Is the choice of epsilon delta free in limit definition?

Question

We use limit definition as: $$ \lim_{x \to a} f(x)=L $$ if $$ \forall \varepsilon > 0 \quad \exists \delta > 0 \quad (|x−a| < \delta \implies |f(x)−L| < \varepsilon) $$ But why we cannot use: $$ \lim_{x \to a} f(x) = L $$ if $$ \forall \delta > 0 \quad \exists \varepsilon > 0 \quad (|x−a| < \delta \implies |f(x)−L| < \varepsilon) $$

Answer 1

Take the function$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x\in\mathbb Q\\0&\text{ otherwise.}\end{cases}\end{array}$$This function has limit nowhere, but according to your alternative definition, it has a limit everywhere.

Answer 2

The second is not equivalent to the definition of limit.

As an example let consider $f(x)=x$ for $x\to 0$ assume $L=10$ and apply the second statement.

Then $\forall \delta > 0$ we can find $\varepsilon > 0$ such that $|x-10|<\varepsilon$ but we know that the limit is $0$.

Answer 3

If you choose $$\forall \delta > 0 \quad \exists \varepsilon > 0 \quad (|x−a| < \delta \implies |f(x)−L| < \varepsilon)$$ for $$\lim_{x \to a} f(x) = L$$

then you can choose your $\epsilon $ large enough to satisfy $$|f(x)−L| < \varepsilon$$ without any requirement that $f(x)$ is actually close to $L.$

Answer 4

Your proposed alternative definition doesn't say anything about $f$ except that it is bounded in every neighbourhood of $a$. It also doesn't uniquely determine the limit (i.e. the value of $L$)

In fact, if we use your definition, then for any function that has a limit (according to your definition) at a point, the limit at that point could be any real number.

To see this, suppose that $$ \lim_{x \to a} f(x) = L $$ according to your definition, and let $M$ be any other real number. We'll show that the limit is also equal to $M$.

To see this, consider any $\delta > 0$. Then your definition tells us that there exists $\varepsilon > 0$ such that $$ |x - a| < \delta \implies |f(x) - L| < \varepsilon. $$

But if $|f(x) - L| < \varepsilon$, then by the triangle inequality, we have that $$ |f(x) - M| \leq |f(x) - L| + |L - M| < \varepsilon + |L - M|. $$

Thus we have that if $\varepsilon^\prime = \varepsilon + |L - M|$, then $$ |x - a| < \delta \implies |f(x) - M| < \epsilon^\prime. $$

It follows from the proposed alternative definition of a limit that $$ \lim_{x \to a} f(x) = M. $$

In other words, your definition does tell us that $$ \lim_{x \to 2} x = 2, $$ but it also tells us that $$ \lim_{x \to 2} x = 63478609. $$

The alternative definition doesn't capture the notion that $f(x)$ must be close to $L$, because we're free to choose any value of $\epsilon$ that we like, and the value of $\epsilon$ can be very large. The usual definition does imply that $f(x)$ should be close to $L$ when $x$ is sufficiently close to $a$, because it requires the statement to be true for every $\varepsilon$, no matter how small. So it tells us that we can make $f(x)$ as close to $L$ as we like by taking $x$ close enough to $a$.

Answer 5

Roughly speaking, the (first, correct) definition says that the closeness of $x$ implies closeness of $f(x)$.

The direction of this implication means that the distance standard $\epsilon$ we use to measure the closeness of $f(x)$ should be dependent on (or is a function of) the distance standard $\delta$ for measuring the closeness of $x$.

This is precisely what the first definition prescribes. For any $\delta$, there exists an $\epsilon$ for that $\delta$, (i.e. $\epsilon = g(\delta)$, for some function $g$) such that if $x$ is within $\delta$ of $a$ then $f(x)$ is within $\epsilon$ of $L$.

The second definition is incorrect, because it defines the dependence between $\epsilon$ and $\delta$ in a wrong order w.r.t. the direction of the implication mentioned above.