$\forall\varepsilon > 0\exists\delta > 0:(|f(x) - \ell| < \varepsilon)\implies(0<|x-a| < \delta )$. $f(x)=x$ satisfies this for all $a$ and $\ell$?

Question

This is to be found in the answer to problem 26 (b) of chapter 5 of Michael Spivak's Calculus 3rd edition. The problem reads as follows:

26 Give examples to show that the following definitions of $\lim_{x\to a}f(x)=\ell$ are not correct.

(b) For all $\varepsilon > 0$ there is a $\delta > 0$ such that if $|f(x) - \ell| < \varepsilon$, then $0<|x-a| < \delta$.

I thought of constant functions because they do not satisfy this condition, but approach a limit "everywhere". This means, let $f(x)=c, c\in \mathbb R$ and let $a\in \mathbb R$ and let $\varepsilon > 0$, then $|f(x) - c|=0 < \varepsilon$ for all $x$ satisfying $0<|x-a| < \delta$ for any $\delta > 0$ and not just some $\delta > 0$. (It is already shown in the text of this chapter that $\lim_{x\to a}c=c$.)

In his answer Spivak also mentions constant functions but then adds the following:

Moreover, the function $f(x)=x$, for example, satisfies this condition no matter what $a$ and $\ell$ are.

What does he mean by that? It seems to me that if $|f(x) - \ell| < \varepsilon$ holds, then $|x - \ell| < \varepsilon$ so $a=\ell$ does hold, which means it does matter what $a$ and $\ell$ are. Unfortunately, I really don't understand what he is getting at.

Answer 1

If $|f(x)- l| < \varepsilon$ holds, then $|x-l|<\varepsilon$

This is true.

so $a = l$ does hold

This is also true, but not necessary. You have assumed that $\delta$ is small (equal to $\varepsilon$), but this need not be the case. We are free to chose $\delta$ as we please, so $|a-l|$ can actually be as big as we like (using this incorrect definition).

In the true definition, we are asking "Can we find $x$ close enough to $l$ such that $|f(x) - l| < \varepsilon$". In the bad definition above, we are asking "Can we find a $\delta$ big enough such that if $|f(x) - l| < \varepsilon$, then $|x - a| < \delta$".

The important thing to take note of is the order which each section appears. Let's break down the 'definition' given in the question step by step to see where we go wrong.

Step 1 - What is given?

We are trying to show that $\lim_{x\to a} f(x) = l$, so $a$ and $l$ are fixed.

Step 2 - Choose an epsilon

We have been told that what follows must hold for all $\varepsilon > 0$, so we let $\varepsilon > 0$ be arbitrary but, from this point onwards, fixed.

Step 3 - Find a delta

We want to find a $\delta$ that 'works'. By this we mean, can we find a $\delta$ such that, if |$f(x) - l| < \varepsilon$ then $|x-a| < \delta$?

Suppose that $|f(x) - l| < \varepsilon$. Then, by definition of $f$, $|x-l| < \varepsilon$, so we have $x$ being 'close' to $l$. Can we find a delta such that $|x-a| < \delta$? Sure we can! We just pick delta to be as big as we need. So, using this definition, as long as we pick $\delta$ large enough, it works for all choices of $a$ and $l$.

Example

Let's consider the following example:

Using this (incorrect) definition, show that $\lim_{x\to 5}f(x) = 27$, where $f(x) = x$

Let $\varepsilon > 0$ be arbitrary. We seek $\delta > 0$ such that, if $|x-27|<\varepsilon$, then $|x - 5| < \delta$.

If $|x - 27| < \varepsilon$, and $\delta > 22 + \varepsilon$, then \begin{align*} |x - 5| &= |x - 5 + 27 - 27| \\ &\leq |x - 27| + |-5 + 27| \\ &= |x - 27| + |22| \\ &< \varepsilon + 22 \\ &< \delta \end{align*}

Hence $\lim_{x\to 5}f(x) = 27$.

Clearly, this is nonsense, but it is a good exercise in demonstrating why the order of the statements in the definition is important.

Answer 2

Edit JEdwards and fleablood already explained much of what I added. Sorry! Not trying to poach answers. I'll leave mine up in case it's useful but please don't give me points that should go elsewhere!

I've missed the main thrust of what Spivak's getting at. The bad stuff I pointed out is true, however there's a more general issue.

With $f(x)= x$, no matter what we choose for $\ell$, $a$, or $\varepsilon$, we can find some $x$ such that $0<|x-a|<\delta$ and $|f(x) -\ell| < \varepsilon$. We just have to make $\delta$ big enough.

For example, take $a = 0$, $\ell = 10$.

Now, For any $\varepsilon > 0$ there is a $\delta$ ($\delta \geq 11$ works) such that $|f(x) - 10| < \varepsilon$ and $0 <|x-a|< \delta$.

But this suggests that $\lim_{x\to 0}x = 10.$

In other words, this $f$ is such that it approaches every number somewhere so using the criteria of our bad limit definition, if we make $\delta$ big enough we can show that $f(x)$ approaches any value, anywhere.

Original Post

What does he mean by that? It seems to me that if $|()−\ell|<\varepsilon$ holds, then $|−\ell|<\varepsilon$ so $=\ell$ does hold, which means it does matter what $$ and $\ell$ are.

This is true, however, the bad limit condition also says that for all $x$ satisfying $|()−\ell|<\varepsilon$, that $0 < |x-a|$.

For $f(x) = x$, $\lim_{x\to a} x = a$.

For $x = a$, we have $|f(x) - \ell| = |a - a| = 0 < \varepsilon$, but, $0 \not<|x-a|< \delta$

The same thing happens for constant functions. They satisfy $|f(x) - \ell| < \varepsilon$ at $x = \ell$, which is not allowed.

As an aside, this bad limit definition also runs into problems with functions that have the same limit at multiple different points. For example, $f(x) = x^2$ approaches $1$ at both $x = 1$ and $x = -1$.

Here we're forced to make $\delta$ big enough to cover both of them. Consequently, the $\delta$ neighborhood includes many $x$ that do not satisfy the original $\varepsilon$ condition. For cases like this, the $\delta$ neighborhoods stop being useful (even setting aside the earlier $x=a$ problem.)

Edit: Another point: using this definition of limits and considering constant functions, if $f(x) = c$ then then for every $\varepsilon > 0$ and all $x$, $|f(x) - c| < \varepsilon$. The fact that every $x$ satisfies this $\varepsilon$ requirement means that for the limit $c$ to exist at any point $a$, we require some $\delta > 0$ such that $|x-a|< \delta$ for all $x$. No such $\delta$ exists. Constant functions cannot approach limits.

Answer 3

Oh, okay:

Consider this. Using the weird wrong definition.

$\lim_{x\to 3} x = 591$.

Proof: For any $\epsilon > 0$ let $\delta = 588 + \epsilon$

If $|f(x) - 591| < \epsilon$ then $|x-591| < \epsilon$ so $591 -\epsilon < x < 591 + \epsilon$. So $588 -\epsilon < x -3 < 588+ \epsilon$.

So $-\delta = -588 - \epsilon < 588-\epsilon < x-3< 588 + \epsilon=\delta$

So $|x-3| < \delta$.

So $\lim_{x\to 3} x = 591$.