Limit of Greatest Integer Function...7

Discussion in 'Calculus' started by nycmathguy, Apr 17, 2022.

  1. nycmathguy

    nycmathguy

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    Extra practice not from the textbook.

    Find the limit of

    f(x) = [[ x^2 ]]^(2x - 1) + [[ 1/x ]]^([[ 1/x^2 ]]

    as x tends to zero.

    How is this done? Can you show the graph of f(x)?
     
    nycmathguy, Apr 17, 2022
    #1
  2. nycmathguy

    MathLover1

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    lim( [[ x^2 ]]^(2x - 1) + [[ 1/x ]]^([[ 1/x^2 ]]) as x->0

    lim( [[ x^2 ]]^(2x - 1)) + lim([[ 1/x ]]^([[ 1/x^2 ]])

    substitute x
    lim( [[ 0^2 ]]^(2*0 - 1)) + lim([[ 1/0 ]]^([[ 1/0^2 ]])

    lim((0^2)^( - 1)) + lim([[ 1/0 ]]^( 1/0^2)=1/0^2 +(1/0)^(1/0)=∞

    so,
    upload_2022-4-17_18-37-33.gif
     
    MathLover1, Apr 18, 2022
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Thank you. This one threw me into a loop-o-plane.
     
    nycmathguy, Apr 18, 2022
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