Let [math: f(x)] be the function that is equal to 0 for all rational [math: x] and 1 for every irrational [math: x]. Is [math: f(x)] continuous at [math: x=π]x=\pi?Single choice
A
a. No.
B
b. Yes.

Question

Let [math: f(x)] be the function that is equal to 0 for all rational [math: x] and 1 for every irrational [math: x]. Is [math: f(x)] continuous at [math: x=π]x=\pi?

BlackTom AI · Accepted Answer

This question asks about the continuity of a classic example: the Dirichlet function, defined as f(x) = 0 if x is rational and f(x) = 1 if x is irrational.

Option a: 'No.' If we examine continuity at x = π, we note a crucial fact: every open interval around π contains both rational and irrational numbers. Consequently, as x approaches π through rational values, f(x) = 0, and as x approaches π through irrational values, f(x) = 1. The limit would have to be the same along all sequences approaching π, but we obtain two different limit values (0 and 1) depending on the path of approach. Thus the limit as x → π does not exist. Since π is itself irrational, f(π) = 1. For continuity, the limit as x → π must equal f(π); however, the non-existent limit already fails this prerequisite, so the function is not continuous at π.

Option b: 'Yes.' This statement claims continuity at π. The above reasoning shows that the limit does not exist due to the alternating values 0 and 1 in every neighborhood of π, which already violates the basic requirement for continuity. Moreover, even if one tries to apply an epsilon-delta criterion, choosing ε = 1/2 would force any δ-neighborhood around π to contain points x with f(x) = 0 and points with f(x) = 1, making |f(x) − f(π)| equal to 1 for some x within δ, which cannot be kept below ε. Therefore, this option is incorrect since the function fails the continuity test at π.

In summary, the key idea is that the Dirichlet function oscillates between 0 and 1 in any interval, so no point, including π, can be a point of continuity. The correct assessment is that the function is not continuous at π, which contradicts the statement in option b.

Let [math: f(x)] be the function that is equal to 0 for all rational [math: x] and 1 for every irrational [math: x]. Is [math: f(x)] continuous at [math: x=π]x=\pi?Single choice

Similar Questions

Find the intervals on which the function is continuous. y =

Which of the following statements are TRUE? (Select all that apply.)

For f(x) given below, find the numbers at which f is discontinuous. At which of these values is f continuous from the right, from the left, or neither? f(x)={2x if x≤1 3−x if 1<x≤4 √ x if x>4

Which of the following trees corresponds to a potential parse of the ambiguous sentence below, with correct syntactic categories? Some diagnostics are provided.

Which of the following sentences contain two non-finite verbs? Select all that apply.

Analyze the following sentence: The sentence contains [ Select ] 8 6 5 7 3 4 functional words, [ Select ] 2 3 1 4 5 count nouns, [ Select ] 4 5 2 1 3 mass nouns, and [ Select ] 1 5 2 3 4 adjectives.

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler