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Proof of continuity of f(x) = 1/x

  1. Jan 5, 2007 #1

    radou

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    I just spent about an hour going through the proof that f(x) = 1/x is continous at every point in R\{0}, and I'm still not completely sure if I understood the proof. I wonder if someome could perhaps present a more elegant and easy way to proove this, or is this the only way? I should actually present the proof here, but I'm interested in how someome else would proove that.

    P.S. If there exists some proof using limits, I'm not really interested in it, since the textbook defines continuity of a function without using limits. :smile:
     
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  3. Jan 5, 2007 #2

    George Jones

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    For f(x) = 1/x, the inverse image of an open interval (that doesn' contain zero) is an open interval.
     
  4. Jan 5, 2007 #3
    You could try proving that your function is differentiable, hence continuous.
     
  5. Jan 5, 2007 #4
    you can represent this function as integral like \int_{-\infty}^{x}1/z^2 dz
    After that we can use geometric meaning of integral - square under curve.
    It seems obvious that if we integrate over R/{0} this square will be finite.
    This method works if function is not step-like.
     
  6. Jan 5, 2007 #5

    Hurkyl

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    It's hard to help if we don't know what definition of continuity you want to use, what proof you're looking at, and where you're having trouble with it. :tongue:


    Are you using the "f is continuous iff the inverse image of an open set is open" definition? You know that if this condition holds for neighborhoods in the range, it's true for all open sets in the range, right? There are only two kinds of neighborhoods in R: open intervals containing 0 and open intervals not containing 0.
     
  7. Jan 5, 2007 #6

    radou

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    Ok, here it goes.

    As said, we want to prove that f : x --> 1/x is continous at every point of R\{0}.

    A real function defined on an open interval I = <a, b> is said to be continuous at a point c from I if for every [itex]\epsilon > 0[/itex] there exists at least one [itex]\delta > 0[/itex] such that for every x from I, for which [itex]|x-c|<\delta[/itex] holds, [itex]|f(x)-f(c)|<\epsilon[/itex] holds, too. Further on, a function is continuous on the interval I if it is continuous at
    every point c from I.

    Now, let [itex]\epsilon>0[/itex] be given. We have to find some [itex]\delta>0[/itex] such that, for [itex]x\neq 0[/itex], we have [itex]|x-c|<\delta \Rightarrow |f(x)-f(c)| = |\frac{x-c}{x\cdot c}|<\epsilon[/itex].

    Further on, the proof says than now we consider all real numbers x such that [itex]|x-c|<\frac{1}{2}|c|[/itex]. Now, this relation does not hold for x = 0, so we assured that x cannot be 0. Further on, it is equivalent to [itex]c-\frac{1}{2}|c|<x<c+\frac{1}{2}|c|[/itex], so it is obvious that for every [itex]c\neq 0[/itex], any x lies in an interval which does not contain 0, hence x cannot be 0. The first thing that troubles me: could we also consider all the reals x such that [itex]|x-c|<\frac{1}{9}|c|[/itex], for example? Is that formulation chosen just because it is 'easy' to calculate with 2? (Sorry if it's a dumb question, but I want to understant everything correctly.)

    Further on, the proof says that [itex]|x-c|<\frac{1}{2}|c|[/itex] implies [itex]|x|>\frac{1}{2}|c|[/itex], which is obvious, and, as a further consequence, we have [itex]|x\cdot c|>\frac{1}{2}|c|\cdot |c|=\frac{1}{2}|c|^2[/itex] (1). Now, the key part of the proof suggests that all above implies:

    [tex]|x-c|<\frac{1}{2}|c|\Rightarrow |f(x)-f(c)|=|\frac{x-c}{x\cdot c}|\leq \frac{2|x-c|}{|c|^2}[/tex]. (2)

    I'm not really sure where that comes from. Obviously the point was to write the implied term in terms of |x-c| somehow to be able to set up a relation between [itex]\delta[/itex] and [itex]\epsilon[/itex]. But where did that come from? Is it from the fact that (1) implies [itex]\frac{2|x\cdot c|}{|c|^2}> 1[/itex], so when we multiply [itex]|\frac{x-c}{x\cdot c}|[/itex] with something greater than 1, it must become greater or equal to [itex]|\frac{x-c}{x\cdot c}|[/itex] (equal in the case x=c)?

