Continuity of piecewise function undefined for 1<x<=2

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The piecewise function f : [0,1] U (2,3] defined by f(x)= x for 0<=x<=1 and f(x)= x-1 for 2 PREREQUISITES

  • Understanding of piecewise functions
  • Knowledge of continuity in mathematical analysis
  • Familiarity with limits and convergence
  • Basic understanding of function domains
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  • Learn about the formal definition of continuity in calculus
  • Explore examples of functions with undefined points and their continuity
  • Investigate the concept of limits and their application in different domains
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Students of mathematics, educators teaching calculus, and anyone interested in the properties of functions and continuity in mathematical analysis.

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My math book claims that the piecewise function f : [0,1] U (2,3] --> R defined by

f(x)=
x for 0<=x<=1
x-1 for 2<x<=3

is continuous. But it's undefined for 1<x<=2 so how can it be continuous? According to the definition of continuity, a function is at a point x0 if for a sequence x_n converging to x0 the image limit converges to f(x0), that is (the limit taking n to infinity)

lim(f(x_n))=f(x0)

But I cannot really figure out how this would work at x0=2 since I would then have to take f(2) which I can't since it is undefined there.

Thanks in advance!
 
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It is NOT continuous on R and your math book doesn't say it is. Note that, for example [itex]f:R\rightarrow R[/itex], defined by f(x)= 1/x, is a different function than [math]f:(-\infty, 0)\cup (0, \infty)[/math]. The first is not a continuous function because it is not continuous at x= 0. The second is a continuous function because it is continuous at every point in it given domain.

"lim(f(x_n)= f(x0)" doesn't have to work at x= 2 because 2 is not in the declared domain of the function.
 

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