Delta-Epsilon Definition of a Limit...5

Discussion in 'Calculus' started by nycmathguy, Apr 24, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 1.7

    How about 24? Please, show every step along the way. Screenshot_20220422-230705_Samsung Notes.jpg
     
    nycmathguy, Apr 24, 2022
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  2. nycmathguy

    MathLover1

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    24.

    Theorem: For any constant c and any real number a, upload_2022-4-24_14-56-57.gif .

    Proof:

    Let ϵ > 0 be given, now we need to choose a δ > 0 such that for any x, that satisfies |x − a| < δ, the condition
    |f(x) − c| < ϵ is satisfied.

    Well this one is an easy one to prove because f(x) = c for any x.

    So |f(x) − c| = |c − c| = 0.

    So there is no game here at all, because for any given ϵ > 0, it doesn’t matter what δ I choose, for any x, in any (a−δ, a+δ) interval corresponding |f(x) − c| = |c − c| = 0 < ϵ.
    So without a difficulties in this case is proved that upload_2022-4-24_15-0-0.gif .
     
    MathLover1, Apr 24, 2022
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    This is simply amazing work. Great study notes.
     
    nycmathguy, Apr 24, 2022
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