Limits: I can solve them, but epsilon-delta feels like a magic trick. What am I missing?

10 answers & replies

• [+][–]Arjun Nair12thSample· 4d ago

  Think of it as a game with an opponent. The opponent says: "I want $f(x)$ to be within $\varepsilon$ of $L$", where $\varepsilon$ is any tolerance they like, as tiny as they want. You reply: "Fine. If $x$ is within $\delta$ of $a$, I guarantee it." The limit exists if you can always reply, whatever $\varepsilon$ they pick. So the opponent chooses $\varepsilon$, you choose $\delta$, and $\delta$ will usually depend on $\varepsilon$.

  For $3x+1$: $|f(x)-7| = |3x-6| = 3|x-2|$. To make that less than $\varepsilon$ you need $|x-2|<\varepsilon/3$. So $\delta=\varepsilon/3$. And yes, working backwards is exactly how you find $\delta$ (scratch work). The proof is the same steps written forwards: "Given $\varepsilon>0$, take $\delta=\varepsilon/3$. If $0<|x-2|<\delta$, then $|3x+1-7| = 3|x-2| < \varepsilon$." Done.

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• [+][–]Harsh Gupta12thSample· 4d ago

  so basically the limit is just f(a), the value at that point? like for x² at 3 it's 9, because you just put x = 3. why make it so complicated lol

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  • [+][–]Ishaan Bose11thSample· 4d ago

    no wait, that's the thing it's NOT about. take $(x^2-1)/(x-1)$ at $x=1$. It's $0/0$, not even defined there. But the limit is 2 because for every other $x$ it equals $x+1$. That's why the definition has "$0<|x-a|$": it deliberately ignores $x=a$ itself. Limit is about what happens near $a$, not at $a$. For nice functions like $x^2$ it happens to match $f(a)$ (that's continuity).

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• [+][–]Meera Patel11thSample· 4d ago

  On your $x^2$ attempt: $\delta=\varepsilon$ does not work in general. Test $\varepsilon=1$, so $\delta=1$. Then $x=3.5$ is allowed, but $|3.5^2-9| = 3.25 > 1$. Plugging in a couple of convenient values cannot prove it, only one bad value can disprove it.

  The right way: $|x^2-9| = |x-3|\,|x+3|$. The factor $|x-3|$ is the one you control. You need to bound the other factor. Suppose you first agree to take $\delta\le1$. Then $2<x<4$, so $|x+3|<7$. Hence $|x^2-9| < 7|x-3|$. To make that less than $\varepsilon$, you want $|x-3|<\varepsilon/7$. So choose $\delta=\min\!\left(1,\ \varepsilon/7\right)$. That handles both requirements.

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  • [+][–]Dev Agarwal12thSample· 4d agoAuthor

    ohh. so the min is because I need two conditions at once, one to keep $x$ near 3 so the other factor stays bounded and one to make the product small. okay that finally looks like something I can do myself. let me try $x^2$ at 5 on my own

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• [+][–]Rohan Sharma11thSample· 4d ago

  why "for every" ε though? why not just "there exists some ε"? seems like it's making the definition harder than it needs to be

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  • [+][–]Vivek Mishra12thSample· 4d ago

    If it is only for one ε, then many wrong answers also pass. For example take ε = 100. Then a lot of different numbers L are within 100 of f(x), so you cannot say which one is the limit. You need every small ε, even the tiny ones, so only one L can survive. That is what I understood.

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• [+][–]Riya Singh12thSample· 4d ago

  I'm a poetry person so take this as feeling and not as maths: a limit is a promise you can keep without ever arriving. "Within δ of a" is how close you're willing to stand, and the definition says: however near the answer asks me to be, I can always get there. And you never have to touch the point. It's the closest thing maths has to a love letter.

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• [+][–]Aman YadavDropperSample· 3d ago

  On usefulness for JEE. Directly, almost none. They give limits to compute, not to prove. Indirectly it helps you understand continuity and differentiability questions, because those are defined with limits. One evening on it is fine. Don't spend a week.

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  • [+][–]Dev Agarwal12thSample· 3d agoAuthor

    got it all. summary for myself: ε = opponent's tolerance, δ = my reply, usually δ depends on ε, limit ignores x = a itself, and for x² near 3 it's min(1, ε/7). thanks all, this was way better than the book

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