Why isn't the ε-δ definition $\mid f(x) - L\mid < ε \Rightarrow 0 < \mid x - a\mid < δ$? Why can't you reverse δ,ε in the implication?

Question

Here's the definition of a limit that I was taught: $\displaystyle \lim_{x \to a} f(x) = L$ means:
$\forall \varepsilon > 0, \exists \delta > 0$ such that $0 < \mid x - a\mid < \delta \to \mid f(x) - L\mid <\varepsilon $

1. In the following picture, why doesn't the limit exist at $a_1$ based on the definition of a limit?

Simply because $f(x)$ is not defined on the given interval?

2. Can the definition be written as: $\mid f(x) - L\mid < \varepsilon \Rightarrow 0 < \mid x - a\mid < \delta$?

Basically, why is the equality one way and not two ways (iff)? If you want to specify how close you want $f(x)$ to be to $L$, then find out how close $x$ needs to be to $c$ in order to guarantee it, why does the delta part imply the epsilon part? This seems a little un-intuitive.

Answer 1

You have two questions here:

1. Why doesn't the limit exist in your graph?
2. Why is a limit defined using implication and not equivalence.

Question 1

To say $\lim_{x\to a} =L$ we need $|x-a|\leq \delta \implies |f(x) -L|\leq \epsilon$ for all $0<|x-a|<\delta$.

Let's define $x_l, x_r$ as the closest values of $x$ to $a$ where $f(x)$ is defined, when approaching from the left and right, respectively.

If we want to say $L=f(x_l)$ then for any $\epsilon > 0$ we need to find a $\delta >0$ so that $|f(x)-L| < \epsilon$ if $0<|x-a|<\delta$. However, we can see that if $\delta < |a-x_l|$ then $f(x)$ is not defined for any $x \in |x-a|<\delta$. If $\delta>|x_l-a|$ then $f(x)$ is not defined for any $x\in (x_l,a)$ so we fail to satisfy the definition (there is no "get out of logic free" card for undefined points ;-)

Question 2

If we changed the definition to be

$$\lim_{x\to a} =L \text{ when }$$ $$ \forall \epsilon > 0\;\exists \delta > 0: 0<|x-a|\leq \delta \iff |f(x)- L| \leq \epsilon$$

Then any function that is not strictly monotonic would fail to have a limit, since $|f(x)-L|< \epsilon$ could happen at two disjoint intervals of $x$.

Answer 2

The limit in your picture doesn't exist at $a_1$ by the definition

So the definition of limit is technically two parts. The definition you cited is only the second part. The first part requires that there is an open interval $I$ containing $a$ such that $I\setminus \{a\} \subseteq \operatorname{dom} (f)$ which in your picture will not be satisfied since $(a_1-\delta, a_1+\delta) \cap \operatorname{dom} (f)=\emptyset$ where $\delta:=\operatorname{dist}(a_1, \operatorname{dom} (f))$. So the limit at $a_1$ doesn't exist in your figure because the first part of the definition is not satisfied.

Note: Sometimes people bake in the first part of the definition such as how proofwiki did by stipulating the domain of $f$ to be $(a,b) \setminus \{c\}$.

Example of why the converse is not a part of the definition

Consider the constant function $f(x)=1$ on $\mathbb R$.

Let $\varepsilon>0$ be given. Since $f$ is constant, we trivially have $|f(x)-1|<\varepsilon$ BUT notice there is no $\delta \in (0,\infty)$ that guarantees $0<|x-a|<\delta$ for all $x \in \mathbb{R}$ (consider $x=\delta+|a|$). Moreover, when the function $f$ is continuous at $a$ there is no delta that somehow implies $x \neq a$ since $x=a$ will always satisfy $|f(x)-f(a)|<\varepsilon$.