Disproving Continuity of f(x) = {sin(1/x), 0} Using Epsilon-Delta Proof

[rman144]

Show that:

...{sin(1/x), x not zero
f(x)={
...{0, x=0

Is not a continuous function using epsilon-delta.


EDIT: I honestly haven't a clue. I figured I could just show that regardless of how small you make delta, there is always a value of f(x) that equals one, but I don't know how to write that as an epsilon-delta format.

EDIT: I understand that; the issue I'm having is writing this as a formal epsilon-delta proof, not the mathematics behind it.

 

[dx]

Show your attempt.

 

[dx]

Consider the sequence of points x_n = 1/[(π/2) + 2πn]. What is sin(1/x_n)?

 

[dx]

The epsilon-delta proof is what I'm trying to lead you to. Can you find a δ such that for all x with |x - 0| < δ, |f(x) - 0| < 1/2? Use my previous hint.

The thread would be easier to read if you post your responses as new posts rather than editing the original post.

 

[Elucidus]

Delta-epsilon proofs for non-convergence to a limit usually follow this pattern:

(1) Let L be the value to which the function is being shown to not converge to as x approaches a.

(2) Suppose epsilon is a specified value (one you know is going to fail).

(3) Then for any delta > 0, we can find an x closer to a so that |f(x) - L| > epsilon and this is how.

(4) Hence f cannot converge to L.

For your puroses epsilon could be any number strictly between 0 and 1 (1/2 would be fine) and the sequence that dx mentioned always contains a term that is within delta of a for any delta.

--Elucidus

 

[roam]

\lim_{x \to 0} sin \frac{1}{x} when a=0. In this case the limit of f near a=0 does not exist, when it approaches zero the function just oscilates between -1 and 1 and never really has a limit. We need to show that for \forall L \in R , L is not the limit of f near a.

One can divide the proof in three cases:

1) L>0
2) L<0
3) L=0

 

[union68]

Most of the time when a person says that a function "is continuous" they mean it is continuous at all points in its domain. To prove that a function is "not continuous" the first step is to find a point of discontinuity (obviously, ha!). Using the graph and your knowledge of the \sin function, you might venture a guess that the function is continuous on all points in \mathbb{R} - \{0\}, so we should look at the function's behavior around x=0.

The definition of continuity at a is

\lim_{x\to a} f \left(x\right) = f \left(a\right).

We wish to show that this is false at a=0. So, since f \left(0\right) = 0, we must prove

\lim _{x\to 0} f \left(x \right) \neq 0.

Negating the epsilon-delta definition means that we must show that there exists an \epsilon>0 such that for all \delta >0 there exists an x so that \left|x\right| < \delta and \left| f \left(x\right) \right| \geq \epsilon. In more intuitive terms, if we pick a specific \epsilon we must show that no matter how close we make x to 0, f \left(x\right) will sometime be outside \left( -\epsilon, \epsilon \right).

Now, the graph of \sin 1/x should help. What happens as x gets closer and close to 0? The function ALWAYS bounces between 1 and -1. Using this and dx's generous hint, you should be able to see what's going on. So, if you understand the math and intuition behind it (which you said you did...hence why I'm skipping some parts here), you can write the proof extremely concisely...

Proof:

Pick \epsilon = 1/2. For any \delta >0 there exists

x = \frac{2}{\pi \left( 4n-3 \right)}, \quad n \in \mathbb{N},

such that \left|x\right| < \delta and \left| f\left(x\right)\right| = 1 \geq 1/2.

Q.E.D.

Notice how my proof EXACTLY matches the negated epsilon-delta definition? To write a proper epsilon-delta proof you must have that thing MEMORIZED. I don't mean being able to roboticaly recite it on command, I mean live and breathe the thing.