TITLE: $\epsilon$ - $\delta$ definition of a limit - smaller $\epsilon$ implies smaller $\delta$?
QUESTION [14 upvotes]: The definition in my book is as follows:

Let $f$ be a function defined on an open interval containing $c$ (except possibly at $c$) and let $L$ be a real number. The statement $$\lim_{x \to c} f(x) = L$$
means that for each $\epsilon>0$ there exists a $\delta>0$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.

With the definition the way it is, I don't see how choosing a smaller and smaller $\epsilon$ implies a smaller and smaller $\delta$. 
To me, in order to produce that implication, we would need to restrict $\epsilon$ to be small enough to force $f(x)$ to be strictly increasing/decreasing on $(L-\epsilon, L+\epsilon)$, and define increasing/decreasing without the use of derivatives. However, that is not part of the definition.
P.S. Please refrain from using too much notation for logic, I am not familiar with most of the symbols such as the upside down A and such.

REPLY [2 votes]: "With the definition the way it is, I don't see how choosing a smaller and smaller ϵ implies a smaller and smaller δ."
Not necessarily so. It does not have to be monotone.
Delta-epsilon just says: There exists a step in the ongoing appliance of the function which leads to a result nearer to the limit, the distance is smaller than any certain number ϵ. No matter how small this number might be.