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I do have an issue understanding this definition (I know there have been multiple questions around this, but I couldn't find a direct answer to my question, so please bear with me).

The informal definition of a limit states that we need to find a number $L$, the value of the limit, such that when $x$ approaches some $a$, $f(x)$ approaches $L$.

The epsilon-delta definition states that for all $\epsilon>0$ there exists a $\delta>0$ such that, whenever $|x−a|<\delta$ then $|f(x)−L|<ϵ$.

What I do not understand is how the "$x$ approaches $a$ implies $f(x)$ approaches $L$" is implied. Intuitively, for arbitrary small $\epsilon, \delta$ should get arbitrary small too, according to the intuitive definition.

But I do not see any reason why, as $\epsilon$ gets small, $\delta$ cannot get larger.

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  • $\begingroup$ Intuitive/informal way of saying epsylon-delta definition is "$f(x)$ gets arbitary close to $L$ as $x$ gets arbitary close to $a$". $\endgroup$ – Юрій Ярош Jun 29 '18 at 8:53
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"$x$ approaches $a$" corresponds to $|x-a|<\delta$ (or really $0<|x-a|<\delta$), while "$f(x)$ approaches $L$" corresponds to $|f(x)-L|<\epsilon$.

The idea (or one idea, at least) behind $\epsilon$-$\delta$ is the following: We want to know that we can force $f(x)$ to get as close as we want to $L$ just by restricting $x$ to be close to $a$. If you read the $\epsilon$-$\delta$ definition of $\lim_{x\to a}f(x) = L$ carefully, that is exactly what it says: $\lim_{x\to a}f(x) = L$ if, by only using some restriction of the form $0<|x-a|<\delta$ (with $\delta$ positive), you can force $|f(x)-L|$ to be smaller than any given positive number (often denoted $\epsilon$).

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Of course $\delta$ can be large sometimes; for example, if $f(x)=L$ for all $x$, then any $\delta > 0$ will do, regardless of the value of $\epsilon>0$. But the only requirement is that there is some $\delta > 0$ such that bla bla, and this only depends on what $f(x)$ is doing for values of $x$ “close to $a$”.

(And most of the time you will in fact need to make $\delta$ smaller as $\epsilon$ is made smaller.)

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$\delta$ is dependent of a choice of $\epsilon \gt 0$, and $\delta$ is often small, in case we choose a small $\epsilon \gt 0$. The reason, I think, is as follows.

The statement (1) "$|x-a| \lt \delta \implies |f(x)-L|\lt \epsilon$" can be understood in terms of sets. Let $A_{\delta}:=\{x\in \mathbb{R} : |x-a|<\delta \}$ and $B_{\epsilon}:= \{y\in \mathbb{R} : |y-L|<\epsilon \}$.

Then, the statement (1) is equivalent to "$x\in A_{\delta} \implies f(x)\in B_{\epsilon}$"

Now it is evident that if $\epsilon$ gets smaller, then the set $B_{\epsilon}$ also get smaller. Thus, in order that $B_{\epsilon}$ contains the image of $A_{\delta}$ by the function $f$ (i.e. $f(A_{\delta})\subset B_{\epsilon} $) , we should sufficiently decrease the value of $\delta$, thereby increasing the chance such that $f(A_{\delta})\subset B_{\epsilon} $.

Therefore, we usually try to find small $\delta \gt 0$ for the above reason.

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