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Math Help - Continuity

  1. #1
    Super Member Showcase_22's Avatar
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    Continuity

    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
    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 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?
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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  (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
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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  (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 .
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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 \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?
    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 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?
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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 \epsilon between <br /> <br />
0 < \varepsilon  < 3<br />
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?
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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 \epsilon between <br /> <br />
0 < \varepsilon  < 3<br />
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 .
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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 \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 \epsilon?
    You put  \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 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??
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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)| \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 .
    Last edited by Sampras; May 30th 2009 at 06:31 PM.
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