# 2 Epsilon-Delta Definition

## Limits of Functions

We can easily understand the limits of functions if we clearly understand the limits of sequences. The intuitive picture of the limit of a sequence is that

the value of $a_n$ infinitely approaches a constant value $\alpha$ as the index $n$ increases.

As was explained in the previous chapter, this picture is precisely characterized by the following definition:

For any $\varepsilon > 0$, there exists a natural number $N(\varepsilon)$ such that for all $n \geq N(\varepsilon)$, we have $| \, a_n \, - \, \alpha \, | < \varepsilon$.

On the other hand, the intuitive picture of the limit of a function is that

the value of $f(x)$ infinitely approaches a constant value $\alpha$  as the value of $x$ with $x \neq a$ approaches $a$.

If we regard the variable $x$ as corresponding to the index $n$, we find that this picture of the limit of a function is essentially the same as that of a sequence.

The difference between these pictures is in the phrase “as the value of $x$ with $x \neq a$ approaches $a$.”

In the case of sequences, “as the index $n$ increases” is expressed by the phrase “for all $n \geq N(\varepsilon)$” since the value of $N(\varepsilon)$ is generally large.

In contrast, in the case of functions, “as the value of $x$ with $x \neq a$ approaches $a$” is expressed by:
$$0 < | \, x \, - \, a \, | < \delta(\varepsilon),$$
where $\delta$ denotes the Greek letter called “delta.”

Notice that $x$ is close to $a$ if $\delta(\varepsilon)$ is small. In fact, recalling that $| \, x \, - \, a \, |$ means the distance between $x$ and $a$, the inequality $| \, x \, - \, a \, | < \delta(\varepsilon)$ implies that $x$ is in a ball (interval) with radius $\delta(\varepsilon)$ and center $a$. In addition, we have $x \neq a$ because of $0 < | \, x \, - \, a \, |$.

Thus, we arrive at the following definition of the limit of a function:

For any $\varepsilon > 0$, there exists a positive real number $\delta(\varepsilon)$ such that for all $x$ satisfying $0 < | \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, \alpha \, | < \varepsilon$.

When $f(x)$ satisfies this condition, we say that $f(x)$ converges to $\alpha$ as $x \to a$, which is denoted by $\displaystyle\lim_{ x \to a} f(x) = \alpha$. Moreover, $\alpha$ is called the limit value of $f(x)$ as $x \to a$. This is the definition of the limit of a function by the epsilon-delta argument. The key point of this definition is that $\delta$ is a function of $\varepsilon$.

The existence of the function $\delta(\varepsilon)$ guarantees the convergence of a function in the same way as that of a sequence.

I understand that $\delta$ is a function of $\varepsilon$, and that the existence of $\delta(\varepsilon)$ guarantees the convergence of a function. However, I wonder whether the value of $\delta(\varepsilon)$ decreases as $\varepsilon$ decreases. In the case of sequences, the value of $N(\varepsilon)$ increases as $\varepsilon$ decreases.

In general, the value of $\delta(\varepsilon)$ decreases as $\varepsilon$ decreases. I think that the following figure is helpful for understanding this property intuitively. Suppose that the value of $\varepsilon$ in Figure 1 is smaller than that in Figure 2. Then, the value of $\delta(\varepsilon)$ in Figure 1 is smaller than that in Figure 2.  In order to understand the definition of the limit of a function, we will show $\displaystyle\lim_{x \to 2} f(x) = 4$, where $f(x) = x^2$. That is,

for any $\varepsilon > 0$, there exists a positive real number $\delta(\varepsilon)$ such that for all $x$ satisfying $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, 4 \, | < \varepsilon$.

Note the last inequality $| \, f(x) \, - \, 4 \, | < \varepsilon$ in the above condition. Since
$$| \, f(x) \, - \, 4 \, | = | \, x^2 \, - \, 4 \, | = | \, x \, - \, 2 \, | | \, x \, + \, 2 \, |,$$
we seek the range of $x$ where $| \, x \, - \, 2 \, | | \, x \, + \, 2 \, |< \varepsilon$ holds. We consider that the value of $| x + 2 |$ can be approximated by $4$ because we take $x$ in a small neighborhood of $2$ for examining the limit of $f(x)$ as $x \to 2$. Therefore, we assume that $$| x + 2 | < 5$$ holds, where we add $1$ to $4$ with a margin. Then, for all $x$ satisfying $| x \, - 2 | < \displaystyle\frac{\varepsilon}{5}$, we have $$|f(x) \, - \, 4| = |x+2| |x-2| < 5 | x-2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon$$
The above argument holds under the assumption $| x + 2 | < 5$. This inequality is equivalent to $-5 < x + 2 < 5$; i.e., $-7 < x < 3$. Hence, we can obtain $| x + 2 | < 5$ as long as we restrict the range of $x$ to satisfy $1 < x < 3$; i.e., $| x \, - 2 | < 1$. Thus, if $$| x \, - 2 | < 1 \ \ \ {\rm and} \ \ \ | x \, - 2 | < \displaystyle\frac{\varepsilon}{5},$$ then, noting $|x +2 | < 5$ by $|x -2 | <1$, we have $$|f(x) \, - \, 4| = |x+2| |x-2| < 5 | x-2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$ Now, we define $\delta(\varepsilon)$ as follows:
$$\delta(\varepsilon) = \min( \frac{\varepsilon}{5}, \ 1 ),$$
where $\min(x, y)$ denotes the minimum value of $x$ and $y$. We note that $\delta(\varepsilon) \leq \displaystyle\frac{\varepsilon}{5}$ and $\delta(\varepsilon) \leq 1$ hold.

