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2 lmit problems

  1. Feb 6, 2009 #1
    Two basic limit problems just solve the limit if it exists. I know the answers but I don't understand why the answers are what they are.
    1.) lim (x^2-x-2) / (x-2)^2
    x->2
    2.) lim ((1/x)-(1/2)) / (x-2)
    x->2

    Answers were:
    1.) DNE
    2.) -1/4

    I kind of understand that the first problem the denominator would be 0, but I thought if you factored the top and bottom you could find the limit of (x+1).

    The second one I am completely stuck the only thing I could think to do was use the division rule but I would still have 0 for the denominator. If anyone could explain this it would be a big help.
     
  2. jcsd
  3. Feb 6, 2009 #2
    1. Simplifying this expression by factoring the numerator is the correct approach. However, the behavior of (x+1)/(x-2) does not approach any value at x=2. To see this, try values of x slightly less than 2 and values of x slightly greater than 2. The graph is pretty much the graph of y = 1/x shifted 2 units to the right on the x-axis and 1 unit up on the y-axis. You can then add a rigorous epsilon-delta proof that the limit does not exist if necessary

    2. Hint: Simplify the numerator by adding the two fractions.
     
    Last edited: Feb 6, 2009
  4. Feb 6, 2009 #3
    I can not believe I did not think of adding the fractions together.

    And I found out why I couldn't get the answer for the first problem. On my sheet I forgot to write the square on the denominator. And I wasn't thinking when I typed the problem. When I saw that you wrote (x+1)/(x-2) I went back to my sheet and saw that I did in fact forget the square.

    Thank you so much for the help. Kind of ticked I did not see my mistake earlier though, spent a lot of time on the first problem trying to figure it out.
     
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