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Thread: Why is this limit proof wrong?

  1. #1
    Super Member Deadstar's Avatar
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    Why is this limit proof wrong?

    This may be a stupid question thats gonna easily be proved but i just wondered at which step is this wrong. This is taken from a proof in my uni class that limit x->3 (7x) = 21 but just changing one number. why is it wrong and where does it go wrong?


    Prove using the $\displaystyle \epsilon - \delta$ definition of a limit that $\displaystyle \lim_{x\to 3} 7x = 28$ (was originally 21 of course)

    Rough Work:

    |f(x) - L| = |7x - 28| = 7|x - 4|

    |x - p| we can control with $\displaystyle \delta$

    If $\displaystyle |x - 4| < \delta$, then $\displaystyle |f(x) - L| < 7|x - 4| < 7\delta$

    Proof.

    Let $\displaystyle \epsilon > 0$ be given. Choose $\displaystyle \delta = \frac{\epsilon}{7}$, then if $\displaystyle 0 < |x - 4| < \delta$ => $\displaystyle |f(x) - L| < 7\delta = 7\frac{\epsilon}{7} = \epsilon$

    From the $\displaystyle \epsilon - \delta$ definition of a limit, $\displaystyle \lim_{x\to 3} 7x = 28$

    Can somebody point out the mistake in that? surely there must be some step thats wrong!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Deadstar View Post
    This may be a stupid question thats gonna easily be proved but i just wondered at which step is this wrong. This is taken from a proof in my uni class that limit x->3 (7x) = 21 but just changing one number. why is it wrong and where does it go wrong?


    Prove using the $\displaystyle \epsilon - \delta$ definition of a limit that $\displaystyle \lim_{x\to 3} 7x = 28$ (was originally 21 of course)

    Rough Work:

    |f(x) - L| = |7x - 28| = 7|x - 4|

    |x - p| we can control with $\displaystyle \delta$

    If $\displaystyle |x - 4| < \delta$, then $\displaystyle |f(x) - L| < 7|x - 4| < 7\delta$

    Proof.

    Let $\displaystyle \epsilon > 0$ be given. Choose $\displaystyle \delta = \frac{\epsilon}{7}$, then if $\displaystyle 0 < |x - 4| < \delta$ => $\displaystyle |f(x) - L| < 7\delta = 7\frac{\epsilon}{7} = \epsilon$

    From the $\displaystyle \epsilon - \delta$ definition of a limit, $\displaystyle \lim_{x\to 3} 7x = 28$

    Can somebody point out the mistake in that? surely there must be some step thats wrong!
    if there was no typo error, then it is really wrong..
    because x approcahes 3, not 4.. and if it approaches 3, then the limit cannot be 28..
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Deadstar View Post
    This may be a stupid question thats gonna easily be proved but i just wondered at which step is this wrong. This is taken from a proof in my uni class that limit x->3 (7x) = 21 but just changing one number. why is it wrong and where does it go wrong?


    Prove using the $\displaystyle \epsilon - \delta$ definition of a limit that $\displaystyle \lim_{x\to 3} 7x = 28$ (was originally 21 of course)

    Rough Work:

    |f(x) - L| = |7x - 28| = 7|x - 4|

    |x - p| we can control with $\displaystyle \delta$

    If $\displaystyle |x - 4| < \delta$, then $\displaystyle |f(x) - L| < 7|x - 4| < 7\delta$

    Proof.

    Let $\displaystyle \epsilon > 0$ be given. Choose $\displaystyle \delta = \frac{\epsilon}{7}$, then if $\displaystyle 0 < |x - 4| < \delta$ => $\displaystyle |f(x) - L| < 7\delta = 7\frac{\epsilon}{7} = \epsilon$

    From the $\displaystyle \epsilon - \delta$ definition of a limit, $\displaystyle \lim_{x\to 3} 7x = 28$

    Can somebody point out the mistake in that? surely there must be some step thats wrong!
    Where you take $\displaystyle |x - 4| < \delta$, means you are constructing the limit as $\displaystyle x \to 4$ not $\displaystyle 3$.

    RonL
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