Question

70-72. Proving that \(\lim _{x \rightarrow a} f(x) \neq L\) Use the following definition for the nonexistence of a limit. Assume \(f\) is defined for all x near a, except possibly at a. We write \(\lim f(x) \neq L\) if for some \(\varepsilon>0,\) there is no value of \(\delta>0\) satisfying the condition $$|f(x)-L|<\varepsilon \text { whenever } 0<|x-a|<\delta$$ Let $$ f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \text { is rational } \\ 1 & \text { if } x \text { is irrational. } \end{array}\right. $$ Prove that \(\lim _{x \rightarrow a} f(x)\) does not exist for any value of \(a\). (Hint: Assume \(\left.\lim _{x \rightarrow a} f(x)=L \text { for some values of } a \text { and } L, \text { and let } \varepsilon=\frac{1}{2} .\right)\)

Short answer

Based on our analysis, provide a short answer explaining why the function \(f(x)\) does not have a limit for any value of \(a\). The function \(f(x)\) does not have a limit for any value of \(a\) because it alternates between the values 0 (for rational x) and 1 (for irrational x). In any neighborhood of \(a\), there will always be both rational and irrational values of \(x\), resulting in contradictions when examining the nonexistence condition with a chosen \(\varepsilon = \frac{1}{2}\). As it is impossible to satisfy this condition for any \(a\) and \(L\), the limit does not exist for any value of \(a\).

Step-by-step solution

Define the function

To prove that the limit of \(f(x)\) does not exist for any value of \(a\), we will examine the given function and its behavior: $$ f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \text { is rational } \\\ 1 & \text { if } x \text { is irrational. } \end{array}\right. $$

Assume limit exists

To get the contradiction, let's assume that the limit exists for some value of \(a\) and \(L\). Specifically, let's assume that \(\lim_{x \rightarrow a} f(x) = L\).

Set the value of \(\varepsilon\)

Based on the hint, we will now choose the value of \(\varepsilon = \frac{1}{2}\). This value will be used to show that we cannot find a suitable \(\delta\) satisfying the nonexistence of a limit condition.

Examine the neighborhood of \(a\)

To fulfill the nonexistence condition, we need to find a suitable \(\delta>0\) for our given \(\varepsilon\). Observe that in any interval centered around \(a\) (i.e., \((a-\delta, a+\delta)\)), there will always be rational and irrational values of \(x\).

Show contradiction

Analyzing the function behavior, if \(x\) is rational, we have \(f(x)=0\), and if \(x\) is irrational, we have \(f(x)=1\). In both cases, we examine the nonexistence condition: - If \(x\) is rational, then \(|f(x)-L|=|0-L|=|L|\) - If \(x\) is irrational, then \(|f(x)-L|=|1-L|=|(1-L)|\) No matter the values of \(a\) and \(L\), in any interval \((a-\delta, a+\delta)\), there will be both rational and irrational \(x\) values. This means that it is impossible to satisfy \(|f(x)-L|<\varepsilon=\frac{1}{2}\) for all \(x\) in the interval, since we cannot simultaneously make both \(|L|<\frac{1}{2}\) and \(|(1-L)|<\frac{1}{2}\) for every \(x\). This is a contradiction.

Conclude

Since we have shown that it is impossible to satisfy the nonexistence condition for any \(a\) and \(L\), we conclude that the limit \(\lim _{x \rightarrow a} f(x)\) does not exist for any value of \(a\).

Key concepts

Epsilon-Delta Definition of a Limit

Understanding the concept of a limit is fundamental in calculus, and the epsilon-delta definition is a formalization of the idea of a limit. This definition provides a rigorous mathematical way to define what we mean precisely when we say that a function approaches a certain value as its input approaches some point.

The definition starts with two values, \(\varepsilon\) (epsilon) and \(\delta\) (delta). \(\varepsilon\) represents an arbitrarily small positive distance away from the function value we are approaching, and \(\delta\) represents an arbitrarily small distance from the point to which the input of the function is approaching. In simpler terms, for every \(\varepsilon > 0\) we choose, there should be some \(\delta > 0\) such that if the input \(x\) is within \(\delta\) of the point \(a\) (without being exactly at \(a\)), then the function's value is within \(\varepsilon\) of the limit \(L\).

If such a \(\delta\) cannot be found for even one positive \(\varepsilon\), then the function does not have a limit \(L\) at the point \(a\). In the exercise, this epsilon-delta relationship is used to demonstrate that no matter how small a \(\delta\) we choose, the function's values cannot be contained within the \(\varepsilon\) interval owing to its discontinuous nature at every point.

Proof of Nonexistence of a Limit

Proving the nonexistence of a limit may at first seem unintuitive. Instead of demonstrating something is true, you are tasked with showing something cannot possibly be true under the given conditions. You begin by assuming that the limit does exist, and then try to arrive at a contradiction with this assumption.

Using the epsilon-delta definition, if a function's limit as \(x\) approaches \(a\) is assumed to be \(L\), you should theoretically be able to find a \(\delta\) for every positive \(\varepsilon\) chosen. However, if you can show that for a particular \(\varepsilon\), no matter how small of a \(\delta\) you choose, you cannot keep the function's values within \(\varepsilon\) of \(L\), then you've proven the limit does not exist.

In our exercise, by choosing \(\varepsilon = \frac{1}{2}\), we illustrate there is no \(\delta\) that can satisfy the necessary condition with a function that flips between two values based on whether \(x\) is rational or irrational.

Rational and Irrational Numbers

In the context of the exercise, understanding the difference between rational and irrational numbers is key. Rational numbers are the numbers that can be expressed as the quotient of two integers, where the denominator is not zero. They include integers, fractions, and repeating or terminating decimals. Examples include \( \frac{1}{2} \) and \( -3 \).

In contrast, irrational numbers cannot be written as a simple fraction. They are real numbers with non-repeating, non-terminating decimal expansions. Some famous examples are \( \pi \) and \( \sqrt{2} \). Because of their distinct qualities, between any two real numbers, there are infinitely many rationals and infinitely many irrationals.

The significance in our exercise is that no matter how close we get to any point \(a\), we will always find both rational and irrational numbers. This ensures that the function \(f(x)\), which assigns different values based on the number type, will never settle down to one value as \(x\) approaches \(a\), thereby making the limit nonexistent. This characteristic of rational and irrational numbers is vital in constructing the proof of nonexistence of the limit.