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Calculus Done Right

# Working with Epsilons and Deltas

In this optional quiz, we’ll look at the rigorous definition of a limit, the so-called “epsilon-delta” definition. It’s a beautiful concept, although notoriously tricky for students when they see it for the first time. This material won’t be needed anywhere else in the course.

Before beginning, read carefully through the first section of Brilliant’s page on this definition. Remember when reading it that $$\left| a - b \right|$$ has the geometric interpretation as the distance between $$a$$ and $$b$$ on the numberline.

The later sections of that document will be less useful to us. We’re not going to talk much about finding deltas algebraically from epsilons. Our focus here is on getting an intuitive understanding of what this definition, and others like it, are really saying.

Let’s frame the definition in terms of a game. You’re trying to prove that $$\lim\limits_{x \to a} f(x)$$ exists, and you have a particular candidate $$L$$ for the value of that limit. Your opponent challenges you with $$\varepsilon$$’s. Every time she challenges you, you have to respond with a $$\delta$$ successfully. If you can respond successfully to every one of her challenges, you win, and you’ve proven that $$\lim\limits_{x \to a} f(x) = L.$$ (You correspond to Bob in Brilliant's writeup on the definition, and your opponent corresponds to Alice.)

Now think of the opponent’s $$\varepsilon$$ challenge as a vertical target around $$L.$$ And your $$\delta$$ response is a horizontal shooting range around $$a.$$ What does it mean for your $$\delta$$ to be successful? It means that whenever you stand in your shooting range (except for standing at the point $$a$$ itself), and you shoot, you make it into the $$\varepsilon$$ target, as in the picture.

The smaller an $$\varepsilon$$ your opponent chooses, the smaller the target, and the harder your job is: you may have to pick a correspondingly smaller $$\delta.$$ But if you can convince someone that no matter what $$\varepsilon$$ you’re challenged with you can always find some $$\delta$$ response that works, you’ve won the game, and proven that $$\lim\limits_{x \to a} f(x) = L.$$

Let’s look at a simple example: $$f(x) = 2x.$$ We would like to prove that $$\lim\limits_{x \to 3} f(x) = 6.$$

Suppose the opponent challenges us with $$\varepsilon = 1.$$

Which of these is a successful $$\delta$$ response? Which of these defines a horizontal “shooting range” around 3 such that, when we stand anywhere in that range, we will hit inside the $$\varepsilon=1$$ target?

In the previous problem, we saw that when we were challenged with $$\varepsilon = 1,$$ we could make a successful response with $$\delta = 0.5.$$

True or false: We would have also made a successful response with $$\delta = 0.1.$$

In the last two questions we looked at handling a single challenge of $$\varepsilon = 1.$$ To prove that $$\lim\limits_{x \to 3} f(x) = 6,$$ we have to show we can meet the challenge for every $$\varepsilon > 0.$$ So we have to handle the general case.

For an arbitrary $$\varepsilon > 0,$$ which of these is a successful $$\delta$$ response?

To wrap things up, here is what the full argument looks like.

Proposition: For $$f(x) = 2x,$$ $$\lim\limits_{x \to 3} f(x) = 6.$$

Proof: Let $$\varepsilon > 0.$$ Choose $$\delta = \frac{\varepsilon}{2}.$$ Note that $$\delta > 0$$, because $$\varepsilon > 0.$$ Now, for every $$x$$ satisfying $$0 < \left| x - 3 \right| < \delta,$$ we must show that $$\left| f(x) - 6 \right| < \varepsilon.$$

If $$0 < \left| x - 3 \right| < \delta,$$ then by definition of $$\delta,$$ $$\left| x - 3 \right| < \frac{\varepsilon}{2}.$$ Multiplying both sides by 2 gives $$\left| f(x) - 6 \right| < \varepsilon,$$ which is what we wanted. QED.

In general, finding a $$\delta$$ that works for a given $$\varepsilon$$ isn’t as obvious as it was in this example, and often involves algebraic trickery. But the idea behind what you’re trying to do is always the same in proving a limit exists: find a $$\delta$$ shooting range that works for a given $$\varepsilon$$ target.

To prove that a candidate $$L$$ is not the limit of a function, we switch roles. Now, we’re playing the role of the challenger, and we have to find one value of $$\varepsilon$$ for which our opponent will be unable to find a $$\delta$$ that works!

For example, consider the function $f(x) = \begin{cases}\begin{array}{rl} -1, & x < 0 \\ 1, & x \geq 0 \\ \end{array}\end{cases}$

Let’s show that the $$\lim\limits_{x \to 0} f(x)$$ is not 1. What challenge $$\varepsilon$$ should we pick? What vertical target should we place around 1 so that our opponent will be unable to come up with any shooting range around 0 that works?

Once you understand the language of $$\varepsilon$$’s and $$\delta$$’s in terms of challenge and response, you have the ability to rigorously formulate many new ideas. For example, in a previous quiz we looked at the statement $$\lim\limits_{x \to a} f(x) = \infty.$$ Intuitively this means that as $$x$$ approaches $$a,$$ the values $$f(x)$$ get arbitrarily large.

$$\lim\limits_{x \to a} f(x) = \infty$$

We can give a precise definition by formulating the idea in challenge-response terms. Which of the following should we take as our definition? (This might be tricky! Think what each proposed definition is trying to say.)

A) For every $$\varepsilon > 0$$ there exists $$\delta > 0$$ such that for all $$x$$, if $$0 < \left| x - a \right| < \delta$$ then $$\left| f(x) - \infty \right| < \varepsilon$$

B) For every $$M > 0$$ there exists $$\delta > 0$$ such that for all $$x$$, if $$0 < \left| x - a \right| < \delta$$ then $$f(x) > M$$

C) For every $$\varepsilon > 0$$ there exists $$M > 0$$ such that for all $$x$$, if $$x > M$$ then $$\left| f(x) \right| < \varepsilon$$

Let’s try another concept. Let’s capture the notion of limits at infinity.

$$\lim\limits_{x \to \infty} f(x) = L$$

Which of these statements corresponds to $$\lim\limits_{x \to \infty} f(x) = L$$? (Again, each proposed definition is trying to say something in terms of challenge and response… which statement corresponds to the picture?)

A) For every $$\varepsilon > 0$$ there exists $$N > 0$$ such that for all $$x$$, if $$x > N$$ then $$\left| f(x) - L \right| < \varepsilon$$

B) For every $$M > 0$$ there exists $$\delta > 0$$ such that for all $$x$$, if $$0 < \left| x - L \right| < \delta$$ then $$f(x) > M$$

C) For every $$\varepsilon > 0$$ there exists $$\delta > 0$$ such that for all $$x$$, if $$\left| x \right| < \delta$$ then $$\left| f(x) - L \right| < \varepsilon$$

Here’s a new $$\varepsilon - \delta$$ definition for you to parse. Let’s say that a pair of functions $$f$$ and $$g$$ are friendly if they satisfy the following definition:

For every $$\varepsilon > 0$$ there exists $$N > 0$$ such that, for all $$x,$$ if $$x > N$$ then $$\left| f(x) - g(x) \right| < \varepsilon.$$

Take some time to understand what this definition is really saying. (Remember that $$\left| f(x) - g(x) \right|$$ means the distance between $$f(x)$$ and $$g(x)$$.) Can you think of some functions that are friendly with, say, $$f(x) = 1$$?

Which of the following is a friendly pair of functions? You don’t have to give a rigorous proof; you should understand what the definition is saying visually, and then apply that to the graphs of these functions.

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