Definition of a limit of a function confusion

[fogvajarash]

Homework Statement

Show that $\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to 0} f(x+a) = L$

Homework Equations

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The Attempt at a Solution

For the forward direction (ie $1 \Rightarrow 2$), I tried to first assume that 1. holds true (ie $\forall \epsilon>0, \exists \delta>0, \forall x \neq a, x \in A, |x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$). Then, I let $x+a=t$, and then by our choice of $t$ we will have $|x|=|t-a|$, then if we choose $\delta$ such that $|t-a|<\delta$, it follows $|f(t)-L|<\epsilon$, so $|f(x+a)-L|<\epsilon$ proving statement 2.

However, I'm not quite sure if my argument is correct. The main concern I have is the chance that $t \notin A$, so we couldn't apply the definition to this case. If the above can't be done, which would be a better way to solve the problem?

Thank you for your help.​

 

[Fredrik]

Your solution is fine. You just need to rewrite your proof in a way that makes it a bit more clear how you're using the assumption and the definition of limit. For example like this: (Since you have solved the problem, I don't think it's against the rules to show you my version).

Let $\varepsilon>0$ be arbitrary. Suppose that $\lim_{x\to a} f(x)=L$. Let $\delta>0$ be such that for all $x\in A$,
$$0<|x-a|<\delta\ \Rightarrow\ |f(x)-L|<\delta.$$ For all $x\in A$ such that $0<|x|<\delta$, we have $0<|(x+a)-a|<\delta$, and therefore $|f(x+a)-L|<\varepsilon$. This implies that $\lim_{x\to 0}f(x+a)=L$.