Problem 39 - Calculus Early Transcendentals

Question

If the function \(f\) is defined by $$f(x)=\left\\{\begin{array}{ll}{0} & {\text { if } x \text { is rational }} \\\ {1} & {\text { if } x \text { is irrational }}\end{array}\right.$$ prove that $$\lim _{x \rightarrow 0} f(x)$$ does not exist.

Step-by-Step Solution

Answer

The limit does not exist because the values approach 0 for rationals and 1 for irrationals as \( x \to 0 \).

Step 1: Recall the Definition of Limit at a Point

The limit of a function \( f(x) \) as \( x \) approaches a point \( c \) exists if and only if \( f(x) \) approaches the same single value \( L \) from both sides as \( x \) tends to \( c \). This can be written as \( \lim_{{x \to c}} f(x) = L \).

Step 2: Analyze the Behavior of \( f(x) \) for Rational Inputs

Consider rational numbers \( x \) approaching 0. For any rational \( x \), the function \( f(x) = 0 \). Hence, the function approaches 0 from the rational side, i.e., \( \lim_{{x \to 0, x \in \mathbb{Q}}} f(x) = 0 \).

Step 3: Analyze the Behavior of \( f(x) \) for Irrational Inputs

Now, consider irrational numbers \( x \) approaching 0. For any irrational \( x \), the function \( f(x) = 1 \). Hence, the function approaches 1 from the irrational side, i.e., \( \lim_{{x \to 0, x \in \mathbb{R} \setminus \mathbb{Q}}} f(x) = 1 \).

Step 4: Compare Limits from Rational and Irrational Sides

Since the limits from the rational side \( \lim_{{x \to 0, x \in \mathbb{Q}}} f(x) = 0 \) and the irrational side \( \lim_{{x \to 0, x \in \mathbb{R} \setminus \mathbb{Q}}} f(x) = 1 \) do not agree (0 \( eq \) 1), the limit \( \lim_{{x \to 0}} f(x) \) does not exist.

Key Concepts

Rational and Irrational Numbers

Rational and irrational numbers play a crucial role in understanding various mathematical concepts, including limits.

• Rational Numbers: These are numbers that can be written as a fraction, such as \( \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b eq 0 \). Examples include 1/2, 3/4, and -7.
• Irrational Numbers: These cannot be expressed as a simple fraction. They have non-repeating, non-terminating decimal expansions. Classic examples are \( \sqrt{2} \) and \( \pi \).

These two kinds of numbers fill the entire number line. Any real number is either rational or irrational. When approaching a limit, inputs can be rational or irrational depending on the function being evaluated. For the function in the exercise, rational inputs yield a different result compared to irrational inputs. This distinction leads to differences in behavior, which is crucial for determining whether a limit exists.

Limit Does Not Exist

A limit "does not exist" if a function approaches different values from the left and right as you come close to a specific point. For the function \( f(x) \) presented in the exercise:

• From Rational Side: As \( x \) approaches 0 through rational numbers, \( f(x) = 0 \).
• From Irrational Side: As \( x \) approaches 0 through irrational numbers, \( f(x) = 1 \).

Since the limits from rational and irrational sides as \( x \to 0 \) differ, the overall limit does not exist. This is because the essential condition for a limit to exist - converging to the same value from both sides - is not met.

Definition of Limit

The definition of a limit captures the idea of a function approaching a particular value as its input approaches some point. Mathematically, for a function \( f(x) \), the limit as \( x \) approaches \( c \), if it exists, is \( L \) if the values of \( f(x) \) can be made arbitrarily close to \( L \) by taking \( x \) sufficiently close to \( c \). There are specific conditions that must be satisfied for the limit to exist:

• The function \( f(x) \) must approach the same value \( L \) from both the left and the right as \( x \) approaches \( c \).
• The value \( L \) is the point the function tends towards; it need not be the actual value of the function at that point.

In this exercise's context, since \( f(x) \) does not approach the same value both from rational and irrational sides, \( \lim_{{x \to 0}} f(x) \) does not exist based on this definition.