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    @user added the comment! And thank you for the clarification! It makes sense now! – DevRS Sep 27 '24 at 23:14
  • You are welcome! Bye – user Sep 27 '24 at 23:15

1 Answers1

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According to the definition of right (left) limits:let E and X are metric space ,d is the metric function of E$f:E \to X$, for all $\varepsilon$>0,exist $\delta$, so that , for all $x\in E$ that satisfy $0<x-x^0<\delta (0<x^0-x<\delta)$ , $|f(x)-f(x^0)|<\varepsilon$, so $\lim\limits_{x\to1^-}f(x)$does not exist (precisely,it is not defined, but if we admit $x=x^0$, the limit is 1 ),so do others. On the other side,the definition of continous at $x^0$ is: for all $\varepsilon$>0,exist $\delta$, so that , for all $x\in E$ that satisfy $d(x,x^0)<\delta$, $d(f(x),f(x^0))<\varepsilon$,so both of the answers are yes.

wille zur macht
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