Continuety of splitted function

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Function
Click For Summary
SUMMARY

The discussion focuses on the continuity of a piecewise defined function, specifically proving that it is continuous at every irrational point and discontinuous at every rational point. The function is defined as f(x) = 0 for irrational x and f(x) = 1/n for rational x, where m/n is in lowest terms. The limit of f(x) as x approaches any irrational number is established as 0, while the limit along rational numbers must also equal 0 for continuity. The provided hint emphasizes the finite nature of possible denominators in rational numbers to demonstrate that the limit exists and equals 0.

PREREQUISITES
  • Understanding of piecewise defined functions
  • Familiarity with limits and continuity in calculus
  • Knowledge of sequences and convergence
  • Basic understanding of rational and irrational numbers
NEXT STEPS
  • Study the definition of continuity in the context of piecewise functions
  • Learn about the epsilon-delta definition of limits
  • Explore sequences and their convergence properties in real analysis
  • Investigate the properties of rational and irrational numbers in mathematical proofs
USEFUL FOR

Mathematics students, educators, and anyone studying real analysis or calculus, particularly those interested in the properties of continuity and limits of functions.

transgalactic
Messages
1,386
Reaction score
0
prove that this function continues on every irrational point
and discontinues in every rational point?
http://img136.imageshack.us/img136/5985/87260996wu4.png
 
Last edited by a moderator:
Physics news on Phys.org
The English for "splitted function" is "piecewise defined" function although here it is more "pointwise" defined:
f(x)= 0 if x is irrational, 1/n if x is ration, x= m/n reduced to lowest terms (and n is assumed to be positive).

As I said in a previous response, [itex]lim_{x\rightarrow a} f(x)= L[/itex] if and only if [itex]lim_{n\rightarrow\infty} f(a_n)= L[/itex] for any sequence [itex]{a_n}[/itex] converging to x. Obviously for any sequence, [itex]a_n[/itex]., of irrational numbers converging any x, that limit is 0.

So in order that this be continuous, the limit as we approach along rational numbers must also be 0 and the function value must be 0.

Here's a hint. If x= m/n, given any [itex]\epsilon> 0[/itex] there are only a finite number of possible N such that [itex]N< 1/epsilon[/itex] (so [itex]1/N> 1/\epsilon[/itex]) and for each such N there are only a finite number of M such that M/N is within [itex]\delta[/itex] of m/n. Use that to prove that the limit always exists and is always equal to 0.
 
i know the definition of bound and of continuity
i can't understand what the last hint means and how to use it
??
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K