WHY are we always given $\epsilon > 0$ first, then solving for a $\delta>0$? This is in the limit definition.
I want to ask:
Can we say "given $\delta>0$, there exists $\epsilon>0$"? Since we can always solve for one given the other.
I found three counterexamples, but I don't understand them:
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Let $f(x) = \sin x$, let $L$ and $\delta$ be arbitrary real numbers. Then $\epsilon = |L| + 2$ satisfies your definition. (from post)
Q: What's wrong with setting $\epsilon = |L| + 2$? It's big, but it's not wrong!
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Let $f(x) = 1/x$, and let $a = 1$. The definition fails for $\delta \ge 1$, since for any $\epsilon$ we can choose $x=1/(L+\epsilon)$ if $L+\epsilon > 1$, so that $f(x)-L \ge \epsilon$. (from post)
Q: What are they saying here? At $x=1$, the definition fails for $\epsilon \ge 1$ too! The problem is not $\delta$. The problem is the function is undefined for $x \le 0$.
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Counterexample: $\lim\limits_{x \to 0} f(x) = L$
$f(x) = \begin{cases} \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$
Given any $\delta > 0$, we can find $\epsilon > 0$ such that $|f(x) – L| < \epsilon$ whenever $|x| < \delta$. For instance, set $\epsilon = 2$; then any choice of $L \in (-1,1)$ will satisfy this "reversed" situation. (from post)
Q: I don't see how setting $\epsilon = 2$ violates any definition. I mean, we did find a $\epsilon$ for a given $\delta$.
Thanks all for the pouring answers, I'll get back to each one personally. If I did not choose an answer, that means all submissions are still welcomed! The best answer will be chosen based on # of upvotes (50%) and if I understood it and agree it's the best (50%).
Best Answer
The point of the definition of limit is to capture the idea that we can force the values of $f$ to be close enough to $L$, provided only that the values of $x$ be sufficiently near $a$: if you tell me how close you want $f(x)$ to be to $L$, I can guarantee this outcome by telling you how close $x$ should be to $a$.
Reversing the logical dependency between $\epsilon$ and $\delta$ makes the logical dependency run the "wrong" way: you are saying that you will tell me how close you want $x$ to be to $a$, and then I will be forced to tell you how close I can guarantee $f(x)$ to be to $L$.
It might seem like this would be good enough, but it doesn't work. It seems to be the same because you may be thinking of the limit as saying "the closer you get to $a$, the closer the values will get to $L$"; but saying the limit is $L$ is more than that: it says that values get arbitrarily close to $L$, and that all values get close to $L$ near $a$, not just some.
If you are allowed to pick the value of $\epsilon$, then you are not guaranteeing that the values get arbitrarily close to $L$, just that they get "sufficiently" close to $L$.
So for example, you want the limit of $f(x)$ to approach at most one thing, not two or more. But say that $f(x)$ always takes values between $-1$ and $1$, as $f(x)=\sin(x)$ does. If I take $L$ to be any value between $-1$ and $1$, and then let $\epsilon=3$, then regardless of what your $\delta$ is, we will indeed satisfy that $|f(x)-L|\lt \epsilon$ whenever $|x-a|<\delta$. So every number between $-1$ and $1$ is a limit. And worse, any number is a limit: if you give me $L=10$, then provided I let $\epsilon>11$, every value of $f(x)$ will be within $\epsilon$ of $L$.
That means this definition doesn't really capture the notion we want the definition of limit to capture.
Remember: to convince me that the limit is $L$, you challenge me to fall within an arbitrarily thin horizontal band around $L$. The challenge lies in how thin the horizontal band it; if you let me pick how thin that band is, then I can make it really fat and have absolutely no challenge at all.