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

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 L. It does not matter whether the limiting value L exists or not.

$Solution.$ Let us consider the function $f:\mathbb{R}\to\mathbb{R}$ for $f(x)=x$. As per our definition for a limit, if $f$ has a $L$ as $x\to0$ then for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in\mathbb{R}$ $$0<|x|<\delta\ \implies\ |x-L|<\varepsilon.$$ 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 a contradiction since $0<|x|<\delta$ implies $|x-L|<\varepsilon$, which would need to hold for $x=0$ since $0\in\mathbb{R}$. Therefore $L=0$ if $L$ exists.

Final comments: My primary concern is that the definition of the function itself is not invoked when I’m manipulating the inequalities to produce some $L$. I had hoped that looking at the domain would be sufficient. So if we looked at $1/x$ instead this approach would correctly show that as $x\to0$ the function is undefined. However, the working would be identical and also redundant, since we would only be looking at the domain. Is there a legal way to do this or is this just a lost cause?

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