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Continuity and Limits

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    1) 5ee5o7.jpg

    2) 27y226c.jpg

    2. Relevant equations



    3. The attempt at a solution

    1) I have done plenty of these, but this one is stumping me. I tried plugged in 0 approach for h and I got 11-11=0. With h on the bottom as 0. I know this isn't the right answer. I also know the limit does exist. If anyone could help me with HOW you come to the conclusion I would greatly appreciate it.

    2) I have pasted the problem with my attempts at an answer. I am having a little bit of trouble here trying to determine exactly what intervals are continuous and why.


    Thanks guys for any help you can give!
     
  2. jcsd
  3. Sep 6, 2010 #2
    Try using an Algebraic Conjugate.
     
  4. Sep 6, 2010 #3

    eumyang

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    1) Rationalize the NUMERATOR.

    2) By my count, there are two more intervals where g is continuous. What is happening at x = 4?
     
    Last edited by a moderator: Sep 15, 2010
  5. Sep 6, 2010 #4
    1) I got 11 out of the square root of 121 which cancels with the other -11. Confused about the remaining h.

    2) I was confused at x = 4. Are my current 3 answers correct?
     
  6. Sep 6, 2010 #5

    eumyang

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    1) No, you can't do that. You have a square root of a SUM, and you can't split into a sum of two square roots:
    [tex]\sqrt{121 + h} \ne 11 + \sqrt{h}[/tex]

    You rationalize the numerator by multiplying top and bottom by the numerator's conjugate. For example, if the numerator of a fraction is
    [tex]a + \sqrt{b}[/tex],
    then you multiply top and bottom by
    [tex]a - \sqrt{b}[/tex],
    because in doing so, you make the numerator a rational number (hence, rationalizing).
    [tex](a + \sqrt{b})(a - \sqrt{b}) = a^2 - b[/tex].
     
    Last edited by a moderator: Sep 15, 2010
  7. Sep 6, 2010 #6
    Ok, let me figure this out.

    So I multiply the numerator and the denominator by

    sqrt (121+h)+11?
     
    Last edited: Sep 6, 2010
  8. Sep 7, 2010 #7

    eumyang

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    Yes. Keep going. What happens next?
     
    Last edited by a moderator: Sep 15, 2010
  9. Sep 7, 2010 #8
    If you substitute h=0 you get 0/0 ie., an indeterminate form. So you can use L' Hospital's rule for this problem.
    i.e., differentiate numerator and denominator separately and find the limit of their ratios.
     
  10. Sep 7, 2010 #9
    I got h on the top. (sqrt (h + 121)) h + 11h on bottom. Is that correct?
     
    Last edited by a moderator: Sep 15, 2010
  11. Sep 8, 2010 #10
    That's right. Now notice that each term in the denominator has a factor of h. After you factor out the h, what can you do next?
     
  12. Sep 8, 2010 #11

    eumyang

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    This may not be helpful if the OP is beginning Calc. I at a uni. in the U.S. He/she may have just started to study limits and so he/she would not know what L'Hôpital's rule (not L'Hospital :wink:) is.
     
    Last edited by a moderator: Sep 15, 2010
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