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Math Help - 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 \epsilon - \delta definition of a limit that \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 \delta

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

    Proof.

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

    From the \epsilon - \delta definition of a limit, \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 \epsilon - \delta definition of a limit that \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 \delta

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

    Proof.

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

    From the \epsilon - \delta definition of a limit, \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 \epsilon - \delta definition of a limit that \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 \delta

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

    Proof.

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

    From the \epsilon - \delta definition of a limit, \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 |x - 4| < \delta, means you are constructing the limit as x \to 4 not 3.

    RonL
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