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Math Help - continuity

  1. #1
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    continuity

    let f(x) = x^(2) sin (1/x^2) for 0<x< or = 1 and f(0)=0. prove that f is continuous on [0,1].

    the answer said that since lx^(2) sin (1/x^2)l < or = l x^2 l, then f is continuous on 0 and hence continuous on [0,1]..but i do not see how this statement proves the question.

    i thought to prove continuous, it would be assume that f is cont at c where c is between [0,1]
    then
    l x^(2) sin (1/x^2) - c^(2) sin (1/c^2)l < l x^2 - c^2l < 2 l x-cl
    thus continuous?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    We have:

    0\leq |f(x)|\leq x^2\quad (\forall x\in [0,1])

    Applying the Sandwich Theorem:

    \displaystyle\lim_{x \to 0}|f(x)|=0

    Hence:

    \displaystyle\lim_{x \to 0}f(x)=0=f(0)


    Fernando Revilla
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    let f(x) = x^(2) sin (1/x^2) for 0<x< or = 1 and f(0)=0. prove that f is continuous on [0,1].

    the answer said that since lx^(2) sin (1/x^2)l < or = l x^2 l, then f is continuous on 0 and hence continuous on [0,1]..but i do not see how this statement proves the question.

    i thought to prove continuous, it would be assume that f is cont at c where c is between [0,1]
    then
    l x^(2) sin (1/x^2) - c^(2) sin (1/c^2)l < l x^2 - c^2l < 2 l x-cl
    thus continuous?
    The only x where f(x) may not be continuous is x=0 (due to the zero in the denominator).
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  4. #4
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    why does showing that the limit of the function = 0 imply that the function on continuous from 0 to 1?

    i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?

    thanks!
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    why does showing that the limit of the function = 0 imply that the function on continuous from 0 to 1?
    Implies that is continuous at 0.

    i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?
    I don't know why do you thought such a thing.


    Fernando Revilla
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  6. #6
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    Quote Originally Posted by alexandrabel90 View Post
    why does showing that the limit of the function = 0 imply that the function on continuous from 0 to 1?
    As you said before, "The only x where f(x) may not be continuous is x=0 (due to the zero in the denominator)." So in order to prove f is continuous on the interval [0, 1] it is only necessary to show that f is continuous at x= 0. And " \lim_{x\to 0}f(x)= f(0) is the definition of "continuous at x= 0".

    i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?
    Actually, showing [tex]\lim_{x\to a} f(x)= f(a)[/itex] only shows f is continuous at x= a, not on any interval at all. For example, the function f, defined by "f(x)= 1/n if x is a rational number, equal to m/n reduced to lowest terms, f(x)= 0 if x is irrational" is continuous only at x= 0, not on any interval around 0.

    But, again, as you said, continuity everywhere except x= 0 is obvious.

    thanks!
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