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Using the definition of continuity prove a function is continuous.?

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Using the definition of continuity, prove that the function f(x) = sin x
    is continuous.

    Hint: sin a − sin b = 2 sin (a-b)/2 . cos (a+b)/2


    2. Relevant equations


    3. The attempt at a solution

    Using the idea that:

    |sin(x)| ≤ |x|

    |cos(x)| ≤ 1

    along with the hint: sin a − sin b = 2 sin (a-b)/2 . cos (a+b)/2

    We can write:

    |sin(a) - sin(b)| ≤ |a - b|

    I'm assume we could maybe use this to show how f(x) = sin x is continuous..

    What do you think?

    Regards as always
    Tam
     
    Last edited: Nov 4, 2011
  2. jcsd
  3. Nov 4, 2011 #2

    Deveno

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    well you have to use the definition of continuity somewhere, eh?

    let "a" be the variable x, and let "b" be the point x0 where you are testing for continuity.

    given ε > 0, how can you choose a δ > 0 so that:

    |x - x0| < δ will force |sin(x) - sin(x0)| < ε?

    (use the inequalities you have to write a relationship between:

    |sin(x) - sin(x0)| and |x - x0|).
     
  4. Nov 4, 2011 #3
    I understand what you mean but unsure of what to do...

    Could you take δ = 1 .... or could you take it as δ≥ε ?

    Attempt (1):

    Hence, |sin(x) - sin(x0)| ≤ ε ≤ |x - x0| ≤ δ :/

    Attempt (2)

    Hence, since |sin(a) -sin(b)| ≤ |a-b| then
    |sin(x) - sin(x0)| ≤ |x - x0|

    Hence continuity...

    Am I missing something?

    Regards
     
  5. Nov 4, 2011 #4

    Deveno

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    here's the rules: i pick epsilon, you pick delta.

    you have to convince me, your delta works for my epsilon (and you don't know what my epsilon is).

    the only "help" you get, is that you get to assume that x and x0 are closer together than δ, and you want to leverage that into showing sin(x) and sin(x0) are also close together.

    your "attempt one" is just nothing. you can't start a proof by just claiming what you want to prove is true. otherwise, i'd have proven myself wealthy by now.

    this is what you have: you know that you control delta.

    you also know, that |sin(x) - sin(x0)| is less than "something".

    (what is that something? well, you're the one who first stated it, don't you know?).

    perhaps if that something was less than (some expression in epsilon), that would mean that perhaps the absolute value of the differences of the two sines was less than that something, and thus less than (some expression in epsilon).

    think.

    you want to make the difference of the sines small. is there something that 'bounds" it, that also can be made small (yes, smaller than small, counts as small)?
     
  6. Nov 4, 2011 #5
    OK, here goes,

    so we could take δ = 1 assuming x and x0 are closer together than δ

    so What you are wanting is:

    |sin(x) - sin(x0)| ≤ |x-x0| ≤ (some expression of ε)

    Now,

    That 'expression of ε' is: 1 ?????

    Thanks for your help.. I'm getting there slowly!
     
  7. Nov 4, 2011 #6

    Deveno

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    is δ = 1 going to work for ANY ε > 0 i pick?

    what if my choice for ε was 1/1,000,000?

    surely π/2 - π/4 is less than 1 (which you picked for delta),

    but sin(π/2) - sin(π/4) = 1 - √2/2 = (2 - √2)/2, which is around 0.29, and certainly NOT less than 1/1,000,000.
     
  8. Nov 4, 2011 #7

    SammyS

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    Two facts that may help:
    1. |cos(θ)| ≤ 1 .

    2. For θ in radians, |sin(θ)| ≤ |θ| .​
     
  9. Nov 4, 2011 #8

    Deveno

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    indeed the thread starter listed this very inequalities in the first thread.

    i think what is going on, is that the thread starter does not have a clear idea of how epsilon-delta limit proofs work.

    because once he picks the right delta (and this isn't hard), he has all the machinery in place to get 'er done in 2 lines.
     
  10. Nov 5, 2011 #9
    Okay, I have read up on how the proofs work..

    Am I right in saying that if:

    0 < | x – c | < δ
    then
    | f(x) – L | < ε

    So basically,

    As we restrict x to being within δ units of c, then, as a result of that restriction, f(x) becomes restricted to being within ε units of L ??

