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How to prove there is no limit of cos1/x using theorm of limit

  1. Oct 12, 2004 #1
    How to prove there is no limit(x->0) of cos1/x using theorm of limit?
    Anybody can give me some hints?
    thanks
     
    Last edited: Oct 12, 2004
  2. jcsd
  3. Oct 12, 2004 #2
    I'm just learning this myself, but here is how I would approach it.

    [itex]\lim_{x\to 0}f(x)=\cos(\frac{1}{x})[/itex]
    You know what happens to [itex]\lim_{x\to 0}f(x) = \frac{1}{x}[/itex]

    You also know what a graf with [itex]f(x) = \cos(x)[/itex] looks like.

    Now consider

    [itex]y = \frac{1}{x}[/itex]

    and

    [itex]\lim_{y\to\infty}f(y)=\cos(y)[/itex]
     
    Last edited: Oct 12, 2004
  4. Oct 12, 2004 #3
    thanx

    Choose an arbitrary number called "L". (Think of it as between -1 and 1 if you like)
    Show that there exists a neighbourhood U of 0, so that given |L|<1, there will always be some x in U so that |L-cos(1/x)|>1/2
    but I don't know how to continue....

     
  5. Oct 12, 2004 #4
    my question is how to find x to makes |L-cos(1/x)|>1/2

    my question is how to find x to makes |L-cos(1/x)|>1/2
     
  6. Oct 13, 2004 #5

    matt grime

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    Use the easier, and equivalent, definition of limit.

    If you can find two sequences a(n) and b(n) tending to zero such that cos(a(n)) and cos(b(n)) tend to different numbers you're done.
     
  7. Oct 13, 2004 #6
    thanx,but I want to know if I can prove it by E-Delta therom? Hope U can Help me..

    thanx,but I want to know if I can prove it by E-Delta therom? Hope U can Help me..
     
  8. Oct 13, 2004 #7
    Take limit as x approaches 0 from the right (lim->0+) of cos(1/x). Using direct substitution, you would get cos(infinity). cos(infinity) does not approach a single value, because it is not a monotonic function over the required interval. So, no limit exists, because no single value is approached. If one side of a limit does not exist, the limit does not exist.

    This is the way I learned to do limits. Not sure if it applies here though.
     
  9. Oct 13, 2004 #8

    matt grime

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    To prove the counter example it suffices to show that given any d there is a e such that there is some x with |cos(x)-L| > e and |x|<d

    but this is trivial. firstly L must be between -1 and 1, and let x be some sufficiently large solution to cos(x)=1 or -1, then one of |1-L| and |-1-L| must be greater than 1 (which we can choose to be e)

    nb e:=epsilon, d:=delta
     
  10. Oct 13, 2004 #9
    thanx

    I got it now,thanx a lot
    Actually,I thought it should hold for all x ,my mistake!

     
  11. Oct 14, 2004 #10

    matt grime

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    In the negation of propositions never forget that 'for all' is changed to 'there exists'
     
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