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l'hopital's rule to find limit watch

1. quick explination:
i have

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

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.

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 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
2. 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.
3. (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?
4. 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
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:

derived:
(a) in terms of z only:

(b) or deriving everything:

which is the correct way? either way i'm still not getting

substituting z=0 in (a):

(i can see i'm close here but not quite there!)

or substituting z=0 in (b):

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!!
6. You need to differentiate using the product rule.
7. f= z => f'=1

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

then use algebra of limits
8. YES i got it, was easy after all, thank you all!!
9. (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.
10. (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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