Prove Limit Using Epsilon-delta Definition...1

Discussion in 'Calculus' started by nycmathguy, May 7, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.4

    I don't need to know this stuff to succeed in Calculus. However, I would like to have your math work reply file away as future study notes and/or reference notes. How about working out 22?

    Screenshot_20220507-174246_Samsung Notes.jpg
     
    nycmathguy, May 7, 2022
    #1
  2. nycmathguy

    MathLover1

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    22.

    upload_2022-5-9_18-2-21.gif

    We need to show that for every positive ε, there is a positive δ such that if

    0<|x-(-1.5)| < δ we get |(9-4x^2)/(3+2x) -6|<δ

    Note first that |x-(-1.5)| =|x+1.5|


    For every x other than −1.5, we have:

    |(9-4x^2)/(3+2x) -6|= |(3-2x) -6| ( you get that when you simplify (9-4x^2)/(3+2x) =3-2x, then subtract -6 which is 3-2x-6=-3-2x)


    then |-3-2x|=| -4|(x+1.5) | =|-4| *|(x+1.5) | =4||(x+1.5) |

    We want this to be less than ε and we control, through δ, the size of |x+1.5|

    if we make |x+1.5| <ε /4, then we will have

    4||(x+1.5) | < 4(ε /4)

    4||(x+1.5) | < ε as desired


    Now we are ready to write the proof:

    Given ε>0, let δ=ε/4. Observe that this δ is also positive, as required.

    Now if x is chosen so that 0<|x+1.5|<δ, we will have:

    |(9-4x^2)/(3+2x) -6| =4||(x+1.5) |

    4||(x+1.5) | <4δ

    then 4δ=ε

    that is:

    if 0< |(x+1.5) |< δ, then |(9-4x^2)/(3+2x) -6| <ε

    so, by the definition of lim, upload_2022-5-9_18-2-48.gif
     
    MathLover1, May 10, 2022
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Wish I could do this just like you. Why am I struggling with this stuff? Thank God it's not required for me to learn calculus.

    See link below for the last one today.

    Use Graph to Find Delta...2
     
    nycmathguy, May 10, 2022
    #3
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