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Analysis: Find limits

  1. Oct 16, 2011 #1
    1) = 0

    The denominator grows much faster than the numerator.

    2) = 0

    Each of these terms would simply be zero, right?

    For 4) and 5), the f(x) = x3

    I simplified these to

    4) = [(x-2)(x2+2x+4)]/(x-3)

    5) = [(x-3)(x2+3x+9)]/(x-2)

    http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled2.png [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 16, 2011 #2

    HallsofIvy

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    (1) is correct, (2) is not. In (2) your sequence is
    [tex]\frac{1}{n}+ \frac{2}{n}+ \cdot\cdot\cdot+ \frac{n}{n^2}= \frac{1+ 2+ 3+ \cdot\cdot\cdot+ n}{n^2}[/tex]

    Use the fact that [itex]1+ 2+ 3+ \cdot\cdot\cdot+ n= (1/2)n(n+1)[/itex].

    If you are thinking you can take the limit in each term so that you get 0 for each, no that is not correct.

    You can do that but the obvious point should be that denominator goes to 0 while the numerator goes to f(3)- f(2)= 27- 8 which is NOT 0.

    Again, the denominator goes to 0 while the numerator does not.

     
    Last edited by a moderator: May 5, 2017
  4. Oct 16, 2011 #3
    For 2), I didn't know you could rewrite the sum in that manner.

    I then get [(1/2)n(n+1)]/n = (1/2)[(n+1)/n] = 1/2.

    If I had to guess, I would say 4) and 5) go to +/- infinity, respectively.
     
    Last edited: Oct 16, 2011
  5. Oct 18, 2011 #4
    Am I right?
     
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