Continuity

**Showcase_22** · May 30th 2009, 07:18 AM

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

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

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

**Plato** · May 30th 2009, 07:31 AM

Originally Posted by Showcase_22

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

Define a sequence: $\left( {\forall n \in \mathbb{Z}^ + } \right)\left[ {x_n = \frac{2}{{\pi \left( {2n + 1} \right)}}} \right]$ .

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

Does that give you a start?

**Random Variable** · May 30th 2009, 07:51 AM

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

**Sampras** · May 30th 2009, 08:11 AM

Originally Posted by Random Variable

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

He wants to prove that $f(x) = 3 \sin \left(\frac{1}{x} \right)$ is discontinuous at $x = 0$ using the $\epsilon- \delta$ definition of continuity.

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

**Random Variable** · May 30th 2009, 08:15 AM

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?

**Showcase_22** · May 30th 2009, 10:08 AM

But we can probably use Plato's sequence to come up with an appropriate $\epsilon$ .

You can use sequences to come up with an epsilon?

I can already prove it is discontinuous using sequences, I use $a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $b_n=\frac{1}{2n \pi-\frac{\pi}{2}}$ . I tried proving discontinuity using $\epsilon-\delta$ 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 just had another go:

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

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

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

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

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

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

Is this right?

**Plato** · May 30th 2009, 10:20 AM

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

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

**Showcase_22** · May 30th 2009, 10:39 AM

I think I know what you mean.

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

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

**Sampras** · May 30th 2009, 10:55 AM

Originally Posted by Showcase_22

I think I know what you mean.

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

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

No $0 <\epsilon = \frac{3}{\delta} \leq 3$ for all $\delta >0$ . With this $\epsilon$ , the discontinuity at $0$ can be shown. There is no need for a general expression for $\delta$ .

**Showcase_22** · May 30th 2009, 10:58 AM

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

**Sampras** · May 30th 2009, 11:00 AM

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 $\epsilon$ ?

You put $\epsilon = \frac{3}{\delta}$ . And that is correct.

**Showcase_22** · May 30th 2009, 11:01 AM

YAY!

So was that entire proof correct?

**Sampras** · May 30th 2009, 11:01 AM

Originally Posted by Showcase_22

YAY!

So was that entire proof correct?

yes.

**xalk** · May 30th 2009, 05:52 PM

Originally Posted by Showcase_22

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

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

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)| $\geq\epsilon$

AND in symbols:

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

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

Notice the difference with your definition.

Do you agree??

**Sampras** · May 30th 2009, 06:13 PM

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)| $\geq\epsilon$

AND in symbols:

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

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

Notice the difference with your definition.

Do you agree??

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

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