Let $x$ satisfy $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$. Then, noting that $\delta(\varepsilon) \leq 1$, we have $| x \, - 2 | < 1$, which leads to $| x + 2 | < 5$. Therefore, we have
$$|f(x) \, - \, 4| = |x + 2| |x - 2| < 5 | x -2| < 5 \, \delta(\varepsilon) \leq 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$
Thus, for any $\varepsilon > 0$, there exists a real positive number $\delta(\varepsilon) = \min( \displaystyle\frac{\varepsilon}{5}, 1 )$ such that for all $x$ satisfying $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, 4 \, | < \varepsilon$.

Exercise 2  Let $f(x) = \sqrt{x}$. Show that $\displaystyle\lim_{ x \to 1 }f(x) = 1$.

## Continuity of Functions

When the graph of a function is neither cut off nor has a hole, we say that the function is continuous. To be precise, we say that a function $f(x)$ is continuous at $x = a$ if

(A) the value of $f(x)$ is defined at $x = a$,

and

(B) the limit value $\displaystyle\lim_{x \to a} f(x)$ exists, and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds.

For example, the function presented in Figure 1 is continuous at $x = a$ because the limit value $\displaystyle\lim_{x \to a} f(x)$ exists and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds; the graph of the function is neither cut off nor has a hole at $x = a$.

In contrast, the functions presented in Figures 2 and 3 are not continuous at $x = a$. In the case of Figure 2, the limit value $\displaystyle\lim_{x \to a} f(x)$ does not exist; the graph of the function is cut off at $x = a$. In the case of Figure 3, $\displaystyle\lim_{x \to a} f(x)$ exists, but $\displaystyle\lim_{x \to a} f(x) \neq f(a)$ does not hold; the graph of the function has a hole at $x = a$. By using the definition of the limit of a function, the above condition B that the limit value $\displaystyle\lim_{x \to a} f(x)$ exists and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds is given by:

For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $0 < | \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

However, noting that the value of $f(x)$ is defined at $x = a$, the last inequality in this statement $| \, f(x) \, - \, f(a) \, | < \varepsilon$ holds at $x = a$. In fact, we have $| \, f(a) \, - \, f(a) \, | = 0 < \varepsilon$. Therefore, the above statement can be slightly modified as follows:

For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

This is exactly the condition that $f(x)$ is continuous at $x = a$ when the value of $f(x)$ is defined at $x = a$.

In order to understand the definition of the continuity of a function, we will show that $f(x) = x^2$ is continuous at $x = a$. Since it is clear that the value of $f(x)$ is defined for all $x$, our goal is to show the following:

For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

As was explained before, noting the last inequality in the above assertion $| \, f(x) \, - \, f(a) \, | < \varepsilon$, we seek the range of $x$ where this inequality holds. Since
$$| \, f(x) \, - \, f(a) \, | = | \, x^2 \, - \, a^2 \, | = | \, x \, - \, a \,| | \, x \, + \, a \,|,$$
we must seek the range of $x$ where $|x \, - \, a| | x + a | < \varepsilon$ holds.

When $x$ is close to $a$, we can consider that $| x + a |$ is approximated by $2|a|$. Therefore, noting that $x$ is close to $a$, we assume that $$| x + a | < 2|a| + 1$$
holds, where we add $1$ to $2|a|$ with a margin. Then, for all $x$ satisfying $| x \, - \, a | < \displaystyle\frac{\varepsilon}{2|a| + 1}$, we have
$$| \, f(x) \, - \, f(a) \, | = | \, x \, - \, a \,| | \, x \, + \, a \, | < \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1 ) = \varepsilon.$$
This result is obtained under the assumption $| x + a | < 2|a| + 1$. However, we have not yet specified the range of $x$ where this inequality holds.

Suppose $| x \, - \, a | < \rho$, where $\rho$ will be specified later. Then, we have $|x| < |a| + \rho$ because $x$ is in a ball (interval) with radius $\rho$ and center $a$. Therefore, we have $$\begin{array}{l} | x + a | \leq |x| + |a| \\[2ex] \ \ \ \ < |a| + \rho + |a| \\[2ex] \ \ \ \ = 2|a| + \rho \end{array}$$ Hence, when $\rho \leq 1$, we have
$$| x + a | < 2|a| + 1.$$ Consequently, noting the case that $\rho = 1$, we see that $| x \, - \, a | < 1$ implies $| x + a | < 2|a| + 1$.

