Proving Symmetric Continuity of a Function at X0

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SUMMARY

A function f is symmetrically continuous at a point X0 if the limit of [f(X0 + h) - f(X0 - h)] as h approaches 0 equals 0. It is established that if f is continuous at X0, then it is symmetrically continuous at that point. However, the converse is not true, as demonstrated by the function f defined as f(x) = x for x ≠ 0 and f(0) = 1, which is symmetrically continuous at 0 but not continuous.

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  • Understanding of limits in calculus
  • Knowledge of continuity and its definitions
  • Familiarity with symmetric continuity concepts
  • Basic function definitions and properties
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  • Study the formal definition of continuity in calculus
  • Explore examples of symmetric continuity in various functions
  • Investigate the implications of discontinuities on symmetric continuity
  • Learn about theorems related to limits and continuity
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Mathematics students, calculus learners, and educators looking to deepen their understanding of continuity and its properties in real analysis.

mrs.malfoy
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A function f is said to be symmetrically continuous at X0 if

lim [f(X0 + h) - f(X0 - h)]= 0
h-> 0



Show that if f is continuous at X0, it is symmetrically continuous there but not conversely.
 
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This sounds like homework so I'm not going to go into too much detail, but note that if f is continuous at x then: [tex]lim_{h\rightarrow0}f(x+h)=lim_{h\rightarrow0}f(x-h)=f(x).[/tex]

There isn't much more to do.
 
For the converse, take
f(x) =x if x is nonzero ,
f(0) =1.
f is symmetrically continuous at 0, but not continuous.
 

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