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    quick explination:
    i have

    f(z)=\frac{z}{(2z-pi)(sin2z)}

    with a value z=0. differentiating each term, z, (2z-pi), (sin2z) until they equal a non-zero, z is differentiated once, (2z-pi) not differentiated, and (sin2z) differentiated once.

    so

    \frac{1}{(0) + (1)} gives 1/1 which is a removable pole.



    NOW i need to use L'Hopital's rule to find the limiting value of this removable pole.

    the answer should be

    \frac{1}{-pi}  \frac{1}{2} = \frac{1}{-2pi}

    which i'm not sure how to get. i know i need to derive, but would i derive with the original equation with z, or with the 0 value substituted? does the \frac{1}{(0) + (1)} i just found have anything to do with how many times i differentiate? or do i completely forget about what i've done so far as that only had to do with finding which kind of singularity (this case, removable) it is?

    thanks for any input
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    I would simply go: sin 2z = 2z + O(z^3), so (sin 2z)/z = 2 + O(z^2).

    So you have f(z) = 1/(2z-pi)(2 + O(z^2)). At this point you can just set z = 0 to get the limit.
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    (Original post by DFranklin)
    I would simply go: sin 2z = 2z + O(z^3), so (sin 2z)/z = 2 + O(z^2).

    So you have f(z) = 1/(2z-pi)(2 + O(z^2)). At this point you can just set z = 0 to get the limit.
    hmm, i don't know what O is?
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    L'hopital's rule if f(z)->0 and g(z)->0 as z->c and lim f '(z)/g'(z) as z-> c exists and equals L then lim f/g as z->c = L

    So you need to differentiate top and bottom and then take the limit of f'/g' at z=0
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    (Original post by thebadgeroverlord)
    L'hopital's rule if f(z)->0 and g(z)->0 as z->c and lim f '(z)/g'(z) as z-> c exists and equals L then lim f/g as z->c = L

    So you need to differentiate top and bottom and then take the limit of f'/g' at z=0
    thanks, i've tried this but i don't seem to be getting the right answer:

    \frac{z}{(2z-pi)(sin2z)}

    derived:
    (a) in terms of z only:
    \frac{1}{(2-pi)(2cos2z)}

    (b) or deriving everything:
    \frac{1}{(2)(2cos2z)}

    which is the correct way? either way i'm still not getting \frac{1}{-2pi}

    substituting z=0 in (a):
    \frac{1}{(2-pi)(2)} = \frac{1}{4-2pi}
    (i can see i'm close here but not quite there!)

    or substituting z=0 in (b):
    \frac{1}{(2)(2)} = \frac{1}{4}

    looking at there two i'm guessing (a) is the correct way, only deriving in terms of z? but i'm still doing something wrong with it. sorry to be a pain!!
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    You need to differentiate (2z - \pi) \sin 2z = 2z \sin 2z - \pi \sin 2z using the product rule.
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    f= z => f'=1

    g= 2zsin2z -(pi)sin2z => g' = 2sin2z + 4zcos2z - 2(pi)cos2z

    then use algebra of limits
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    YES i got it, was easy after all, thank you all!!
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    (Original post by furryvision)
    hmm, i don't know what O is?
    In this context, O(z^3) means terms of order z^3 and higher.
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    (Original post by DFranklin)
    In this context, O(z^3) means terms of order z^3 and higher.
    ah i remember doing that, great thanks!
 
 
 
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