Thus, if $$| x \, - a | < 1 \ \ \ {\rm and} \ \ \ | x \, - a | < \displaystyle\frac{\varepsilon}{2|a| + 1},$$ then, noting $| x + a | < 2|a| + 1$ by $| x - a | <1$, we have $$| \, f(x) \, - \, f(a) \, | = | \, x \, - \, a \,| | \, x \, + \, a \, | < \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1 ) = \varepsilon.$$

Now, we define
$$\delta(\varepsilon) = \min\left( \frac{\varepsilon}{2|a| + 1}, \ 1 \right).$$
Notice that $| x \, - \, a | < \delta(\varepsilon)$ implies $| x \, - \, a | < 1$ and  $| x \, - \, a | < \displaystyle\frac{\varepsilon}{2|a|+1}$ because $\delta(\varepsilon) \leq 1$ and $\delta(\varepsilon) \leq \displaystyle\frac{\varepsilon}{2|a|+1}$.

Therefore, for all $x$ satisfying $| x \, - \, a | < \delta(\varepsilon)$, we have $| x + a | < 2|a| + 1$ by $| x \, - \, a | < 1$, which leads to $$\begin{array}{l} | f(x) \, - \, f(a) | = | x \, - \, a| | x \, + \, a | \\[2ex] \ \ \ \ < | x \, - \, a| (2|a| + 1) \\[2ex] \ \ \ \ < \delta(\varepsilon) (2|a|+ 1) \\[2ex] \ \ \ \ = \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1) = \varepsilon \end{array}$$ Hence, we see that $f(x) = x^2$ is continuous at $x = a$.

From the definition of the continuity of a function, I understand that $\delta$ is a function of $\varepsilon$ because $\delta$ depends on $\varepsilon$. However, in the above example, we define $\delta = \min\left( \displaystyle\frac{\varepsilon}{2|a| + 1}, \ 1 \right)$. Therefore, I think that $\delta$ can be regarded as a function of $a$ as well as of $\varepsilon$.

That’s a very good point. You are completely right.

It is usual that a function is defined on a set on the real line ${\bf R}$. This set is called the “domain.” In many cases, the domain of a function is given by an interval on ${\bf R}$. For example, when the domain of a function $f(x)$ is the interval $I = [a, \, b]$, the interval $I$ is given by the set
$$[a, \, b] = \{ \, x \, | \, a \leq x \leq b \, \}.$$
In this case, the value of the function $f(x)$ is defined for $x$ satisfying $a \leq x \leq b$. We can define a function on an interval that does not include its endpoints, such as
$$(a, \, b) = \{ \, x \, | \, a < x < b \, \}.$$
Moreover, the real line denoted by ${\bf R}$ can be regarded as an interval, and is often expressed by ${\bf R } = (-\infty, \, \infty)$.

Suppose that the domain of a function $f(x)$ is an interval $I$. Then, we say that $f(x)$ is continuous on $I$ if for any $a \in I$, $f(x)$ is continuous at $x = a$. That is, the definition that $f(x)$ is continuous on $I$ is as follows:

For any $a \in I$ and for any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon ; a) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon ; a)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

It should be noted that $\delta$ is a function of $\varepsilon$ and $a$ in this definition. For example, the function $f(x) = x^2$ is continuous on ${\bf R}$. In fact, as was explained above, for any $a \in {\bf R}$ and for any $\varepsilon > 0$, there exists a real positive number $\delta(\varepsilon ; a)$ defined by
$$\delta(\varepsilon ; a) = \min\left( \displaystyle\frac{\varepsilon}{2|a| + 1}, \ 1 \right)$$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon ; a)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

I understand that when a function $f (x)$ is continuous on an interval $I$, $\delta$ is a function of $a \in I$ and $\varepsilon > 0$. However, I wonder whether there exists a function such that $\delta$ is a function of only $\varepsilon$?

Indeed. When $\delta$ is a function of only $\varepsilon$, we say that $f(x)$ is uniformly continuous on $I$. As seen in the following exercise, there are many uniformly continuous functions. The uniform continuity of functions often plays an important role in advanced mathematical analysis.

Exercise 3　Show that $f(x) = x^2 -x$ is continuous at any $a \in {\bf R}$.

Exercise 4   Show that $f(x) = x^2$ is uniformly continuous on an interval $[a, \, b]$, where $a$ and $b$ are finite real numbers.

## Concluding Remarks

We come to the end of our explanation for the basis of the epsilon-delta argument. The idea of the epsilon-delta argument is to formulate our naive intuition for the limits of functions as the existence problem of a function $\delta = \delta(\varepsilon)$. Although this idea is quite simple, the epsilon-delta argument requires some technical training in using logic symbols, estimating inequalities, negating statements, and so on. We will be happy if this website can help calculus beginners to understand the basic underlying ideas and to overcome the technical barriers to using the epsilon-delta argument.

Can you explain the underlying basic idea of the epsilon-delta definition of the limits of functions (the continuity of functions) through the parable of an airplane approaching an airport in the previous chapter? I think that it is a good exercise to confirm your understanding for the epsilon-delta definition of limits of functions. epsilon-delta
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