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Math Help - Proof - Limit of a function

  1. #1
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    Proof - Limit of a function

    Suppose that f(x) = mx + b with m > 0. Let c be a real number. Prove that
    lim x-> c (f(x)) = mc + b.
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    I know that |mx + b -(mc + b)| < \epsilon. So |mx + b -mc -b| < \epsilon. Then |mx -mc| < \epsilon. This can be factored as |m||x-c|<\epsilon. But I am not sure where to go from h ere.
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  4. #4
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    All that looks good so far. In looking at the definition of limit, you've got

    \displaystyle{\lim_{x\to c}f(x)=L} if for every \epsilon>0 there exists a \delta>0 such that if 0<|x-c|<\delta, then |f(x)-L|<\epsilon.

    Overall, your goal is to find a \delta(\epsilon) such that these conditions are satisfied. So, what do you want \delta to be?
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  5. #5
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    Ah, so if I let \delta = \epsilon/m, then |m(x-c)|<\epsilon which is what I am trying to prove. Right? Thank you!
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  6. #6
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    Close. Note that \delta>0. So that requires what modification?
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  7. #7
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    \epsilon/m > 0.
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  8. #8
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    Oh, you're right. The problem assumed positive slopes. I was led astray by your unnecessary absolute values in post # 3. If the problem allowed negative slopes, you could just slap on absolute values and you'd be good to go.

    So... a final proof looks like what?
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  9. #9
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    Let \epsilon > 0. Choose \delta = \epsilon/m. Now if 0<|x-c|<\delta, then -\delta < x-c < \delta.
    -\epsilon/m < x-c < \epsilon
    -\epsilon < m(x-c) < \epsilon
    |m(x-c)| < \epsilon
    |mx - mc| < \epsilon
    |mx + b - mc - b| < \epsilon
    |mx + b - (mc + b)| < \epsilon
    ==> Lim x->c (mx + b) = mc + b.
    Q.E.D.

    I probably have too many of my initial calculations in there, but I want to make sure it flows from start to finish.
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  10. #10
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    Looks good to me. Your initial calculations in post # 3 are the ones that you have to reverse in the actual proof. So this is a very typical delta-epsilon proof.

    I'd say you're done!
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