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Thread: Continuity

  1. #1
    Super Member Showcase_22's Avatar
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    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
    Last edited by Showcase_22; May 30th 2009 at 10:09 AM.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    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?
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  3. #3
    Super Member Random Variable's Avatar
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    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
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  4. #4
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Random Variable View Post
    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 $.
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  5. #5
    Super Member Random Variable's Avatar
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    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?
    Last edited by Random Variable; May 30th 2009 at 08:28 AM.
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  6. #6
    Super Member Showcase_22's Avatar
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    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 just had another go:

    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?
    Last edited by Showcase_22; May 30th 2009 at 10:30 AM.
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  7. #7
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    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 $\displaystyle 0 < \varepsilon < 3$?
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  8. #8
    Super Member Showcase_22's Avatar
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    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?
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  9. #9
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    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 $.
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  10. #10
    Super Member Showcase_22's Avatar
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    Didn't I do that in one of my previous posts? Or do I have to explicitly assign a value to $\displaystyle \epsilon$?
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  11. #11
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    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.
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  12. #12
    Super Member Showcase_22's Avatar
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    YAY!

    So was that entire proof correct?
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  13. #13
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    YAY!

    So was that entire proof correct?
    yes.
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  14. #14
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    Quote Originally Posted by Showcase_22 View Post
    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??
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  15. #15
    Senior Member Sampras's Avatar
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    Quote Originally Posted by xalk View Post
    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 $.
    Last edited by Sampras; May 30th 2009 at 06:31 PM.
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