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##.