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Is this function continuous?

  1. Jun 10, 2005 #1
    Is this function continuous? Edit: Fixed - function should load now

    f\left( x \right) = e^{\left[ x \right]}

    Where the argument of the exponential is the greatest integer less than or equal to x.

    For the function to be continuous at a point x = a we need [tex]\mathop {\lim }\limits_{x \to a} f(x) = f(a)[/tex]. For this particular function, f(x) at x = a is just f(a) where a is an integer? But what about the limit? As far as I can see this function is like a sequence so that if I looked at the graph I would just see some dots. Is it possible to take any limits with this function? For example, can I actually take lim(x->3)f(x) and get a finite value? Further, could I take lim(x->2.5)f(x) for this particular function. Any help appreciated.

    Edit: Fixed f(x)...it should look right now.
    Last edited: Jun 10, 2005
  2. jcsd
  3. Jun 10, 2005 #2
    Take a sequence x_n in (0, 1) with x_n -> 1. If f was continuous (at 1), then f(x_n) -> f(1) = e, but [x_n] = 0, i.e. f(x_n) = e^0 = 1 -> 1. Contradiction.

    Surely a similar argument can show that f is discontinuous everywhere.
    Last edited: Jun 10, 2005
  4. Jun 10, 2005 #3


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    The graph isn't "dots" it's a step function. You know what the graph of g(x)=[x] looks like? Similar thing. For limits (and continuity) can you answer these questions about g(x)=[x]? (consider left and right handed limits seperately at integers)
  5. Jun 10, 2005 #4
    let [tex]X\subseteq\Re[/tex], [tex]f: X\longrightarrow\Re[/tex] and [tex]x_{0}\in X[/tex] where X is the domain of the function, then f is continuous at [tex]x_{0}[/tex] if [tex]\forall x_{n}\in X[/tex],[tex]x_{n}\longrightarrow x_{0}[/tex] and [tex]f(x_{n})\longrightarrow f(x_{0})[/tex], you will see that if you choose a sequence on the interval [1,2] that converges to the intiger 2 for example [tex]{x_{n}}={2-\frac{1}{n}[/tex] which gives [tex]f(x_{n})\longrightarrow f(2)[/tex], but a sequence on the interval [2,3] which converges to 2 for example [tex]{x_{n}}={2+\frac{1}{n}[/tex] gives [tex]f(x_{n})\neq>f(2)[/tex]

    does anybody know how to show something does not approach something else on latex?
    Last edited: Jun 10, 2005
  6. Jun 10, 2005 #5


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    [tex]\lim_{a \rightarrow b} f(a) \neq c[/tex]

    Last edited: Jun 10, 2005
  7. Jun 10, 2005 #6


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    a more interesting function would be e^(2pi)i[x]
  8. Jun 10, 2005 #7
    Thanks for the help guys.
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