1. ## Continuity

Prove $\displaystyle f(x)=\begin{cases} 3 \sin \left( \frac{1}{x} \right) \ \ \ \ \ x \neq 0\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0\end{cases}$ is discontinuous at 0 using the $\displaystyle \epsilon-\delta$ definition for continuity.

So I gather that I need to find $\displaystyle \epsilon$ according to the definition $\displaystyle \exists \epsilon>0 \ s.t \forall \delta>0 \ |x-c|< \delta \Rightarrow \left|f(x)-f(c) \right|\geq \epsilon$

Umm....I have no idea how to start this. =S

2. Originally Posted by Showcase_22
Prove $\displaystyle f(x)=3 \sin \left( \frac{1}{x} \right)$ is discontinuous at 0 using the $\displaystyle \epsilon-\delta$ definition for continuity.
Define a sequence: $\displaystyle \left( {\forall n \in \mathbb{Z}^ + } \right)\left[ {x_n = \frac{2}{{\pi \left( {2n + 1} \right)}}} \right]$.

What are values of $\displaystyle f\left(x_n\right)?$

Does that give you a start?

3. To show that f is discontinuous at x=0 (assuming f is defined at x=0) find a sequence $\displaystyle (s_{n})$ in the domain that converges to 0 but where the sequence $\displaystyle (f(s_{n}))$ does not converge to f(0). $\displaystyle s_{n} = \frac {2}{n\pi}$ is a good choice

4. Originally Posted by Random Variable
To show that f is discontinuous at x=0 (assuming f is defined at x=0) find a sequence $\displaystyle (s_{n})$ in the domain that converges to 0 but where the sequence $\displaystyle (f(s_{n}))$ does not converge to f(0). $\displaystyle s_{n} = \frac {2}{n\pi}$ is a good choice
He wants to prove that $\displaystyle f(x) = 3 \sin \left(\frac{1}{x} \right)$ is discontinuous at $\displaystyle x = 0$ using the $\displaystyle \epsilon- \delta$ definition of continuity.

So you want to show: $\displaystyle \exists \epsilon >0, \ \text{s.t.} \ \forall \delta >0, \ |x| < \delta \implies |f(x)-f(0)| \geq \epsilon$. Now $\displaystyle f(0)$ is undefined. But we can probably use Plato's sequence to come up with an appropriate $\displaystyle \epsilon$.

5. I'm used to using the defintion to prove continuity, not discontinuity.

But if f is not defined at x=0, what is there to prove? A function cannot be continuous at a point where the function is undefined, right?

6. But we can probably use Plato's sequence to come up with an appropriate $\displaystyle \epsilon$ .
You can use sequences to come up with an epsilon?

I can already prove it is discontinuous using sequences, I use $\displaystyle a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $\displaystyle b_n=\frac{1}{2n \pi-\frac{\pi}{2}}$. I tried proving discontinuity using $\displaystyle \epsilon-\delta$ and I kept getting stuck!

This is a question one of my friends posed me so i'll change the initial question so it's defined at 0.

I need to find $\displaystyle \epsilon$ s.t $\displaystyle |x|<\delta \Rightarrow 3 \left| \sin \left( \frac{1}{x} \right) \right| \geq \epsilon$.

Using $\displaystyle \left| \sin \left( \frac{1}{x} \right) \right| \geq \left| \frac{1}{x} \right|$:

$\displaystyle \left| 3 \sin \left( \frac{1}{x} \right) \right| \geq \epsilon \Leftrightarrow \left| \frac{3}{x} \right| \geq \epsilon$

set $\displaystyle \epsilon= \frac{3}{\delta}$

Therefore since $\displaystyle |x|< \delta \Rightarrow \frac{1}{\delta}< \frac{1}{|x|}$

Hence $\displaystyle \left| \frac{3}{x} \right| \geq \frac{3}{\delta}=\epsilon$ as required.

Is this right?

