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I want to prove the statement that: $\lim_{x \to c} f(x)=L$ if and only if $\lim_{x \to 0} f(x+c)=L$. There are some questions with this problem on this forum, but some of them are solved with substitution, what I don't like or without reverse direction. So my proof.

$"\Rightarrow"$ Since $\lim_{x \to c} f(x)=L$ we have that $\forall \ \epsilon>0 \ \exists \ \delta>0,$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.**

We can also use that $|x-0|<\delta$. So $|x-0|=|(x+c)-c|<\delta$. Now apply ** and we get $\lim_{x \to 0} f(x+c)=L$.

I have a problem with reverse direction. It's easy to prove with contrapositive argument. But can someone help me to prove it direct?

mathguruu
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  • Your forward proof needs work as well. You must start with the statement you want to prove, let $\epsilon\gt 0$ be given and then find the appropriate $\delta$ given that the limit you are assuming to be true has a $\delta$. You assume the left limit is true and prove that the right limit is true. You then assume the right limit is true and prove that the left limit is true. Also, it is $\lvert f(x)-L\rvert\lt\epsilon$, not $f(x)-L\lt\epsilon$. – John Douma Dec 31 '22 at 12:50
  • @JohnDouma I edited it, thank you. But I proved forward direction – mathguruu Dec 31 '22 at 12:56
  • "We can also use that $|x-0|<\delta$." I think this is incorrect. In general we do not have $\vert x \vert < \delta.$ – Adam Rubinson Dec 31 '22 at 13:45
  • What is wrong with substitution? – Li Kwok Keung Dec 31 '22 at 13:47
  • @LiKwokKeung OP probably doesn't understand why substitution works, which is what this question should have been about. – Adam Rubinson Dec 31 '22 at 13:48
  • @AdamRubinson no ) the reason is, I think, that direct proof is better for understanding) – mathguruu Dec 31 '22 at 13:50
  • @AdamRubinson why is it incorrect? It's hypothesis of the second part. We can assume it. – mathguruu Dec 31 '22 at 13:52
  • Substitution is not part of "indirect proof". However, in order to use substitution, you do have to use an axiom I think it's known as the indiscernibility of identicals. See here: https://math.stackexchange.com/questions/607312/axioms-of-equality/607350#607350 This might be the wrong axiom as I'm not an expert in first-order logic. But I do think there is some axiom of substitution you technically need in order to justify making one. Also in your proof, haven't you basically done a substitution of replacing $x$ by $y=x+c$ in ** ? – Adam Rubinson Dec 31 '22 at 13:58
  • @AdamRubinson actually yes. I made a substitution too. You are right. – mathguruu Dec 31 '22 at 14:02

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Seems that the reverse direction proof is basically the same as the forward proof.

$"\Leftarrow"$

Since $$\lim_{x \to 0}f(x+c)=0$$

$$$$

we have $\forall \epsilon \gt 0, \exists \delta \gt 0$ s.t. $|x| \lt \delta \implies |f(x+c)-L| \lt \epsilon.$ $$$$ Hence $\forall \epsilon \gt 0, \exists \delta \gt 0$ s.t. $|x-c| \lt \delta \implies |f\left((x-c)+c\right)-L| \lt \epsilon.$ $$$$ Hence $\forall \epsilon \gt 0, \exists \delta \gt 0$ s.t. $|x-c| \lt \delta \implies |f(x)-L| \lt \epsilon.$ $$$$ Hence $$\lim_{x \to c}f(x)=L$$

Li Kwok Keung
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