# Question regarding $\varepsilon$-$\delta$

Consider the following, I try to prove a limit, and get $$\frac \varepsilon 3$$=$$\delta$$. So, the length $$\delta$$ $$<$$ $$\varepsilon$$. My question is how will every $$\varepsilon$$ correspond with a single and unique $$\delta$$?

• The $\delta$ is not unique
– user801306
Commented Feb 14, 2021 at 16:28
• Suppose we have $\delta > 0$ so that $x\in (x_0-\delta, x_0 + \delta )$ implies $f(x)\in (L-\epsilon, L+ \epsilon)$. Now take $\delta^*=\delta/2$ . Then $$x\in(x_0-\delta^*,x_0+\delta^*) \implies x\in (x_0-\delta, x_0+\delta) \implies f(x)\in(L-\epsilon,L+\epsilon)$$ This is a good question. Asking these kinds of questions allows you to understand the formal definition of a limit.
– user801306
Commented Feb 14, 2021 at 16:38
• @MatthewPilling Quite impressive. Excellent. Would you prefer to copy-paste the argument as an answer? Commented Feb 14, 2021 at 16:48
• @MatthewPilling The argument is one of the neatest I have seen in quite a bit. Thanks. Commented Feb 14, 2021 at 16:50
• @Karl HaHa. I knew that before I posted the quesion, I wanted a neater argument, and see, I got one of them... Commented Feb 14, 2021 at 16:58

If you have $$\lim_{x \to a} f(x) = l$$ then for every $$\epsilon$$ there is a $$\delta$$ such that for every $$x \in D$$ (where D is the domain of $$f$$) with $$|x-a|\lt \delta$$, you have that $$|f(x) - l|<\epsilon$$ by definition. Take one of these $$\epsilon$$'s and one $$\delta$$ that does the trick for that $$\epsilon$$. Now, take any $$0<\delta_{o}<\delta$$. Then, for any $$x \in D$$ with $$|x-a|<\delta_{0}<\delta$$, we have by definition of $$\delta$$ that $$|f(x)-l|<\epsilon$$. With this we conclude that there is actually never uniqueness of a $$\delta$$ correspondent to some $$\epsilon$$, since in the real numbers, there is always a $$0<\delta_{0}<\delta$$. And for each of these $$\delta_{0}$$, the condition is also satisfied. So the possible $$\delta$$'s are always infinite for each $$\epsilon$$
• Great answer. I note that some books, particularly "intro to proof" books (as distinguished from analysis books), do not communicate this point very well, and do sometimes imply that finding the "best" (i.e. largest) $\delta$ that will work, as a function of $\epsilon$, is somehow part of proving that a limit exists. It is true that in some applications you want to know about the size of a $\delta$ that will work, but definitely not when only proving that a limit exists. Commented Feb 14, 2021 at 16:47
• @Daàvid On a side-note, does there exists many $\varepsilon$ for each $\delta$? Commented Feb 14, 2021 at 17:15
• Well, there is in the sense that a $\delta$ that works for an $\epsilon$ also works for any $\epsilon < \epsilon_{0}$ Commented Feb 14, 2021 at 17:26
• In general, no. To see this, you could take, for instance, the constant function $f(x) = a$ with $a \in \mathbb{R}$. Since the function is continuous, for each $\delta >0$, there is an $\epsilon >0$ such that the condition is satisfied around some $x_{0}$. Fix such an $\epsilon_{0}$. Then $a \in (f(x_{0}) - \delta, f(x_{0}) + \delta)$ but all $x$ outside our interval are mapped to $a$. Commented Feb 22, 2021 at 16:40
Daàvid has already answered the question at hand, but I'll do my best to explain the motivation behind why we define limits in this way. Recall that we say $$\lim_{x \to a}f(x) = L$$ if and only if
For every $$\varepsilon > 0$$ there exists $$\delta > 0$$ such that, for all $$x$$, if $$0<|x-a|<\delta$$, then $$|f(x)-L|<\varepsilon$$.
This definition looks pretty complicated, but the idea that it is trying to capture is actually not too tricky. When we say, for instance, that the limit as $$x$$ approaches $$0$$ of $$\sin x/x$$ equals $$1$$, what we mean is that as $$x$$ gets closer and closer to $$0$$, the value of $$\sin x/x$$ gets arbitrarily close to $$1$$. In other words, the difference between $$\sin x / x$$ and $$1$$ eventually becomes smaller than any tiny number (which we often denote as $$\varepsilon$$). More precisely, we can make $$|\sin x/x - 1|$$ smaller than any positive $$\varepsilon$$ by requiring that $$0<|x|<\delta$$. If we want $$\sin x / x$$ to get really close to $$1$$, then we might have to require that $$x$$ is really close to $$0$$ (but not equal to $$0$$). This shows that depending on how small the $$\varepsilon$$ is, the value of $$\delta$$ might have to be small, very small, or extremely small. The point is that no matter how close we want $$\sin x / x$$ to get to $$1$$, it is possible to get there by requiring that $$\delta$$ is sufficiently small. How small that $$\delta$$ needs to be is dependent on the particular $$\varepsilon$$.