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Following limit is to be evaluated

  1. Sep 22, 2005 #1
    the following limit is to be evaluated as x-->infinity

    [tex](\frac{x-1}{x+4})^{3x+1}[/tex]

    here is the work i've done

    taking the natural log:

    [tex]3x+1 ln(\frac{x-1}{x+4})[/tex]

    to make it an indeterminant form i write it in the following way:

    [tex]\frac{ln\frac{x-1}{x+4}}{\frac{1}{3x+1}}[/tex]

    applying l'hopital's rule to this yields:

    [tex]\frac{\frac{x+4}{x-1}}{\frac{-3}{(3x+1)^2}}[/tex]

    simplifying:

    [tex]\frac{x+4(3x+1)^2}{-3(x-1)}[/tex]

    now noticing that the degree of the numerator is 3 and the denominator is 1, the limit as x--> infinity is infinity.. exponentiating gives e to the infinity....the answer to this is apparently 0 as i found when i graphed the function...where did i go wrong and how to correct this?
     
    Last edited: Sep 22, 2005
  2. jcsd
  3. Sep 22, 2005 #2

    Tom Mattson

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    That step is wrong. If you differentiate [itex]\ln(\frac{x-1}{x+4})[/itex], you get:

    [tex]\frac{d}{dx}\ln(\frac{x-1}{x+4})=\frac{x+4}{x-1}\frac{d}{dx}(\frac{x+4}{x-1})[/itex].

    You forgot about the Chain Rule.
     
    Last edited: Sep 22, 2005
  4. Sep 22, 2005 #3
    Just for thought....are you to evaluate it using the limiting process and l'hopital's rule or are you permitted to just eyeball it? Cuase if you only need to eyeball it, I got a neat pointer for ya. If not, no worries
     
  5. Sep 23, 2005 #4

    lurflurf

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    That is much like this limit
    [tex]\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^x=e^a[/tex]
    try to make use of this
     
  6. Sep 23, 2005 #5

    dextercioby

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    I have a hunch it's something like [itex] e^{-15} [/itex].

    Hint

    [tex] \frac{x-1}{x+4}=1-\frac{5}{x+4} [/tex]

    and the one in post #4.

    Daniel.
     
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