Question

Use the following definition for the nonexistence of a limit. Assume \(f\) is defined for all values of \(x\) near a, except possibly at a. We write \(\lim _{x \rightarrow a} f(x) \neq L\) if for some \(\varepsilon>0\) there is no value of \(\delta>0\) satisfying the condition $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0<|x-a|<\delta.$$ 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 \(\lim _{x \rightarrow a} f(x)=L\) for some values of \(a\) and \(L\) and let \(\varepsilon=\frac{1}{2}\).)

Short answer

Given the piecewise function \(f(x)\) where \(f(x) = 0\) if \(x\) is rational and \(f(x) = 1\) if \(x\) is irrational, prove that the limit of \(f(x)\) does not exist for any value of \(a\). Answer: By assuming the limit exists and using the given hint of \(\varepsilon = \frac{1}{2}\), we demonstrate that for any assumed limit at \(a\), it is impossible to satisfy the condition \(|f(x) - L| < \varepsilon\) for both rational and irrational \(x\) within a given interval. This contradiction proves that the limit of \(f(x)\) does not exist for any value of \(a\).

Step-by-step solution

Assume the limit exists

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

Use the given ε

We are given a hint to use \(\varepsilon = \frac{1}{2}\). Now, we have \(|f(x) - L| < \frac{1}{2}\).

Find a rational and irrational number within the δ interval

Since we assumed that the limit exists, there must be a \(\delta>0\) such that the condition \(|f(x)-L|<\varepsilon\) holds for any \(x\) satisfying \(0<|x-a|<\delta\). Choose a rational number \(x_r\) and an irrational number \(x_i\) such that both are in the interval \((a-\delta, a+\delta)\). In other words, for both \(x_r\) and \(x_i\), we have \(0<|x-a|<\delta\).

Evaluate the function at \(x_r\) and \(x_i\)

As \(x_r\) is rational, \(f(x_r) = 0\). As \(x_i\) is irrational, \(f(x_i) = 1\).

Check the ε condition

Now let's check if the condition \(|f(x) - L| < \varepsilon\) holds for both \(x_r\) and \(x_i\): For \(x_r\): $$|f(x_r) - L| = |0 - L| = |L|$$ For \(x_i\): $$|f(x_i) - L| = |1 - L| = |-L + 1|$$ Since we used \(\varepsilon = \frac{1}{2}\), we should have \(|L| < \frac{1}{2}\) and \(|-L + 1| < \frac{1}{2}\).

Prove the contradiction

However, having both \(|L|<\frac{1}{2}\) and \(|-L+1|<\frac{1}{2}\) is impossible, as it would mean: $$L-\frac{1}{2}<0$$ $$L+\frac{1}{2}>1$$ The above inequalities mean that \(L\) is less than the midway between \(0\) and \(1\), and simultaneously that it is greater than the midway between \(0\) and \(1\). This is a contradiction, and hence our assumption that the limit exists for some value of \(a\) and \(L\) is not true. Therefore, we can conclude that \(\lim_{x \rightarrow a} f(x)\) does not exist for any value of \(a\).

Key concepts

Delta-Epsilon Proof

In calculus, the delta-epsilon proof is a formal method for proving the limit of a function at a particular point. This method verifies that the limit exists and is equal to a certain value by employing precise conditions. The core of the delta-epsilon proof lies in showing that for every epsilon (ε), no matter how small, there exists a delta (δ) such that whenever the distance between x and the limit point a is less than δ (but not zero), the distance between f(x) and the limit value L is less than ε.

To demonstrate this concept, let's consider a hypothetical function f(x) and a limit point a. The proof requires us to choose an ε>0 and find a corresponding δ>0 that satisfies the following condition for all x within the range (a-δ, a+δ) but not including a itself:
$$|f(x) - L| < \text{ε whenever } 0 < |x-a| < \text{δ}.$$
If we can locate such a δ for every positive ε, then by definition, the limit as x approaches a of f(x) is L. If we cannot find such a δ even for a single ε, as in the exercise provided, then we must conclude that the limit does not exist.

Limits Involving Rational and Irrational Numbers

When considering limits involving rational and irrational numbers, it's critical to understand the nuanced behavior of functions around these types of numbers. Unlike continuous functions that behave predictably, functions defined differently for rational and irrational numbers can produce oscillating values near any point.

The exercise given involves a function f(x) that assigns the value 0 for all rational inputs and 1 for all irrational inputs. When attempting to find the limit of f(x) as x approaches any value a, the function's value alternates between 0 and 1, depending on whether x is rational or irrational. Since rational and irrational numbers are densely interwoven on the number line, between any two numbers, there are infinitely many rationals and irrationals. This interplay ensures that the function's value has no consistent behavior as x gets close to a, and thus the limit does not exist.

Even within an infinitesimally small interval around a, both rational and irrational numbers exist, resulting in a function that jumps between 0 and 1 without settling on any particular value. Hence, no matter how small the ε chosen, we cannot define a δ interval that maintains the function's values within an ε range of a purported limit L. This characteristic highlights a key aspect of limits involving such piecewise-defined functions.

Nonexistence of Limit

The nonexistence of a limit occurs with functions whose values near a point do not converge to a specific number. In our exercise, we dealt with a function which exhibits different behaviors for rational and irrational numbers, causing the function's value to oscillate as it approaches any point a.

To demonstrate the nonexistence of a limit, one can use a particular ε, in this case suggested as ε = 1/2, and illustrate that no δ can be found to satisfy the necessary conditions for any value L. This is evident in the step by step solution where it was shown that both |L| < 1/2 and |-L + 1| < 1/2 cannot be true simultaneously; thus, exposing a contradiction. This contradiction implies that our initial assumption, the existence of a limit L as x approaches a, is false.

Understanding the nonexistence of a limit is crucial, especially when dealing with discontinuous functions or functions behaving in a non-standard manner. It reveals the intricacies of a function's behavior around a given point and emphasizes the importance of rigor in mathematical proofs, ensuring that one does not mistakenly attribute a limit to a function where none exists.