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Thread: Need help solving for delta part 2

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    Need help solving for delta part 2

    I tried to mimic what the person did on the previous question but still do not understand what I am supposed to do
    Find a δ for which the following statement is true. Start by stating the definition of the limit as applied to this question: lim x→3 x^2 + 5 = 14.
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    Re: Need help solving for delta part 2

    Quote Originally Posted by kidgt23 View Post
    I tried to mimic what the person did on the previous question but still do not understand what I am supposed to do
    Find a δ for which the following statement is true. Start by stating the definition of the limit as applied to this question: lim x→3 x^2 + 5 = 14.
    To get help, please show your complete work. Otherwise, we are just guessing as to what you do not understand.
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    Re: Need help solving for delta part 2

    so what I did was |x-2|>1 so that -1<x<-2 then 1<x<3
    |(x^2+5x)-4|=|x^2+5-1|=|x-2| |x+7|
    Then 1<x<3 so 4<x+7<10 Then
    |x+7|<10
    if epsilon>0 let Delta = min(1,Epsilon/10) So if |x-2|<Delta
    (|x^2+5|-4)=|x-2| |x+7|<Delta (10) < or equal to epsilon

    If this is wrong please show me the right way it is done
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    Re: Need help solving for delta part 2

    Quote Originally Posted by kidgt23 View Post
    so what I did was |x-2|>1 so that -1<x<-2 then 1<x<3
    |(x^2+5x)-4|=|x^2+5-1|=|x-2| |x+7|
    Then 1<x<3 so 4<x+7<10 Then
    |x+7|<10
    if epsilon>0 let Delta = min(1,Epsilon/10) So if |x-2|<Delta
    (|x^2+5|-4)=|x-2| |x+7|<Delta (10) < or equal to epsilon
    I tried to mimic what the person did on the previous question
    You did no such a thing. You did not start as I did.

    $x\to 3$ $x$ approaches three so bound that: $|x-3|<1\text{ or }2<x<4$ that is step one.

    Step two: we think the limit is 14 so write $|(x^2+5)-14|$, the function minus the limit.
    In the case of polynomials it should factor: $|x^3-9|={\color{red}{|x-3|}}|x+3|$ there is our control.

    $2<x<4\Rightarrow 5<x+3<7\Rightarrow |x+3|<7$ now you tell why all of that is done. Why +3??

    If $\varepsilon>0$ then let $\delta=\min\{1,\frac{\varepsilon}{7}\}$ that forces $|(x^2+5)-14|<\varepsilon$ WHY?
    Thanks from kidgt23
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