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Showing the continutity and differnetiability of a function

  1. Jun 5, 2013 #1
    Let f: R-->R be defined by f(x) = x(sin(1/x)) if x ≠ 0, f(0) = 0

    Show that f is continuous at every point and diffeentiable at every point except x = 0

    Attempt:

    So I must be having a brain freeze, but to show continuity of a function without going through the delta-epsilon method which I don't think they want, I should just take the limit of an arbitrary number "a" and then:

    lim x-->a f(x) = a(sin(1/a), but this doesn't feel like that is all that I'm doing.

    Now for differentiability, I'm suppose to use the definition:

    if there exists a value "m" and E(h) such that:

    f(a+h) = f(a) + mh + E(h) then the function is differentiable.

    Using that am I suppose to just solve for m and take the limit as h-->0:

    m = [(a+h)(sin(1/(a+h)) - a(sin(1/a))]/ h ?

    If that's the case then I'm stuck on solving for "m"
     
  2. jcsd
  3. Jun 5, 2013 #2

    Dick

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    You know that function is continuous and differentiable at every point except x=0, just because the parts that make it up are continuous and differentiable there. The ONLY point you should be worrying about is x=0. I strongly suggest you sketch a graph.
     
  4. Jun 5, 2013 #3
    But the question asked me to show continuity and differnetiability everywhere except at 0. Am I suppose to use a delta-epsilon proof to show continuity then? I don't see how a graph can help me.
     
  5. Jun 5, 2013 #4

    Dick

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    Compute the derivative using calculus. You'll get something that exists everywhere except at x=0. That it's differentiable automatically shows it's continuous. Trust me, that's enough. Or use theorems that tell you if f(x) and g(x) are continuous and differentiable so are f(x)*g(x) and f(g(x)) if you want to be more formal. All of the parts of your function are perfectly well behaved away from x=0. The interesting place is x=0. Concentrate your efforts with epsilon/deltas and difference quotients there. The graph will help you figure out what going on at x=0.
     
  6. Jun 5, 2013 #5


    Well taking the derivative in the ordinary fashion I get:

    sin(1/x) - (1/x)(cos(1/x) which makes sense as being continuous every where except at 0.

    Now at x = 0 I know that this function is the sine wave curve (or whatever the name is) but it oscillates as it approaches 0 and never actually reaches 0.

    Now I could say that in words....the issue now is actually translating that into mathematical language....

    Looking at the delta-epsilon relation for continutity was a nightmare that I couldn't figure out, but I think I may have something for differentiability with the difference quotient:

    using the form: m = lim h-->0 [f(a+h) - f(a)]/h - E(h)/h

    I get: [(a+h) sin(1/(a+h)) - a(sin(1/a))]/h - E(h)/h

    now taking the limit as h-->0 I know that x(sin(1/x) --> 0 in general. So that would mean my numerator goes to 0 as well, but I am not sure if the numerator goes to 0 faster than h....
     
  7. Jun 5, 2013 #6

    Dick

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    Do continuity first. If it's not continuous it's not differentiable. sin(1/x) is BOUNDED and x approaches 0. Use the sandwich theorem. Then go back to differentiable. If you put a=0 into the difference quotient (f(a+h)-f(a))/h remember f(0) is DEFINED to be 0. You have to show the limit as h->0 does not exist.
     
  8. Jun 5, 2013 #7
    I think I got it:

    It is continuous because sin(1/x) is bounded as mentioned, I can't fully remember the bound, but I believe it is bounded by |1|. So with that being the case:

    x (sin(1/x)) ≤ |x| now taking the limit of |x| as x-->0 will yield 0. Therefore the function is continuous.

    Now to differentiability: putting in a = 0 as you suggested yields:

    f(h)/h - E(h)/h = h(sin(1/h)/h - E(h)/h = sin(1/h) which taking the limit as h-->0 D.N.E
     
  9. Jun 5, 2013 #8

    Dick

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    Yes, you've pretty much got it. Now clean it up a little, -1<=sin(1/x)<=1 is the bound you want for continuity and the difference quotient is just f(h)/h=sin(1/h) as you said, but I'm not sure what this E(h) you are throwing in is. And maybe a few words about why sin(1/h) doesn't have a limit as h->0 wouldn't hurt.
     
  10. Jun 5, 2013 #9

    E(h)/h I'm trying to talk about is the error term. But that also goes to 0 so essentially it disappears.


    Now I have a more general question with regards to analyzing functions in this form. Is it safe to say that the only times I need to use the fundamental definition of continuity and differentiability is when I am analyzing a "troubled point" in the function? Everywhere else I could just use the simple rules of calculus learned from first year?
     
  11. Jun 5, 2013 #10

    Dick

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    Yes, it is. Like I said in post 4, you should have theorems telling you things like the product or quotient of differentiable functions is differentiable, and the composition of differentiable functions is differentiable. You don't have to go back to the basics to prove it at every point. When something breaks down at a "troubled point", like x=0 here, that's when you need to go back to the definitions.
     
  12. Jun 5, 2013 #11


    Ah. Now it's starting to make sense of when to use what tools. Thanks. Really took a load off of my head.
     
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