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 $\delta$?
 A: 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 $\epsilon$
A: 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 $\varepsilon$.
