# What's wrong with this “backwards” definition of limit?

Can anyone please give an example of why the following definition of $\displaystyle{\lim_{x \to a} f(x) =L}$ is NOT correct?:

$\forall$ $\delta >0$ $\exists$ $\epsilon>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$

I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work.

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Every function defined in a neighborhood of $a$ satisfies that property. Can you see why? –  anon Sep 8 '12 at 20:11
Yes, you are correct. If f(x) is defined near a, then you can choose an epsilon as big as needed so that abs(f(x)-L)<e will be true for every L. Thanks a lot. –  Sebastian Sep 8 '12 at 20:17
You've got $\varepsilon$ and $\delta$ reversed! Other than that, your definition is right. –  Michael Hardy Sep 8 '12 at 20:18
Michael Hardy: thanks, but I think anon got it right. There's more that having epsilon and delta reversed here. The quantifiers are reversed too, and I think that is a game changer. –  Sebastian Sep 8 '12 at 20:21
@anon: even if $f$ is unbounded? To me, this statement basically says that $f$ is locally bounded (regardless of what $L$ might exactly be, so long as it's finite). –  tomasz Sep 9 '12 at 0:01
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There are two problems with your "backwards" definition, which I'll illustrate with examples:

1. Let $f(x) = \sin x$ and let $a$, $L$ and $\delta$ be arbitrary real numbers. Then $\epsilon = |L| + 2$ satisfies your definition.

2. Let $f(x) = 1/x$ (for $x \ne 0$, and let $f(0) = 0$, just to make $f(x)$ defined everywhere), and let $a = 1$. Then, for any $L$ (including $L = 1$), your definition fails for all $\delta \ge 1$, since for any $\epsilon$ we can choose $x=1/(L+\epsilon)$ if $L+\epsilon > 1$ and $x=1$ otherwise, so that in either case $f(x)-L \ge \epsilon$.

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Thanks! I wasn't aware of the second problem with the backwards definition but it does makes sense. –  Sebastian Sep 8 '12 at 21:09

Your condition can be translated into words as:

$f$ is bounded in every open ball around $a$

This is quite different from being continuous!

PS Not every function satisfies this condition, but the ones that don't are rather fierce. (In fact, in a strict sense, almost no function satisfies the condition, but the functions arising in analysis normally do.)

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I'm a little confused when you ask wether it is correct or not. Do you ask as opposed to the well known definition? That is, as opposed to

We say that $\lim\limits_{x\to a}f(x)=\mathscr L$ if for every $\epsilon >0$ there exists a $\delta >0$ such that, for all $x$, if $0<|x-a|<\delta$ then $|f(x)-\mathscr L|<\epsilon$.

If so, you can just see $\epsilon$ and $\delta$ are reversed.

However, suppose this definition was not given, and we want to define what we mean by limit. First, we clearly want to understand that we're concerned about what happens near $a$, but not at $a$. The idea of a "limit" near $a$ is then that of a value $\mathscr L$ a function $f(x)$ approaches when $x$ approaches $a$, So the idea we want to capture is that a function $f(x)$ has a number $\mathscr L$ as a limit when $x$ approaches $a$ if we can make $\mathscr L$ and $f(x)$ as near as we wish, by taking $x$ sufficiently near $a$. So when do we say that two number are near? We need to formalize our idea of proximity of numbers.

$\bf A$. So, we can associate to each $x,y\in \bf R$ the real number $d(x,y)=|x-y|$ and call it the distance from $x$ to $y$. Note that this distance has some properties we really want any notion of distance to have:

$(1)$ The distance is symmetric. $$d(x,y)=d(y,x)$$

$(2)$ It is always positive, unless $x=y$. That is, the distance of two numbers is zero if and only if they are the same number. $$d(x,y)>0 \text{ and } d(x,y)=0\iff x=y$$

$(3)$ The distance from a point $x$ to a point $y$ will always be less than or equal to the distance from $x$ to another $z$ plus the distance from that $z$ to $y$. We're just saying the shortest distance from $x$ to $y$ is precisely the straight line that joints them (this generalizes to higher dimensions). $$d(x,y)\leq d(x,z)+d(z,y)$$

$\bf B$. Now, consider this silly theorem:

Suppose $x,y\in \Bbb R$. Then, for every $\epsilon>0$, $d(x,y)<\epsilon$ if and only if $x=y$.

