Find the Limit...6

Discussion in 'Calculus' started by nycmathguy, Apr 10, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 1.6

    Screenshot_20220408-184703_Samsung Notes.jpg

    IMG_20220410_104426.jpg

    IMG_20220410_104441.jpg

    IMG_20220410_104453.jpg
     
    nycmathguy, Apr 10, 2022
    #1
  2. nycmathguy

    MathLover1

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    18 and 20 are correct

    22. correction:

    lim((sqrt(4x+1)-3)/(x-2),x->2)

    rationalize numerator:
    ((sqrt(4x+1)-3)(sqrt(4x+1)+3))/((x-2)(sqrt(4x+1)-3))

    =((4x+1)-9)/((x-2)(sqrt(4x+1)+3))

    =(4x-8)/((x-2)(sqrt(4x+1)+3))

    =(4(x-2))/((x-2)(sqrt(4x+1)+3)).......simplify

    = 4/(sqrt(4x+1)+3)

    then

    lim(4/(sqrt(4x+1)+3),x->2)=4/(sqrt(4*2+1)+3)=4/(sqrt(9)+3)=4/(3+3)=4/6=2/3
     
    MathLover1, Apr 10, 2022
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    I see my error. The denominator should include positive 3 not negative 3.
     
    nycmathguy, Apr 11, 2022
    #3
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