Proof of Odd functions' Continuity

In summary, the problem states that an odd function g(x) is right-continuous at x = 0. The goal is to prove that g(x) is also continuous at x = 0 and that g(0) = 0. The key is to show that \lim_{x \to 0^{-}} g(x) exists and equals to \lim_{x \to 0^{+}} g(-x). This can be proven using the definition of odd functions and applying epsilon-delta proofs. Therefore, g(x) is continuous at x = 0 and g(0) = 0.
  • #1
bjgawp
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Homework Statement


If an odd function g(x) is right-continuous at x = 0, show that it is continuous at x = 0 and that g(0) = 0. Hint: Prove first that [tex]\lim_{x \to 0^{-}} g(x) [/tex] exists and equals to [tex]\lim_{x \to 0^{+}} g(-x)[/tex]


Homework Equations





The Attempt at a Solution


Suppose [tex]\lim_{x \to 0^{-}} g(x) = M[/tex] Let [tex]\epsilon[/tex] > 0. We must find [tex]\delta[/tex] > 0 such that whenever [tex]-\delta[/tex] < x < 0, it follows that |g(x) - M | < [tex]\epsilon[/tex]. We know that -x < [tex]\delta[/tex] but relating this to f(-x) is where I'm stuck. Like the other problem I posted, I can't see what the end result is supposed to be. Any help would be very much appreciated!
 
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  • #2
I think your attempted solution takes it too far! For an odd function f(-x) = -f(x), that should be enough to get your proof.
 
  • #3
Hmm I think I do have to incorporate epsilon delta proofs in this one. I was thinking that since -g(x) = g(-x) then |-g(x)| = |g(-x)|. And since we're proving that the limit of odd continuous functions have a limit at 0 (I think that's a safe assumption) then:
Given [tex]\epsilon>0[/tex]. [tex]\exists\delta >0[/tex] such that whenever 0 < |x| < [tex]\delta[/tex] it follows that |g(x) - M| < [tex]\epsilon[/tex]. Since M = 0, then [tex]|g(x)|<\epsilon[/tex]. (All this is the function approaching 0 from the left)

Then looking at g(-x), for the same epsilon and delta (since the definition of odd functions implies symmetry about the origin) we arrive at the same conclusion |g(x)| < [tex]\epsilon[/tex].

Do I seem to be on the right track or is this just dead wrong? It's difficult making the transition from concrete, repetitive epsilon-delta proofs to more general and abstract ones. The methods don't quite seem to be the same ...
 
Last edited:

What is an odd function?

An odd function is a mathematical function that satisfies the property f(-x) = -f(x), meaning that when the input is multiplied by -1, the output is also multiplied by -1. This results in the function being symmetric about the origin.

How do you prove the continuity of an odd function?

To prove the continuity of an odd function, you must show that it is continuous at 0 and at any other point in its domain. This can be done by using the definition of continuity, which states that a function is continuous at a point if the limit of the function as it approaches that point is equal to the value of the function at that point.

What is the importance of proving the continuity of an odd function?

Proving the continuity of an odd function is important because it allows us to determine whether or not the function is well-behaved and predictable. A continuous function is one that can be graphed without any breaks or jumps, making it easier to analyze and use in mathematical calculations.

Can an odd function be discontinuous?

Yes, an odd function can be discontinuous. While odd functions are typically continuous at all points, there are exceptions, such as the function f(x) = 1/x, which is discontinuous at x = 0. It is important to carefully analyze the behavior of an odd function to determine its continuity.

How does the continuity of an odd function affect its derivatives?

The continuity of an odd function does not have a direct effect on its derivatives. However, if a function is both odd and continuous, it can be shown that its derivative is an even function. This means that the derivative is also symmetric about the y-axis.

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