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When proving the existence of a limit, I know about 2 methods. One is showing that the Right hand limit is equal to the left hand limit. Another one is using the epsilon delta definition of limit. Both are presented, but which one is used for what purpose? Is there a specific reason to prefer one over other? Don't they portray the same idea?

轻型八神
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  • $\epsilon, \delta$ is part of definition of limit and this is used to prove theorems which are then used for evaluation of limits. Note that the definition itself can't be used directly to evaluate limits. See https://math.stackexchange.com/a/2674529/72031 – Paramanand Singh Jan 15 '23 at 01:29
  • Further existence of limit is essentially based on many theorems (depending on specific scenarios, the left right limits are one such technique). $\epsilon, \delta $ again can't be used to prove existence of a limit. – Paramanand Singh Jan 15 '23 at 01:31
  • @ParamanandSingh "ϵ,δ again can't be used to prove existence of a limit." I disagree with this statement. If I show that for some $c$ and $L$ the following statement is true: "$\forall \epsilon > 0, \exists \delta > 0$ s.t. $\forall x, 0<|x-c|<\delta \Rightarrow |f(x) - L| < \epsilon$", then I can say that the limit exists. Now how to find that the right value of $L$ for a given $c$ is a separate matter. But $\epsilon$-$\delta$ can be used to prove existence of a limit. $\epsilon$-$\delta$ can't help with finding the value of $L$, but this is not what you said. – Josh Jan 15 '23 at 02:05
  • @Josh: Existence of a limit is usually proved without the knowledge of $L$. The definition is meant is check if any given number $L$ is the limit or not. It is not for guaranteeing whether such an $L$ exists or not. – Paramanand Singh Jan 15 '23 at 04:03
  • @Josh: you should compare the scenario with the folllowing: proving that there exists a real root of $x^3-2=0$ is a very different matter from checking whether $x=1.1$ is a root or not. – Paramanand Singh Jan 15 '23 at 04:05
  • I understand what you mean, but the statement "ϵ,δ again can't be used to prove existence of a limit" is false. A successful application of $\epsilon$-$\delta$ reveals two things, among other things:
    1. the limit exists
    2. the limit is equal to $L$

    so $\epsilon$-$\delta$ can be used to prove the existence of a limit!

     – Josh Jan 15 '23 at 04:16

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For the most part, if the functions are well-behaved the two methods are equivalent. However, $\epsilon$-$\delta$ definition is definitely more rigor, and usually introduced in more advanced classes. The right-hand, left-hand approach is more intuitive.

To see why they are equivalent, note that in the $\epsilon$-$\delta$ defintion we are showing that:

For any given $\epsilon > 0$, $\exists \delta$ such that $\forall x$ $$0<|x-c|<\delta \Rightarrow |f(x)-L| < \epsilon$$

The right-hand, left-hand approach, in the language of $\epsilon$-$\delta$ definition, states that $\forall \epsilon > 0, \exists \delta > 0 \text{ such that }$ $$0 < x - c < \delta \implies f(x) - L < \epsilon$$ and $$0 < c - x < \delta \implies L - f(x) < \epsilon $$ This method is based on the idea that the function gets arbitrarily close to a particular value $L$ as the input gets arbitrarily close to a particular point $c$, and it is useful in situations where a more intuitive approach is needed.

In short, both methods are acceptable. Which one to use is a context-sensitive question. The right-hand, left-hand approach is more intuitive and might be an easier, but $\epsilon$-$\delta$ approach is rigorous and considered formal.

Josh
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  • This might be a bit off topic, but why is rigorous math preferred over intuition? Does rigorous definitions make something more "correct" than intuitive definitions? – 轻型八神 Jan 14 '23 at 18:49
  • Those rigorous definitions and theorems are the ones that come from the mathematical axioms. Then we can use those rigorous ideas to prove theorems that make our lives easier. Left-hand, right-hand approach is an example of it. It is easier to teach someone who is learning for the first-time about the left-hand, right-hand limit than epsilon-delta which is more abstract. We do not have to be rigor always, but if you are proving something for the first time (in terms of advanced math), you have to rely on rigorous ideas. Intuition may mislead us, but you cannot go wrong with rigorous proofs. – Josh Jan 14 '23 at 18:59
  • For an example, see http://jdh.hamkins.org/all-triangles-are-isosceles/ It gives a seemingly accurate proof that all triangles are isosceles (which is obviously not true). The problem is that the proof follows a diagram, which is based on intuition. Now if there is a flaw in our diagram, it would be carried over to the proof.

    The main point: We can make mistakes, if we rely on intuition. It is a good starting place, but eventually, it needs to be formalized.

     – Josh Jan 14 '23 at 19:03
  • Even Ramanujan's inscrutably powerful intuition could be mistaken, as in supposing $li(x)\le \pi(x)$ for all $x\ge 2$, which was disproved by Littlewood. – DanielWainfleet Jan 14 '23 at 20:22
  • The use of left and right limits is based on theorems and is perfectly rigorous. Your answer seems to indicate that it is less rigorous than $\epsilon, \delta$. Moreover the definition of limit can't be used directly to infer existence of a limit. For that you need various theorems. – Paramanand Singh Jan 15 '23 at 01:34
  • Yes it's rigor. I did not say it is not rigor! But I would also say that the use of left-right limit is more aligned with human intuition than $\epsilon$-$\delta$.That's why it is easier to prove limit questions with left-right limits. Now this is a subjective matter, and you might find that $\epsilon$-$\delta$ is more intuitive. That is fine too. If you treat "rigorness"as a binary value, then the two approaches are equally rigor. But if you look at it as a spectrum (which is a subjective matter), then I'd say $\epsilon$-$\delta$ is more rigor, and it's completely fine to disagree on this one – Josh Jan 15 '23 at 02:19