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

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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?

Calculus Limits

Paxton

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You may use $\epsilon-\delta$ definition of limits to find the limit, but it is usually not necessary and there are much easier ways to do so. But lets focus on your example now.

You say "This means either: (1) $x$ is not $0$ and/or $x$ is not $2L$, or (2) $L=0$" why?! This statement is not accurate and it is not clear how you get that conclusion. I would say your proof is not entirely correct and more complicated that it should be.

This is how I would prove this:

Without loss of generality we may assume that $\delta \leq \epsilon$, otherwise we may define $\delta'=\min\{{\delta, \epsilon}\}$. Suppose $L \neq 0$. Then for
$$0<\epsilon <\frac{|L|}{4}$$
we have
\[|L|-|x|\leq |L-x|<\epsilon \] \[ \Rightarrow |L| \leq |x|+\epsilon \leq \delta+\epsilon\leq \epsilon+\epsilon=2\epsilon.\]
Hence
\[\epsilon < \frac{|L|}{4} \leq \frac{\epsilon}{2} \Rightarrow \epsilon \leq \frac{\epsilon}{2},\]
which is a contradiction. Thus $L=0.$

Note also that this proof does not show that the limit is $0$. It shows that if the limit exists then it must be equal to zero. This is why it is generally a better idea to guess the limit using other methods and use the $\epsilon-\delta$ definition to prove it.

If you want to write a proof for another function, the argument may be different depending on the function.

Nirenberg

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

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