    I am still struggling to grasp this though.


    Lets say I was to choose δ=5, then what this means is that x is within 5 units of x0 .. Is this right?

    I still don't see how you can relate ε and δ.. :(

    regards
     
  11. Nov 5, 2011 #10

    HallsofIvy

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    Since you are given that "sin(a)- sin(b)= 2 sin((a-b)/2)cos((a+b)/2)", you might want to start by looking at [itex]|sin(x)- sin(x_0)|= 2|sin(x-x_0)/2||cos((x+x_0)/2)|[/itex].

    Now, use what you said initially: |sin(A)|< |A| and |cos(A)|< 1.
     
  12. Nov 5, 2011 #11
    We can write:

    |sin(a) - sin(b)| ≤ |a - b|
     
  13. Nov 6, 2011 #12
    anyone?
     
  14. Nov 6, 2011 #13

    SammyS

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    Work through the problem "backwards".

    Start with |sin(x) - sin(x0)| < ε, for some ε, then figure out what δ has to be.

    (This is something one typically does on scratch paper, then reverses to get formal proof.)
     
  15. Nov 9, 2011 #14
    Recap: I forgot about this question.. oops..

    Right so Using the idea that:

    |sin(x)| ≤ |x|

    |cos(x)| ≤ 1

    along with the hint: sin a − sin b = 2 sin (a-b)/2 . cos (a+b)/2

    We can write:

    |sin(a) - sin(b)| ≤ |a - b|


    I now need to let b tend to a and compute the limit of |sin(a)-sin(b)|

    But im not sure how to do this?? Any help would be great

    regards
     
  16. Nov 9, 2011 #15

    Deveno

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    you're still not seeing how this works.

    ε is given. you have to find δ. since ε might be any positive real number, it's highly probable δ will wind up being some function of ε.

    the only functions for which ANY δ works are constant functions (because f(x) - f(x0) is always 0). that's not terribly helpful in the general case.

    i'll show you how the logic works on a different function, maybe you can see how to adapt it for your function.

    problem: prove f(x) = x2 is continuous for all x.

    we want to show that: [itex]\lim_{x \to a} = f(a)[/itex] for all a. what does this mean?

    for any ε > 0, there exists δ > 0 such that 0 < |x-a| < δ implies |f(x)-f(a)| < ε.

    so we want to choose δ to somehow force |f(x)-f(a)| < ε. let's start with where we want to go, and see if we can "work backwards" to find δ.

    |f(x) - f(a)| < ε...well f(x) = x2, so let's write this explicitly:

    |x2 - a2| < ε
    |x + a||x - a| < ε

    hmm...now we have a factor of |x-a|, which we know we can make small by choosing δ small. so we need to somehow get some upper bound for |x+a|.

    well, we know |x + a| ≤ |x| + |a|...if only we knew how far away from "a" x was...

    but we can restrict this however we like, for example, we could insist we'll take the lesser of whatever δ we find otherwise and 1. so if |x - a| < 1, then x is between a - 1 and a + 1.

    suppose a ≥ 0, then the largest |x| can be is |a + 1| = a + 1 = |a| + 1.

    on the other hand, if a < 0, then the largest |x| can be is |a - 1| = -a + 1 = |a| + 1.

    so no matter what a is, if x is "within 1" of a, |x| ≤ |a| + 1. so |x + a| ≤ |x| + |a| ≤ 2|a| + 1.

    so we know that for δ < 1, |x2- a2| ≤ (2|a| + 1)|x - a|.

    if we want (2|a| + 1)|x - a| < ε (and we do), then just a little algebra tells us:

    |x - a| < ε/(2|a| + 1) <---what we need for δ. ok, now we've done our "scratch work", and so we "write it forwards" covering our tracks:

    proof that f(x) = x2 is continuous:

    let a be any real number. let ε > 0. choose δ = min{1, ε/(2|a| + 1)}. let |x -a| < δ.

    then |x2 - a2| = |x + a|| x - a|

    ≤ (|x| + |a|)|x - a| < (2|a| + 1)|x - a| (since |x - a|< δ ≤ 1)

    < (2|a| + 1)(ε/(2|a| + 1)) (since |x - a| < δ ≤ ε/(2|a| + 1))

    = ε
     
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