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Math Help - Proving A Limit

  1. #1
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    Proving A Limit

    Is this proof correct?

    \lim_{x\to0}(4x+7)=7

    Proof:
    \mbox{Choose } \delta=\frac{\epsilon}{4} \mbox{THEN IF} \<br />
0<|x-0|<\frac{\epsilon}{4} \ \mbox{THEN} \<br />
|f(x)-7| < \epsilon
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mike9182 View Post
    Is this proof correct?

    \lim_{x\to0}(4x+7)=7

    Proof:
    \mbox{Choose } \delta=\frac{\epsilon}{4} \mbox{THEN IF} \<br />
0<|x-0|<\frac{\epsilon}{4} \ \mbox{THEN} \<br />
|f(x)-7| < \epsilon
    Yup, that is correct!

    I would state it this way, though (its up to you, but I would show a little more work that that):

    \forall\varepsilon>0,\,\exists\,\delta>0 : 0<\left|x-0\right|<\delta\implies\left|(4x+7)-7\right|<\varepsilon.

    However, \left|(4x+7)-7\right|=\left|4x\right|=4\left|x\right|<\varepsil  on\implies\left|x\right|<\frac{\varepsilon}{4}=\de  lta.

    Thus, take \delta = \frac{\varepsilon}{4}. Therefore, \lim_{x\to0}(4x+7)=7
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Yup, that is correct!

    I would state it this way, though (its up to you, but I would show a little more work that that):

    \forall\varepsilon>0,\,\exists\,\delta>0 : 0<\left|x-0\right|<\delta\implies\left|(4x+7)-7\right|<\varepsilon.

    However, \left|(4x+7)-7\right|=\left|4x\right|=4\left|x\right|<\varepsil  on\implies\left|x\right|<\frac{\varepsilon}{4}=\de  lta.

    Thus, take \delta = \frac{\varepsilon}{4}. Therefore, \lim_{x\to0}(4x+7)=7
    What part of this actually proves the limit? I follow everything, but I am unsure about the implications of how the last line proves that the limit exists. Is this because it has been shown that given any value of epsilon, delta will be less than one quarter of epsilon?
    Last edited by VonNemo19; July 31st 2009 at 06:43 PM.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    What part of this actually proves the limit?
    The proof lies in the fact that \exists\,\delta:0<\left|x-c\right|<\delta\implies\left|f\!\left(x\right)-L\right|<\varepsilon, and we found in the problem that \delta=\frac{\varepsilon}{4}>0 \backepsilon \varepsilon>0\,(\forall\,\varepsilon)

    To find out more about \varepsilon-\delta proofs, look at this thread: http://www.mathhelpforum.com/math-he...ta-proofs.html
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    The proof lies in the fact that \exists\,\delta:0<\left|x-c\right|<\delta\implies\left|f\!\left(x\right)-L\right|<\varepsilon, and we found in the problem that \delta=\frac{\varepsilon}{4}>0 \backepsilon \varepsilon>0\,(\forall\,\varepsilon)

    To find out more about \varepsilon-\delta proofs, look at this thread: http://www.mathhelpforum.com/math-he...ta-proofs.html
    I get it. Thanks man.
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