7. From your edit we see that $\displaystyle f(0)=0$.
Using the sequence I posted, $\displaystyle \delta > 0\, \Rightarrow \,\left( {\exists x_K } \right)\left[ {\left| {0 - x_K } \right| < \delta } \right]$.
BUT $\displaystyle \left| {f(0) - f(x_K )} \right| = \left| {0 \pm 3} \right| = 3$

What happens if $\displaystyle 0 < \varepsilon < 3$?

8. I think I know what you mean.

If we pick an $\displaystyle \epsilon$ between $\displaystyle 0 < \varepsilon < 3$ we can use it to reach a contradiction in the definition of discontinuity.

For example, I could choose $\displaystyle \epsilon=\frac{3}{2}$. If this is the case, then from my previous post, $\displaystyle \delta=2$. Wouldn't I need to find a more general expression for delta though?

9. Originally Posted by Showcase_22
I think I know what you mean.

If we pick an $\displaystyle \epsilon$ between $\displaystyle 0 < \varepsilon < 3$ we can use it to reach a contradiction in the definition of discontinuity.

For example, I could choose $\displaystyle \epsilon=\frac{3}{2}$. If this is the case, then from my previous post, $\displaystyle \delta=2$. Wouldn't I need to find a more general expression for delta though?
No $\displaystyle 0 <\epsilon = \frac{3}{\delta} \leq 3$ for all $\displaystyle \delta >0$. With this $\displaystyle \epsilon$, the discontinuity at $\displaystyle 0$ can be shown. There is no need for a general expression for $\displaystyle \delta$.

10. Didn't I do that in one of my previous posts? Or do I have to explicitly assign a value to $\displaystyle \epsilon$?

11. Originally Posted by Showcase_22
Didn't I do that in one of my previous posts? Or do I have to explicitly assign a value to $\displaystyle \epsilon$?
You put $\displaystyle \epsilon = \frac{3}{\delta}$. And that is correct.

12. YAY!

So was that entire proof correct?

13. Originally Posted by Showcase_22
YAY!

So was that entire proof correct?
yes.

14. Originally Posted by Showcase_22
Prove $\displaystyle f(x)=\begin{cases} 3 \sin \left( \frac{1}{x} \right) \ \ \ \ \ x \neq 0\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0\end{cases}$ is discontinuous at 0 using the $\displaystyle \epsilon-\delta$ definition for continuity.

So I gather that I need to find $\displaystyle \epsilon$ according to the definition $\displaystyle \exists \epsilon>0 \ s.t \forall \delta>0 \ |x-c|< \delta \Rightarrow \left|f(x)-f(c) \right|\geq \epsilon$

Umm....I have no idea how to start this. =S
Showcase to prove discontinuity at x=0 you must show:

After you have chosen ε>0 you must prove:

for all δ:

if δ>0 ,then there exists an ,x such that :

|x|<δ AND |f(x)-f(0)|$\displaystyle \geq\epsilon$

AND in symbols:

$\displaystyle \exists\epsilon$ and $\displaystyle \forall\delta>0\Longrightarrow\exists x$,such that:

|x|<δ AND |f(x)-f(0)|$\displaystyle \geq\epsilon$.

Notice the difference with your definition.

Do you agree??

15. Originally Posted by xalk
Showcase to prove discontinuity at x=0 you must show:

After you have chosen ε>0 you must prove:

for all δ:

if δ>0 ,then there exists an ,x such that :

|x|<δ AND |f(x)-f(0)|$\displaystyle \geq\epsilon$

AND in symbols:

$\displaystyle \exists\epsilon$ and $\displaystyle \forall\delta>0\Longrightarrow\exists x$,such that:

|x|<δ AND |f(x)-f(0)|$\displaystyle \geq\epsilon$.

Notice the difference with your definition.

Do you agree??
Yes that is correct. You need to produce both an $\displaystyle \epsilon$ and an $\displaystyle x_k$ to prove discontinuity. For example, in Plato's sequence, $\displaystyle |f(x_2)-f(0)| = 3$ for $\displaystyle |x_2| < \delta$ for some $\displaystyle \delta$. So choose $\displaystyle \epsilon \in (0,3)$ to prove discontinuity. We can find some $\displaystyle x_k$ for all $\displaystyle \delta>0$.

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