📂Analysis

Epsilon-Delta Argument

Definition¹

Let $I$ be an interval containing $a \in \mathbb{R}$, and suppose that $f$ is a function defined at $I \setminus \left\{ a \right\}$. If for every $\epsilon > 0$, there exists a $\delta>0$ such that

$$ 0 < | x - a | < \delta \implies | f(x) - L | < \varepsilon $$

is satisfied, then we say that $f(x)$ converges to $L \in \mathbb{R}$ as $x \to a$ approaches $a \in \mathbb{R}$.

Explanation

The term “epsilon-delta” comes from, as you can see, epsilon $\varepsilon$ and delta $\delta$ appearing in the definition. It’s an expression first used by Cauchy, the “father of analysis”, where epsilon and delta represent error $\varepsilon$rror and distance $\delta$istance, respectively.

As you can see, the expression is very complex and far from intuitive, which is why it is initially difficult to grasp. Like the limit of a sequence, there is both a reason for redefining it this way and a reason for its complicated definition, but understanding these reasons and truly understanding epsilon-delta are two different matters. In fact, understanding epsilon-delta is not enough; it only becomes useful after getting used to it.

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To get a grip on it, imagine shooting game. In this game, you have a gun $f$ and are shooting at a specified target $L$ from a set position $a$, where whether you hit the target or not is judged within an allowed error $\varepsilon$. Of course, if you couldn’t move at all from $a$, you wouldn’t be able to hit the target. Let’s assume the shooter can determine how much they need to move to hit the target when given an error $\varepsilon$ and thus can propose an allowable distance $\delta$.

The given gun is $f(x) := 2x$, and with it, aiming at $x$ and shooting results in hitting $2x$. To judge if this gun is proper, one might test if it can hit $L=0$ from $a=0$. But can a gun with scattered hit points really hit the target? Let’s examine a few cases in practice.

• **Case 1. $\varepsilon = 12$

  

  The first allowable error is given generously as $\varepsilon = 12$. Since only $| f(x) - L | < \varepsilon$ needs to be satisfied, shooting from $x$ to meet $| f(x) | < 12$ will be considered a hit. So, $x$ must not exceed the absolute value of $6$, meaning as long as it is $| x | < 6$, it will meet $| f(x) | < 12$. Rewriting it as an equation:

  $$ | x | < 6 \implies | f(x) | < 12 $$

  This shows that, with an allowable error of $\varepsilon = 12$, we can identify an allowable distance $\delta = 6$ in which the gun $f$ can hit the target $L = 0$ from $a = 0$. Of course, a smaller distance would also work, but there is no need to make it harder than necessary.

• **Case 2. $\varepsilon = 6$

  

  The second allowable error is given as $\varepsilon = 6$. Just like before, satisfying $| f(x) | < 6$ is all it takes, hence the necessary allowable distance $\delta = 3$ can be presented.

• **Case 3. $\varepsilon > 0$

  

  As we have seen, no matter how the allowable error $\varepsilon > 0$ is set, we can present an allowable distance $\delta = \varepsilon / 2$ to hit the target. Being able to specify $\delta$ for every $\varepsilon> 0$ essentially means the following:

  $$ \forall \varepsilon > 0 , \exists \delta : | x - 0 | < \delta \implies | f(x) - 0 | < \varepsilon $$

  Rewritten in familiar terms, it becomes $\lim_{x \to 0} 2x = 0$. Until now, we demonstrated that when $x \to 0$, it implies $2x \to 0$. The shooting analogy is no longer needed, but to reiterate, it means with the gun $f$, you can hit the target $L = 0$ from $a = 0$.

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For example, when explaining that $\delta (12) =6$ exists, that $\delta (6) = 3$ exists, and so on, you might have felt some understanding. As you read through the explanation, you suddenly proved $\lim_{x \to 0} 2x = 0$, but such an analogy might be forgettable due to its lack of cohesion. Now, let’s consider why epsilon-delta is difficult.

• Intuition: The sensation of using epsilon-delta and the feeling of ‘approaching infinitely close’ like $x \to a$ and $f(x) \to L$ feels different

  In fact, this is the real reason for using epsilon-delta, but for now, it might not be ‘convincing’ why the existence of $\delta$ equates to something like $\lim_{x \to a} f(x) = L$. If this is the only obstacle, it doesn’t mean you failed to understand epsilon-delta; it’s just unfamiliar. Whether it’s $| x - a | < \delta$ or $| f(x) - L | < \varepsilon$, $\delta$ isn’t thought of as a ’large number’. It’s perceived as a sufficiently small positive number that ‘suppresses’ $| x - a |$ and $| f(x) - L |$, ultimately leading to the following thought process:

  $$ | x - a | < \delta \implies \lim_{\delta \to 0} | x - a | = 0 \implies x \to a $$

  $$ | f(x) - L | < \varepsilon \implies \lim_{\varepsilon \to 0} | f(x) - L | = 0 \implies f(x) \to L $$

• Terminology: The phrase ‘$\delta$ exists’ doesn’t quite resonate

  In reality, this isn’t about literally creating $\delta$ but rather about presenting it in relation to $\varepsilon$. If you’ve managed to express $\delta$ as a function $\delta = \delta ( \varepsilon )$ of $\varepsilon$, then since the existence of $\varepsilon > 0$ is already assumed, $\delta$ exists as well.

• Order: The condition is $|x - a| < \delta \implies | f(x) - L | < \varepsilon$, but the order of thought is opposite

  This is a really confusing part because the form of $\implies$ might make it seem like the order should go from front to back. However, as made clear by “for all $\varepsilon > 0$”, $| f(x) - L | < \varepsilon$ must be considered first before $| x - a | < \delta$. If you don’t know what $\varepsilon$ is, then it’s not worth pondering.

Considering these three reasons while re-reading the explanation will be helpful. If you have understood, now a few odd points might appear, such as caring only about $| x - a | < \delta$ instead of $0 < | x - a | < \delta$, suddenly no more mentioning whether $f$ is a proper gun, or not being able to hit the target without moving from $a$, etc. These are merely analogies twisted in various ways to make epsilon-delta as intuitive as possible. Important is not to focus on irrelevant parts but to use concentration on logically necessary proofs.