One direction is easy: if $x=y$, then clearly $d(x,y)=0<\epsilon$ for positive $\epsilon$. Now suppose that for any $\epsilon>0$, we have that $d(x,y)=0<\epsilon$. We aim to prove $x=y$. We will argue by contradiction. Since the distance is symmetric, we might assume $x<y$. Then $d(x,y)=|x-y|>0$. So $|x-y|$ is an $\epsilon>0$, which would mean that $|x-y|<|x-y|$, which is absurd. Thus, it must be that $x=y$.

$\bf C$. Now, we have a formal definition of the notion of near, and we can say that two numbers are near if $d(x,y)$ is small. In particular we just showed that if $x=y$, $d(x,y)$ is a smaller than any positive quantity given, since it is zero. It is but only logical to say that making $\mathscr L$ as near as we wish to $f(x)$ is making $|f(x)-\mathscr L|<\epsilon$ , no matter how small $\epsilon$ is, with $\epsilon<0$.

$\bf D$. We can now try and think about a degree of closeness. Given a number $\delta >0$, we can standarize and say that $x$ and $a$ are sufficiently near if $d(x,a)<\delta$. Think of it as the ruler we use to tell wether you can go in a rollercoaster ride or not. If $d(x,a)\geq \delta$, then $x$ is "bad" and we discard it.

$\bf E$. It is important to note this: we are concerned about making $f(x)$ and $\mathscr L$ close, and we want to succeed in doing so by making $x$ close to $a$. It is not our objective to make $x$ and $a$ close, but our means. So our definition must capture this: given a desired proximity of $f$ and $\mathscr L$, there must be a moment in which any $x$ close enough to $a$ will make $f(x)$ be close to $\mathscr L$. We don't forget that $x\neq a$, which only means $0<|x-a|$. So, let's try and write something down, considering what we discussed in $\bf B,C,D$.

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any prescribed degree of closeness, making $x$ sufficiently close to $a$, but not equal to $a$, will imply that for all those $x$, $f(x)$ and $\mathscr L$ will be within that prescribed degree of closeness.

Applying $\bf B,C$, we can write

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, making $x$ sufficiently close to $a$, but not equal to $a$, will imply that for all those $x$, $|f(x)-\mathscr L|<\epsilon$.

Using $\bf D$, we can write

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, there is a $\delta$ such that for all $x\neq a$, $|x-a|<\delta$ implies $|f(x)-\mathscr L|<\epsilon$.

Finally, since $x\neq a$ is the same as $0<|x-a|$, we can be more succint and write.

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, there is a $\delta$ such that for all $x$, $0<|x-a|<\delta$ implies $|f(x)-\mathscr L|<\epsilon$.

You can prove, or find proofs, that if the limit of $f$ exists for some $a$, then it is unique. And this is important. As others have pointed out, this definition, giving freedom on $d(f(x),\mathscr L)$,for example, makes limits lose their uniqueness.

Now, your definition clearly doesn't capture our original idea: It is somehow saying $f$ has $\mathscr L$ as a limit if for any prescribed degree of closeness $\delta$, there will exist some positive number $\epsilon$ such that making $d(x,a)<\delta$ will imply $d(f(x),\mathscr L)<\epsilon$. But this is giving us a lot of freedom on $d(f(x),\mathscr L)$, which we are most worried about. Of course we can make $x$ and $a$ as near as we wish, but the question is, can we make $f$ and $\mathscr L$ as close as we wish?

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Let's consider a counter example. Let's use your definition to prove that the limit of $x$, as $x$ tends towards 1, is 2. For all $\delta > 0$, we claim that there exists $\varepsilon_{\delta} > 0$ such that:
If $|x - 1| < \delta$ then $|x-2| <\varepsilon_{\delta}$.
We could define $\varepsilon_{\delta} := \delta + 2$. That seems to satisfy your definition.