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Thread: Precise definition of limits

  1. #1
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    Precise definition of limits

    Does anyone knows how to show:

    limit of f(x), x approaches a, equals L

    if and only if

    limit of f(x), x approaches a from the right = to limit of f(x), x approaches a from the left = L

    Using the precise definition of limit.

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  2. #2
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    Quote Originally Posted by fruitcakelover View Post
    Does anyone knows how to show:

    limit of f(x), x approaches a, equals L

    if and only if

    limit of f(x), x approaches a from the right = to limit of f(x), x approaches a from the left = L

    Using the precise definition of limit.

    Let $\displaystyle f$ be defined on an open interval $\displaystyle I$ containing $\displaystyle a$ except possibly at $\displaystyle a$. With domain $\displaystyle D$.

    Let, $\displaystyle \lim_{x\to a}f(x) = L$.
    That means,
    $\displaystyle 0 < |x-a| < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

    We want to show,
    $\displaystyle \lim_{x\to a^+}f(x) = L \mbox{ and }\lim_{x\to a^-}f(x) = L$.
    Meaning,
    $\displaystyle 0 < a - x < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
    $\displaystyle 0 < x - a < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

    Do you see? Note that $\displaystyle |x-a|$ can either be $\displaystyle x-a \mbox{ or }a-x$ depending whether $\displaystyle x$ is larger than $\displaystyle a$ or not. But the point is that bpth of these quantieis $\displaystyle x-a, \ a-x$ are containted in the meaning of $\displaystyle 0<|x-a|< \delta$.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle f$ be defined on an open interval $\displaystyle I$ containing $\displaystyle a$ except possibly at $\displaystyle a$. With domain $\displaystyle D$.

    Let, $\displaystyle \lim_{x\to a}f(x) = L$.
    That means,
    $\displaystyle 0 < |x-a| < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

    We want to show,
    $\displaystyle \lim_{x\to a^+}f(x) = L \mbox{ and }\lim_{x\to a^-}f(x) = L$.
    Meaning,
    $\displaystyle 0 < a - x < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
    $\displaystyle 0 < x - a < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

    Do you see? Note that $\displaystyle |x-a|$ can either be $\displaystyle x-a \mbox{ or }a-x$ depending whether $\displaystyle x$ is larger than $\displaystyle a$ or not. But the point is that bpth of these quantieis $\displaystyle x-a, \ a-x$ are containted in the meaning of $\displaystyle 0<|x-a|< \delta$.
    you forgot the immortal words "let $\displaystyle \epsilon > 0$" ...i don't know why, but i love that line, i'll say it again, "let $\displaystyle \epsilon$ be greater than zero" ...saying it the long way is even nicer

    otherwise, good stuff!
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  4. #4
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    Thanks! I get it!
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