    The rest of the proof is clearer to me, but this parts bothers me somehow, and I feel it is important to (finally) start with completely understanding these proofs, otherwise, I feel I won't ever be able to cope with analysis at a satisfying level. Sorry if I'm boring, but I'd appreciate some confirmation/help/criticism on the presented part of the proof. Thanks in advance. :smile:
     
  8. Jan 5, 2007 #7

    Hurkyl

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    Splitting things in half is a time-honored tradition! :biggrin:

    Yes, you're right: the condition you wrote is just as good for the purposes of this proof. That's one of the nice things about analysis -- the gritty details like this often simply don't matter.

    (That this condition forces x to be nonzero was just an added bonus that makes the proof a little bit more slick)


    That sounds right, I think. This part is just an exercise in algebraic machinery. Personally, I would have thought of it more like dividing

    |x-c|/|c| = |x-c|/|c|

    by the inequality

    |x| > (1/2) |c|.

    Actually, I would have thought "Okay, I'm replacing |x| with something smaller, which makes the fraction bigger".
     
    Last edited: Jan 5, 2007
  9. Jan 5, 2007 #8

    radou

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    Hurkyl, thanks. I don't know if it's just me, but there is something confusing about these 'delta-epsilon' proofs. It sometimes seems like this requires a lot of experience, i.e. to be able to 'set things up' the right way. The problem is, I can read a proof, and understand it to the last detail (probably :rofl: ), but I won't be able to construct my own proof to another problem. I guess one just has to exercise, exercise and exercise.
     
  10. Jan 5, 2007 #9

    Hurkyl

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    Well, of course. Experience is always helpful in learning a subject. :smile: That's why homework exists!

    Algebraically, the core idea in this proof is that in

    |x - c| / |xc|,

    the "problem" is that 1/|x| term. But, if c is nonzero, and x is "near" c, then |x| can't be too small. And so, 1/|x| can't be too big, so we can get rid of it by replacing it with a (larger) constant.
     
    Last edited: Jan 5, 2007
  11. Jan 6, 2007 #10
    The delta/epsilon approach is fundamental to the definition if limits as well, but using limits is much easier :P
     
  12. Jan 7, 2007 #11

    HallsofIvy

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    Please tell us what your textbook's definition of continuity is!
     
  13. Jan 7, 2007 #12

    radou

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    Yes it is, if you assume one knows the basic properties of limits. :wink:

    I have written the definition in some post above, and I am aware that the definition is equivalent to the definition of continuity which states that a function is continuous at some point c if it is defined at that point and if the limit of the function, as x --> c is f(c). :smile:
     
  14. Jan 7, 2007 #13

    matt grime

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    Experience tells us that we want to avoid any argument that explicitly uses eps and delta if possible. Thus we spend a small amount of time showing continuous if and only if the inverse image of an open set is open, and we have a much easier time. Or that the definition is equivalent to sequential continuity, whence the result again follows trivially. (f is sequentially continuous at c if for any sequence a_n converging to c f(a_n) converges to f(c). Given this I would suggest you look up the proof that if a_n converges to c, and nothing is zero, then 1/a_n converges to 1/c.
     
  15. Jan 7, 2007 #14

    radou

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    Thanks, I'll try to investigate these definitions too.
     
  16. Jan 7, 2007 #15

    matt grime

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    Has anyone ever told you how to solve all* analysis questions?

    Fix an x and d. Suppose that |x-y|<d. What can we say about |1/x - 1/y|? Well, it is equal to |y-x|/|xy|. Now, given e, how do we choose d so that this is less than e? Or better, how can we show that this can be made arbitrarily small (which is not quite the same thing)? Now, it is actually quite straightforward from here - there's a delta on top, so all we need to do is get to grips with the bottom, which we can do. d will have to depend on x, but you'd guessed that already from looking at the graph, right?




    * OK, I admit, not quite 'all', but you know what I mean.
     
  17. Jan 7, 2007 #16
    Using point-set topology: A basis element of R is an open interval (in the standard topology of R), and the inverse image of an open interval (that doesn't contain 0) is another open interval. i.e. f((a,b))=(1/a,1/b) is open in R. Thus f is continuous.
     
    Last edited: Jan 7, 2007
  18. Jan 7, 2007 #17

    radou

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    Is it, at any place on this forum, explicitly stated that a homework helper has to be good in all areas of mathematics/physics/engineering?
     
  19. Jan 7, 2007 #18
    Never mind what I said. You are into physics and I forgot that this forum is for physics, engineering, and math.
     
  20. Jan 7, 2007 #19

    HallsofIvy

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    Actually, that is the "limit definition" except that instead of using the word "limit" they write the definition of limit into the definition of "continuous".
     
  21. Jan 7, 2007 #20

    radou

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    Isn't another requirement in the limit definition given with |x-c|>0 ?
     
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