Evaluate the Limit

Discussion in 'Calculus' started by nycmathguy, May 4, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.3

    Note: The following question is not covered by Jenn in any of her limits video lessons. However, I am curious about how this one is done.

    I am interested in number 66. Clearly, I must rationalize the top or bottom as step 1, but which one? If the limit DNE, explain why.

    Screenshot_20220503-193143_Samsung Notes.jpg
     
    nycmathguy, May 4, 2022
    #1
  2. nycmathguy

    MathLover1

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    66.

    lim((sqrt(6-x)-2)/(sqrt(3-x)-1), x->2)=(sqrt(6-2)-2)/(sqrt(3-2)-1)=(sqrt(4)-2)/(sqrt(1)-1)=(2-2)/(1-1)=0/0 which is an undefined form and we can apply L'Hopital rule

    hence we get

    lim((sqrt(6-x)-2)/(sqrt(3-x)-1), x->2)

    =lim(d(sqrt(6-x)-2)/dx)/((sqrt(3-x)-1)/dx), x->2)…

    =lim((-1/(2sqrt(6-x)))/(-1/(2sqrt(3-x))), x->2

    =lim(sqrt(3-x)/sqrt(6-x), x->2)

    =1/2
     
    MathLover1, May 4, 2022
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  3. nycmathguy

    nycmathguy

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    Very good but please use wolfram or LaTex for your reply moving forward. Better yet, write your reply on paper, take a picture with your cell phone and upload your answer.
     
    nycmathguy, May 4, 2022
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  4. nycmathguy

    MathLover1

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    upload_2022-5-6_17-51-2.jpeg evaluate for x=2


    = upload_2022-5-6_17-51-54.gif

    = upload_2022-5-6_17-52-28.gif

    = upload_2022-5-6_17-52-58.gif

    = upload_2022-5-6_17-53-21.gif which is an undefined form and we can apply L'Hopital rule

    hence we get
    upload_2022-5-6_17-51-2.jpeg

    => derivative of upload_2022-5-6_17-55-53.gif and upload_2022-5-6_17-59-26.gif

    so, we have to find lim of upload_2022-5-6_18-7-58.gif , simplified will be (continues)
     

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    MathLover1, May 7, 2022
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  5. nycmathguy

    MathLover1

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    upload_2022-5-6_18-9-26.gif

    and upload_2022-5-6_18-11-36.gif
     
    MathLover1, May 7, 2022
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  6. nycmathguy

    nycmathguy

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    nycmathguy, May 7, 2022
    #6
  7. nycmathguy

    nycmathguy

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    nycmathguy, May 7, 2022
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