Can we use the delta-ep def of a limit to find a limiting value?

Question

I had a random thought recently that we don't really use the delta epsilon definition for evaluating limiting values. So, I decided to have a crack at it just now. Could somebody check whether my reasoning is sound or if I've just assumed some nonsense somewhere?

Example Using only the definition of a limit, evaluate $$\lim_{x\to0}x=L$$ for some limiting value L. It does not matter whether L exists or not.

$Solution.$ Let us consider the function $f:(-a,a)\to\mathbb{R}$ for $f(x)=x$. As per our definition for a limit, there exist a $L$ such that for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in(-a,a)$ $$0<|x|<\delta\ \implies\ |x-L|<\varepsilon.$$ Rewriting our first inequality using the triangle inequality give us $$|x|=|(x-L)+L|\geq||x-L|-|L||.$$ Rearranging $0<|x-L|-|L|<\delta$ gives us $$-\delta<|x-L|-|L|<0 \quad \text{or} \quad 0<|x-L|-|L|<\delta.$$ It follows that $$|L|-\delta<|x-L|<|L| \quad \text{or} \quad |L|<|x-L|<|L|+\delta$$ As such, we know $$|x-L|<|L|\ \text{or}\ |x-L|>|L|\ \implies\ |x-L|\neq|L|$$ This means either: (1) $x$ is not $0$ and/or $x$ is not $2L$, or (2) $L=0$.

However, the former is contradiction since $0<|x|<\delta$ implies $|x-L|<\varepsilon$, which would need to hold for $x=0$ since $0\in(-a,a)$. Therefore $L=0$ if $L$ exists.

Answer 1

As per our definition for a limit, there exist a $L$ such that for every $\varepsilon>0$ ... .

I hope what you intended to say was

As per our definition for a limit, if $f$ has a limit at $0$ then there exist a $L$ such that for every $\varepsilon>0$ ... .

Also ask yourself this: at what point in this proof did you invoke the specific properties of $f$? I have another function $g$ defined on $(-a,a)$ by $$ g(x) = \begin{cases} 1/x & x \neq 0, \\ 0 & x = 0. \end{cases} $$

Does this proof work equally well if we change the definition of $f$ to the definition of $g$? Yes, it does, because no part of the definition of $f$ was ever invoked. Yet obviously $g$ does not have a limit at $0$, so there must be something wrong with the proof.

Other than the missing words right at the beginning, the reasoning seems good to me up to and including this part:

This means either: (1) $x$ is not $0$ and/or $x$ is not $2L$, or (2) $L=0$.

But notice that if $L\neq 0$ then (1) is satisfied for every $x$ due to the "and/or": if $L\neq 0$ and $x = 0$ then (1) is satisfied due to satisfying "$x$ is not $0$", whereas if $L\neq 0$ and $x \neq 0$ then (1) is satisfied due to satisfying "$x$ is not $2L$".