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Thread: Continuity

  1. #1
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    Continuity

    For function $\displaystyle f$ to be continuous at $\displaystyle a$, three conditions must be satisfied:

    (1)$\displaystyle f$ is defined at $\displaystyle a$;
    (2) $\displaystyle \lim_{x \to a} f(x)$ exists;
    (3) $\displaystyle \lim_{x \to a} f(x)=f(a) $ exists;

    Now say a function $\displaystyle g$ is defined as

    $\displaystyle g(x) = \left \{\begin{array}{cc}0,&\mbox{if} x\not \in \mathbb{Z}\\1,&\mbox{if} x\in \mathbb{Z} \end{array}\right$.

    Can we say $\displaystyle g$ is continuous $\displaystyle a$ where $\displaystyle a=2$ and $\displaystyle a \in \mathbb{Z}$?

    Does $\displaystyle g$ satisfy the following?

    (1)$\displaystyle f$ is defined at $\displaystyle a$ where $\displaystyle f(2)=1$;
    (2) $\displaystyle \lim_{x \to a} f(x)$ exists;
    (3) $\displaystyle \lim_{x \to a} f(x)=f(a)=0 $ exists

    ?
    Last edited by novice; Aug 2nd 2010 at 07:47 PM.
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  2. #2
    Senior Member eumyang's Avatar
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    Quote Originally Posted by novice View Post
    Now say a function $\displaystyle g$ is defined as

    $\displaystyle g(x) = \left \{\begin{array}{cc}0,&\mbox{if} x\not \in \mathbb{Z}\\1,&\mbox{if} x\in \mathbb{Z} \end{array}\right$.

    Can we say $\displaystyle g$ is continuous at $\displaystyle a$ where $\displaystyle a=2$ and $\displaystyle a \in \mathbb{Z}$?

    Does $\displaystyle g$ satisfies the following?
    (1)$\displaystyle g(a)$ is defined.
    Yes, $\displaystyle g(2)=1$.

    (2) $\displaystyle \lim_{x \to a} g(x)$ exists;
    Yes, $\displaystyle \lim_{x \to 2} g(x) = 0$.

    (3) $\displaystyle \lim_{x \to a} g(x)=g(a)$
    No. We have in (1) that $\displaystyle g(2)=1$ and in (2) $\displaystyle \lim_{x \to 2} g(x) = 0$. They're not equal, so condition (3) isn't satisfied. So g is not continuous at x = 2.
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  3. #3
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    Now, change point $\displaystyle a$ to $\displaystyle a \not \in \mathbb{Z}$.

    Then $\displaystyle g(x) $ is contiunous at $\displaystyle x= a, \forall a \not \in \mathbb{Z}$.

    Yah?
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  4. #4
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    Yes indeed.
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