Prove Limit Exist For 3 Statements

Discussion in 'Calculus' started by nycmathguy, May 12, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.4

    We end Section 2.4 with this problem.

    Screenshot_20220510-193956_Samsung Notes.jpg
     
    nycmathguy, May 12, 2022
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  2. nycmathguy

    MathLover1

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    44.

    suppose that lim(f(x))=∞ as x->a and lim(g(x)=c as x->a, where c is real number. prove each statement

    a. lim[f(x) +g(x)]=∞ as x->a

    Let M element of R. Then, we need to show that:

    all δ element of R: 0<|x−c|<δ => f(x)+g(x) >M

    Since lim(f(x))=∞ as x→a, it follows that:

    0<|x−c|<δ
    |(f(x) -L)<L/2

    So, that means that:

    0<|x−c|<δ2 =>|f(x) |>L/2

    Define δ=min{δ1,δ2}. Then:

    0<|x−c|<δ =>f(x)+g(x)>2M/L*L/2
    0<|x−c|<δ =>f(x)+g(x)>M which proves that
    lim(f(x) +g(x))=∞as x->a

    so, lim[f(x) +g(x)]=∞ as x->a =infinity +c=∞


    b.
    lim[f(x) *g(x)]=∞ as x->a if c>0

    lim(f(x)*g(x), x→a
    =lim[[f(x)−K][g(x)−L]+Lf(x)+Kg(x)−KL], x→a
    =lim[f(x)−K][g(x)−L], x→a +lim(Lf(x)),x→a +lim(Kg(x)),x→a −lim(KL), x→a
    =0+lim(Lf(x)),x→a+lim(Kg(x)),x→a−lim(KL),x→a
    =LK+KL−KL
    =LK

    in your case L=∞ and K=c, => LK=∞*c=∞

    so, lim[f(x) *g(x)]=∞ as x->a if c>0

    c.
    lim[f(x) *g(x)]= -∞ as x->a if c<0

    all same, just c<0
    in your case L=∞ and K=-c, => LK=∞*-c=-∞

     
    MathLover1, May 12, 2022
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  3. nycmathguy

    nycmathguy

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    Great as usual. Keep in mind that the problems posted are those I didn't quite understand in the textbook. I also don't have sufficient time during the week to show my math (right or wrong) as it may be. It's not laziness. Trust me, the overnight shift is not for everybody.
     
    nycmathguy, May 12, 2022
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