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Inifinity limit with natural log

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  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Limx--> ∞ Ln(x^2-1) -Ln(2x^2+3)

    2. Relevant equations


    3. The attempt at a solution
    Ln(x^2-1)/(2x^2+3)

    Then I divided the top and bottom by x^2 so in the end I got (1/2).

    Is this right?
     
  2. jcsd
  3. Sep 27, 2016 #2

    LCKurtz

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    Science Advisor
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    What happened to the ##\ln##?
     
  4. Sep 27, 2016 #3
    Is this what you did? :

    ##\lim_{n\rightarrow +\infty} {\ln {(x^2 - 1)} - \ln{(2x^2+3)}}##

    ## = \lim_{n\rightarrow +\infty} {\ln ({\frac{x^2 - 1}{2x^2+3}})}##


    ## = {\ln {\lim_{n\rightarrow +\infty}(\frac{1 - \frac{1}{x^2}}{2+\frac{3}{x^2}}})}##

    You got the limit of the inside part as ##\frac{1}{2}## you need to take its ##\ln## to get the right answer.
     
  5. Sep 27, 2016 #4
    Yes thats what I did. So my final answer then should be Ln(1/2) ?
     
  6. Sep 27, 2016 #5

    Mark44

    Staff: Mentor

    Right, or -ln(2)
     
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