Formulate Precise Definition of a Limit

Discussion in 'Calculus' started by nycmathguy, May 30, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.6

    We end Section 6 here.

    Screenshot_20220530-090748_Samsung Notes.jpg
     
    nycmathguy, May 30, 2022
    #1
  2. nycmathguy

    MathLover1

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    formulate precise definition of lim(f(x)=-∞ then use definition to prove that lim(1+x^3)=-∞

    upload_2022-5-30_13-41-29.png
    Definition: Infinite Limit at Infinity (Formal)

    We say a function f has an infinite limit at infinity and write lim(f(x)=infinity as x-> ∞
    if for all M>0, there exists an N>0 such that f(x)>M for all x>N (see Figure).

    We say a function has a negative infinite limit at infinity and write
    limx→∞f(x)=−∞

    if for all M<0 , there exists an N>0 such that f(x)<M for all x>N .

    Similarly we can define limits as x→−∞.


    prove that lim(1+x^3)=-∞

    Apply Infinity Property:
    lim (x->-∞, (ax^n+..... +bx+c )=-∞ , a>0, n is

    in your case a=1 and n=3

    proof:

    Given ϵ >0, we need δ >0 such that if 0<| 1+x|<δ, then |1+x^3|<ϵ.

    Now,
    |1+x^3|=|(x + 1) (x^2 - x + 1)|

    If |x+1|<1, that is, −1<x+1<1, then note that
    −1<x+1<1 <=> -2<x<0 <=> x^2-x+1<0=> 0^2-0+1=1
    and so

    |1+x^3|=|x+1|(x^2-x+1)<1|x+1|
    |1+x^3|=|x+1|(x^2-x+1)<|x+1|

    So if we take δ=min(1,ϵ*1), then

    0<|x+1|<δ <=> |1+x^3|=|x+1|(x^2-x+1)

    since ϵ = -∞ in your case we proved that
    lim(1+x^3)=-∞ as x->-∞
     
    MathLover1, May 30, 2022
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    OMG!!! This is a monster problem. Thank you.
     
    nycmathguy, May 31, 2022
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