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Math Help - limit question

  1. #1
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    limit question

    i would think the lim as x-->0 of f(x)/f(x) would be 1 regardless of what f(x) is. i am looking at a solution in the back of a calculus book and finding a solution of "limit doesn't exist." the function f(x) is sin(1/x). am i missing something?
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  2. #2
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    In considering the function f(x) = \sin \left( {\frac{1}{x}} \right), in any neighborhood of 0 f(x) has infinitely many zeros.
    Therefore, because \frac{0}{0} is undefined so is \frac{{f(\lambda )}}{{f(\lambda )}} where \lambda is a zero of f(x).

    Because there are infinitely many places where \frac{{f(x )}}{{f(x )}} is undefined is any neighborhood of 0, \lim _{x \to 0} \frac{{f(x)}}{{f(x)}} does not exist.
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  3. #3
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    To speak of the limit,
    \lim_{x\to 0} g(x)

    We need to be sure that g(x) is defined on some open interval containing 0 except possibly at 0 it self.

    So if f(x) = \sin \frac{1}{x}.
    And you define,
    g(x) = \frac{f(x)}{f(x)}
    You cannot talk about the limit,
    \lim_{x\to 0}g(x)
    Because as Plato says g(x) is not defined in some "neighborhood".
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  4. #4
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    thanks!
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    g(x) = \frac{f(x)}{f(x)}
    You cannot talk about the limit,
    \lim_{x\to 0}g(x)
    Because as Plato says g(x) is not defined in some "neighborhood".
    Please note: g(x) is not defined at some point in any neighborhood of 0.
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  6. #6
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    noted, thanks!
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