Is some thing wrong with the epsilon-delta definition of limit??

Question

In the epsilon-delta definition of limit which is:

For all $\epsilon>0$ there exists a $\delta>0$ such that, whenever $|x-a|<\delta$ then $|f(x)-L|<\epsilon$ .

Now since $\epsilon$ is given and it should be proven that there is some $\delta$ for the given $\epsilon$, then I think the rest of the definition should have been:

$|f(x)-L|<\epsilon \implies |x-a|<\delta$ .

Please tell me whether I'm right or not.

Answer 1

Your proposed definition

For all $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x)-L| < \epsilon \implies |x-a| < \delta$.

has quite a few problems with it.

For instance, let $f(x) = 0$ and consider $\displaystyle\lim_{x \to 0}f(x)$. In any reasonable definition of a limit, we should have $\displaystyle\lim_{x \to 0}0 = 0$. So, we need to prove:

For all $\epsilon > 0$ there exists a $\delta > 0$ such that $|0-0| < \epsilon \implies |x-0| < \delta$

Unfortunately, $|0-0| < \epsilon$ is true for any $x \in \mathbb{R}$, but $|x-0| < \delta$ is false for any $x > \delta$. Thus, there is never a $\delta > 0$ which makes "$|0-0| < \epsilon \implies |x-0| < \delta$" a true statement. Therefore, by your proposed definition, $\displaystyle\lim_{x \to 0}0$ is not $0$.

Answer 2

The way I used to think about it is this: it is something like a Space Invaders game (if you remember those). The enemy is firing epsilons at you. For each epsilon that comes at you, you have to be able to counter by finding a delta such that for every $x$ such that $|x−a|<δ$, then $|f(x)−L|< \epsilon$

If some epsilon arrives such that you can't find a delta, you lose.

For example, suppose that $f(x) = 1$ for $x < 0.5$ and $f(x) = 2$ for $x \geq 0.5$. (Yes, a step function.)

Look at $a = 0.5$ and Try $L = 1.5$

Suppose $\epsilon = 1$ arrives. Then you're OK because you can reply with $\delta = 0.1$ (for example). For values of $x$ between $0.4$ and $0.6$, the values for $f(x)$ are either $1$ or $2$, so $|f(x) - L|$ is always $0.5$, and it is true that $ |f(x) - L| < \epsilon = 1$.

But if $\epsilon = 0.2$ arrives, then it doesn't matter how small a $\delta$ you reply with, you can't get $|f(x) - L| < \epsilon = 0.2$ for all the $x$ between $0.5 - \delta$ and $0.5 + \delta$.

So you lose, and the limit isn't $L = 1.5$.

With a little bit more effort you can demonstrate that it doesn't matter what value of $L$ you try for the limit, if $\epsilon = 0.2$ arrives, you lose.

So in this example the limit of $f(x)$ as $x$ tends to $0.5$ doesn't exist. Intuitively, of course it doesn't exist, but we want to see how the textbook definition works out for this example.

The definition in the textbooks is correct. It is hard to understand, but maybe the Space Invaders analogy helps.

Answer 3

Firstly, I suggest you learn logic first, otherwise it is very difficult to understand a lot of things rigorously. It seems you do not understand scoping of quantifiers, which is why the English phrasing, although unambiguous, is not clear to you.

Secondly, there is a mistake in the definition as you have written it.

Let me write the correct definition in terms of its logical structure, which I suggest you do for definitions and theorems whose structure you do not understand perfectly.

Limit Definition

$\lim_{x \to a} f(x) = L$ iff:

  For any $ε > 0$:

    For some $δ > 0$:

      For any $x \in Dom(f)$:

        If $0 < | x - a | < δ$:

          $| f(x) - L | < ε$

Let us go through it properly. The limit of a function $f$ at $a$ being $L$ means that no matter what positive $ε$ you give me, I can give you a positive $δ$ such that the pair $(ε,δ)$ has a certain property. That property is that no matter what $x$ you give me from the domain of $f$, the triple $(ε,δ,x)$ has a certain (other) property, which is that if the distance between $x$ and $a$ is less than $δ$, then the distance between $f(x)$ and $L$ is less than $ε$. Note that this last property says nothing at all if the triple currently in consideration is such that $x$ and $a$ are at least $δ$ apart.

So why is it defined this way? Let us give a more intuitive sense of a limit. We can look at a landscape through a window. If we look through at a landscape through a smaller window, we will surely not see more, and for most natural landscapes the region we can see will become more and more 'bland' as the window shrinks (about a fixed centre). (At first we see a tree with branches and leaves, later we see only a leaf, later we see only a patch of a single colour, green.) If the entire region that we see gets closer and closer in colour to green as the window shrinks, then we can say that the limit colour at the 'centre' of the shrinking window is green. This is not quite the mathematical definition yet. Imagine that there is a minuscule dot at the centre of the window, so we cannot see the colour of the landscape there. The limit is still said to exist. But what does "closer and closer" mean? It means that no matter what positive error margin we desire, as the window shrinks to zero size, eventually the colour of the region we can see will always remain near the limit to within that error margin. Consider the landscape to be a function $f$, where $f(x)$ is the colour of the region that can be seen when we look through point $x$ on the window. Call the centre of the window $a$. Consider the window size to be $δ$ and the error margin to be $ε$. Then our above notion means that $f$ has a limit colour $c$ at its centre $a$ if and only if, for any positive $ε$, we can find a positive window size $δ$ such that once it shrinks smaller than that size, all the colours $f(x)$ we can see through it ($| x - a | < δ$) are within $ε$ of the colour $c$ ($| f(x) - c | < ε$).

Now if the window has its centre exactly on the boundary of an object which is of a different colour than its background, the limit colour will not exist because no matter how small we make the window, we will see a little bit of the object and a little of the background. This corresponds to a jump discontinuity at $a$.

Also, we couldn't see the colour of the landscape behind the centre of the window. It could be the same colour as the limit colour, in which case $f$ is continuous at $a$. But it could be a different colour, in which case $f$ has a removable discontinuity at $a$.

Finally, the landscape could have infinite detail that keeps varying widely between colours (within the window) no matter how small we shrink the window. A mathematical example would be if $f(x) = \sin(\frac{1}{x})$ and $a = 0$. In this case $f$ also has no limit at $a$, and the discontinuity is neither a jump discontinuity nor a removable discontinuity.

Lastly, note that this analogy with windows and landscapes immediately applies to the concept of continuity in general metric spaces. After all, we would have to define what distance between two colours means, but the idea is very natural.

Answer 4

The alternative epsilon-delta sentence does describe a notable property of functions, but that property is not continuity or the existence of a limit. The logical formula

$Q(L,a)$ = "For all $\epsilon>0$ there exists a $\delta>0$ such that ($|f(x)-L|<\epsilon \implies |x-a|<\delta$)"

says that "any limit of $f$ at any other point is different from $L$".

Imitating the definition of continuity, by requiring that the formula $Q(L,a)$ holds with $L=f(a)$ at all $a$, characterizes functions $f$ for which the closure of the graph of $f$ is injective. The closure of $f$ could be a relation and not a function, but injectivity continues to make sense (for every $y$, if $aRy$ and $bRy$ then $a=b$).

This characterization is true for functions from one metric space to another.