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Find the domain of definition and then the limit

  1. Sep 14, 2006 #1
    okay i have three limits, i did one and the two others i m stuck...
    well here

    1.
    limit when x tends to zero of
    (x-sin(px)) / (x-sin(qx))

    p and q are positive integers.

    for this one i have no idea what to do, i never worked with p or q...

    2.
    limit when x tends to zero of the function:
    (x(1-cosx)) / (sin3x - 3sinx)
    for this one, i expanded the denominator to get [ sinx(3cos²x-sin²x) ], then i finally have
    (x(1-cosx)) / (sinx(3cos²x-sin²x))
    and the limit is +infinty i think; am i right?

    3.

    f(x)=(x-1)/(x+1-|x|)
    i first need to find the domain of definition, then find the limits in those points.

    first, i need to separate that function into :

    f(x)= (x-1)/(2x+1) [x<0] (x shouldn't equal 1/2)
    f(x)= x+1 [x<o]

    so, the limit in -infinity is -infinity
    the limit in +infinity is 1/2
    the limit in 1/2+ is -infinity
    the limit in 1/2- is +infinity

    is that right ^^
    thanks!
     
    Last edited: Sep 14, 2006
  2. jcsd
  3. Sep 14, 2006 #2

    VietDao29

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    Uhm... this may be a good candidate for L'Hopital's rule, or Taylor's expansion. Have you covered Taylor's expansion yet? Just expand it arround x = 0, and see if you get the answer.
    Nope. :)
    You should check it again:
    sin(3x) = 3 sin(x) - 4 sin3(x), the denominator does not simplify to: sinx(3cos²x-sin²x) :smile:
    x can be 1 / 2, the only value that makes f(x) undefined is x = - 1 / 2.
    You should check the limits for x tends to -, and + infinity again, you seemed to have swapped the two.
    Ok, can you go from here? :)
     
  4. Sep 14, 2006 #3
    L'Hospital's would be great for the 1st one, but since this is in Pre-Calc, I would assume you can't use anything with a deriviative yet.

    I can't see what would be a good candidate, but maybe the Squeeze Theorem would work here.
     
  5. Sep 14, 2006 #4

    arildno

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    As for the first one, do the trick:
    [tex]\frac{x-\sin(px)}{x-\sin(qx)}=\frac{x-px\frac{\sin(px)}{px}}{x-qx\frac{\sin(qx)}{qx}}=\frac{1-p\frac{\sin(px)}{px}}{1-q\frac{\sin(qx)}{qx}}[/tex]
     
    Last edited: Sep 14, 2006
  6. Sep 14, 2006 #5
    okay, for the second one, after simplifying i get -x/4sinx, and the limit of that when x tends to zero is -1/8. is this right?


    the third one, (yes I meant x>o)

    so, the limit in -infinity is 1/2
    the limit in +infinity is 1/2
    the limit in (-1/2)+ is -infinity
    the limit in (-1/2)- is +infinity
    is that right?

    the first one,
    I'd like to know how you went from [tex]\frac{x-\sin(px)}{x-\sin(qx)} to this \frac{x-px\frac{\sin(px)}{px}}{x-qx\frac{\sin(qx)}{qx}}[/tex]

    Thank you very much, I really appreciate your help :)
     
  7. Sep 14, 2006 #6

    arildno

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    Well, px/px=1, isn't it?
    We can multiply any term we wish with 1, since 1 times a number equals that number..
     
  8. Sep 14, 2006 #7

    VietDao29

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    This one is still wrong. :) Re-check it the third time to see why. :rolleyes:
     
  9. Sep 14, 2006 #8
    Got it thanks!

    Oooops!! it's plus infinity... I forgot that there was an absolute value!! Thank you =]
     
  10. Sep 15, 2006 #9
    Okay I have a problem now...

    for the second one,

    (x(1-cosx)) / (sin3x - 3sinx)

    How did you get sin(3x) = 3 sin(x) - 4 sin^3 (x), I didn't get that. and let's suppose it is right... now my function is:

    -x/4sinx(1+cosx)

    i dont know what do afterwards ? :confused::confused:


    EDIT: NEVER MIND! THANKS
     
    Last edited: Sep 15, 2006
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