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Math Help - Delta/Epilson proof

  1. #1
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    Delta/Epilson proof

    Hello,

    After reading the thread's sticky on how to do DE proofs, I attempted one of my home work questions so could someone let me know if I did it right? I think I did but my confidence in my skills at this stage is low so I feel I need lots of reassurance

    Prove lim x^2-4x-12=9
    x->-3

    in my rough work I basically get it to factor to x^2-4x-21 which is |x-7||x-(-3)|
    delta equals epilson/|x-7| and I label |x-7| as M since Delta cannot be defined in terms of x
    x is -4<x<-2 if |x-(-3)|<1
    M is therefore -2
    and Delta becomes E/-2

    I won't write out the full proof but basically at the end I say my delta is min {1, E/-2} and that 0<|x-(-3)| <Delta implies that |x^2-4x-21|<Epsilon

    Thanks!
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  2. #2
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    Re: Delta/Epilson proof

    Quote Originally Posted by funnybabe View Post
    Hello,

    After reading the thread's sticky on how to do DE proofs, I attempted one of my home work questions so could someone let me know if I did it right? I think I did but my confidence in my skills at this stage is low so I feel I need lots of reassurance

    Prove lim x^2-4x-12=9
    x->-3

    in my rough work I basically get it to factor to x^2-4x-21 which is |x-7||x-(-3)|
    delta equals epilson/|x-7| and I label |x-7| as M since Delta cannot be defined in terms of x
    x is -4<x<-2 if |x-(-3)|<1
    You are almost correct to this point.
    Build your other factor: -11<x-7<-9.
    From which you see that |x-7|<11.
    So let \delta  = \min \left\{ {1,\frac{\varepsilon }{{11}}} \right\}.
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  3. #3
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    Re: Delta/Epilson proof

    so when you do the other factor -11<x-7<-9, how come you don't compute that in absolute value? I am assuming you are subbing in x=-4 and then when x=-2. if x=-2, the absolute value is 9 and wouldn't that be the minimum?
    btw, where do you get to use the delta and epsilon characters?
    thanks!
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  4. #4
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    Re: Delta/Epilson proof

    my other question is basically isn't the minimum -2? it sounds like you are saying the minimum is -4. just a little confused. then Delta=E/9?
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    Re: Delta/Epilson proof

    Quote Originally Posted by funnybabe View Post
    so when you do the other factor -11<x-7<-9, how come you don't compute that in absolute value? I am assuming you are subbing in x=-4 and then when x=-2. if x=-2, the absolute value is 9 and wouldn't that be the minimum?
    btw, where do you get to use the delta and epsilon characters?
    thanks!
    If -7<a<-3 then we know that absolutely |a|<7~. Do we not?

    If -4<a<6 then we know that |a|<6~.

    If -4<a<2 then we know that |a|<4~.

    So if -11<x-7<-9 then we know that |x-7|<11~.

    We are using the maximum bound.
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  6. #6
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    Re: Delta/Epilson proof

    I think I get it. Basically just because -4 is lower then -2, the absolute value of -4 is greater then the absolute value of -2 so go with the maximum value of the absolute. is this correct?
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    Re: Delta/Epilson proof

    Quote Originally Posted by funnybabe View Post
    I think I get it. Basically just because -4 is lower then -2, the absolute value of -4 is greater then the absolute value of -2 so go with the maximum value of the absolute. is this correct?
    Given that a<x<b then |x|<\max\{|a|,|b|\}.
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