Proving or Disproving Continuity and Limits for Functions in R

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SUMMARY

The discussion centers on the continuity and limits of a function f: R -> R, specifically addressing two statements. The first statement asserts that if f is continuous at x0, then it is also continuous at x0 + k, which is proven to be true using the epsilon-delta definition of continuity. The second statement claims that if the limit of f at infinity is 0, then f(x) must equal 0 for all x in R, which is also proven true by contradiction, demonstrating that any positive value of f at a point contradicts the limit condition.

PREREQUISITES
  • Understanding of the epsilon-delta definition of continuity
  • Knowledge of limits and their properties in real analysis
  • Familiarity with functions and their behaviors over the real numbers
  • Basic proof techniques, including proof by contradiction
NEXT STEPS
  • Study the epsilon-delta definition of continuity in depth
  • Explore the properties of limits in real analysis
  • Learn about periodic functions and their continuity
  • Investigate proof techniques, particularly proof by contradiction
USEFUL FOR

Mathematics students, particularly those studying real analysis, educators teaching continuity and limits, and anyone interested in formal proofs related to functions in R.

daniel_i_l
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Homework Statement


You have the function f:R->R and f(x) = f(x+k) where k is in R and k>0.
Prove or disprove:
1) If f is continues at x0 then it's also continues at x0 + k
2)If the limit of f at infinity is 0 then f(x)=0 for all x in R.


Homework Equations





The Attempt at a Solution


1)Yes: If for every epsilon>0 there's a lambda>0 so that for every x where 0<|x-x0|<lambda => |f(x) - f(x0)| < epsilon then for every 0<|x-(x0+k)|<lambda => 0<|(x-k)-x0|<lambda => |f(x-k) - f(x0)| < epsilon => |f(x) - f(x0+k)| < epsilon

2)Yes: If we have an x0 in R so that f(x0)>0 then we can choose epsilon=f(x0)/2 and then |f(x0)-0| > epsilon. So then for every N>0 we can find a k so that x0+nk > N and so |f(x0+nk)-0| > epsilon and we've proved that the limit of f at infinity isn't 0 - which is a contradiction to the information given in the question. (if there's some x0 where f(x0)<0 the proof in analogues) So for all x in R, f(x)=0.

Are those right?
Thanks.
 
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