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Proove that a function is cointinuous at one point

  1. Mar 23, 2012 #1
    Hi, I've been trying to proof this with a partner and we couldn't, we tried to apply the definition of limit and continuity but we didn't get anything. This is the problem:

    "Let f a function which satisfies $$|f(x)|\leq|x| \forall{x\in{\mathbb{R}}}$$

    Proof that is continuous at 0.


    Thanks!!


    PD: I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

    $$|f(0)|<=|0|$$ then $$f(0)=0$$ . But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!
     
    Last edited: Mar 23, 2012
  2. jcsd
  3. Mar 23, 2012 #2

    LCKurtz

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    Have you figured out what value f(0) is? Do that and then show us what you get when you apply the ##\epsilon-\delta## definition for continuity at 0.
     
  4. Mar 23, 2012 #3
    Oh yeah! I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

    $$|f(0)|<=|0|$$ then $$f(0)=0$$ . But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!
     
  5. Mar 23, 2012 #4

    LCKurtz

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    Write out the definition, word for word, for a function f(x) to be continuous at ##x=a##. Just copy it from your text. Then put ##a=0## in that statement and see if it is true with what you are given.
     
  6. Mar 23, 2012 #5
    It should be like this:

    $$\forall{\epsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \epsilon}$$

    So:
    $$|x|< \delta$$
    and
    $$|f(x)|< \epsilon$$


    ??
     
  7. Mar 23, 2012 #6

    LCKurtz

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    So given ##\epsilon>0##, you have to find a ##\delta## depending on ##\epsilon## such that if ##|x|<\delta## then ##|f(x)|<\epsilon##. Can you come up with a ##\delta## which works for your function given what you know about it?
     
  8. Mar 23, 2012 #7
    No, I cannot come up with a solution! And I'm sure this is really simple! Can delta and epsilon be zero?
     
  9. Mar 23, 2012 #8
    |f(x)|<=|x|

    So draw two lines.
    y = x and y=-x.


    The only place where f(x) can lie is the area enclosed between these two lines.(How?)

    What do you see?


    The basic idea is that if x<=|a| (let a be positive )

    then, x lies between -a and a. (with -a and a included)

    If x>=|a| then,

    x lies between -infinity to -a and a to infinity
    (with -a and a included)

    So when |f(x)| <=|x|


    what do you get the range of f(x).

    What does this show for
    x---> 0. ?
     
    Last edited: Mar 23, 2012
  10. Mar 23, 2012 #9
    OK, try taking a concrete example and see if you can find a [itex]\delta[/itex] that way. For example, let [itex]f(x) = |x|[/itex] clearly [itex]|f(x)| \leq |x| \forall x \in R[/itex]. Now, can you find a [itex]\delta[/itex] such that [itex]|f(x) - 0| < \epsilon[/itex] whenever [itex]|x| < \delta[/itex]?


    If so, can you see how this can be genralised (it is a very short generalsation) to your problem?


    If you can't come up with a [itex]\delta[/itex] I suggest taking a break from this problem and come back to it in a few hours or tomorrow.
     
    Last edited: Mar 23, 2012
  11. Mar 23, 2012 #10

    Mark44

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    You mean f(0) >= 0, not <= 0.
    This doesn't necessarily imply that |0| = 0. All that you have shown here is that it can't be smaller than 0. You haven't eliminated or dealt with the possibility that it could be larger than 0.

    A simpler explanation uses the idea that the absolute value represents the distance between a given number and zero. Obviously the distance from 0 to 0 is 0, so |0| = 0.
     
  12. Mar 23, 2012 #11
    I made a graphic and analised the function |x| but in that case the delta has to be negative from the left and positive fro the right so the limit is zero. I don't understand what the delta means! You're talking about it as if it was an unique number... I thought these numbers (epsilon and delta) could be chosen freely and one depends on the other but they're not unique, just a range. It's the first time I am proving this kind of things and I know it's basic and will be essential so please don't make me think things I won't realize, I need a demonstration first. I really appreciate your help but I've been trying all day... this is getting me irritated. I cannot think anymore ... lol
     
  13. Mar 23, 2012 #12

    LCKurtz

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    No. You know that ##|f(x)|\le|x|## for your function. How small does ##\delta## need to be so that if ##|x|<\delta##, then ##|f(x)|<\epsilon##? It is really easy if you just look at it.
     
  14. Mar 23, 2012 #13

    LCKurtz

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    No, he means ##|f(0)|\le 0## like he wrote and his argument is correct that ##f(0)=0##.
     
  15. Mar 23, 2012 #14
    Alright.. I made a horrible paint drawing which I attached. Is that representation right? I think I'm not imagining well how it works graphically...
     

    Attached Files:

  16. Mar 23, 2012 #15

    Mark44

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    I stand corrected. I was laboring under the false impression that f(x) = |x|.
     
  17. Mar 23, 2012 #16
    Read each of the posts again.
    Every member has said some really useful point and arranging these like jigsaws will solve your puzzle.

    As i said in my post before,

    You get f(x) as lying between x and -x(how?)

    Meaning f(2) can lie anywhere -2 and 2

    As x tends to 0(from the RHS),

    lets say x is at 0.001
    f(x) can lie between -0.001 and 0.001.

    What do you see when x gets really small as in 0?

    Once you have figured this out graphically and logically you can start working back on your method using delta's which will look much easier.
     
    Last edited: Mar 23, 2012
  18. Mar 23, 2012 #17
    Oh yeah! Thanks! I think I understood the condition graphically... take a look at my attached drawing. Now I will analise it and see if I can come up with something. thanks!
     

    Attached Files:

  19. Mar 24, 2012 #18
    So the delta has to be zero! Why not???
     
  20. Mar 24, 2012 #19
    Isn't part of the definition of continuity "there exists a delta greater than zero"? So delta cannot equal zero.
     
  21. Mar 24, 2012 #20

    LCKurtz

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    You were within one step of being done with this problem. Have you given any thought to the line I re-quoted above?
     
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