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Homework Help: Limits of fractions of polynomials and trig functions

  1. Sep 25, 2007 #1
    I have two...

    1. The problem statement, all variables and given/known data
    The the limit


    2. Relevant equations
    [itex]\lim_{x \rightarrow 1} \frac{1-cosx}{x^2}[/itex]


    3. The attempt at a solution
    I figured to just plug in 1, but I wanted to make sure....

    1. The problem statement, all variables and given/known data
    Find the limit


    2. Relevant equations
    [itex]\lim_{x \rightarrow 3} \frac{\sqrt{x^2-6x+9}}{x-3}[/itex]


    3. The attempt at a solution
    I plugged in the 3, and got 3/0, then I got lost...
     
  2. jcsd
  3. Sep 25, 2007 #2

    EnumaElish

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    In 2, did you try simplifying the numerator? (What are the roots of the polynomial?)
     
  4. Sep 26, 2007 #3
    Yes, I tried doing that.

    [itex](x-3)(x-3)[/itex]

    However, I forgot how to get rid of that radical. Squaring wouldn't work, so I have no idea.


    Also, no thoughts on the first one?
     
  5. Sep 26, 2007 #4

    EnumaElish

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    What is a short hand expression for (x-3)(x-3)?
     
  6. Sep 26, 2007 #5
    (x-3}^2. Oh yeah, so that takes away the square root, and after everything, it leaves 0. thank you.
     
  7. Sep 26, 2007 #6

    Avodyne

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    In the first one, are you sure the problem isn't x->0 instead of x->1 ?
     
  8. Sep 26, 2007 #7
    it is 1, not 0.
     
  9. Sep 26, 2007 #8

    arildno

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    Well, then your book has a typo..
     
  10. Sep 26, 2007 #9
    it's not from my book, it was my teacher.
     
  11. Sep 26, 2007 #10

    Avodyne

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    Well, it's 99% certain that your teacher meant to write 0 instead of 1. With 1, it's trivial, since both the numerator and denominator are finite, nonzero constants in that limit.
     
  12. Sep 26, 2007 #11

    arildno

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    Then he either blundered, or tried to fool you.

    Your function is defined&continuous on all values of x except x=0.

    Your original approach is perfetly valid in the case of x=1.
     
  13. Sep 26, 2007 #12
    allright, thankyou.
     
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