# why its not removable singular point??

#### transgalactic

$$\displaystyle \frac{z-\sin z}{z^4}$$
my singular point is 0
if i will do lhopital 4 times on the lmit i will get -sin z/6 =0
which is singular point because the result is constant

so why its a first order pole?

#### Jose27

MHF Hall of Honor
$$\displaystyle \frac{z-\sin z}{z^4}$$
my singular point is 0
if i will do lhopital 4 times on the lmit i will get -sin z/6 =0
which is singular point because the result is constant

so why its a first order pole?
Expand $$\displaystyle z-\sin (z)$$ in its Taylor series and you'll find it has an order 3 zero at 0

Your reasoning is wrong since applying L'hopital gives you $$\displaystyle \lim_{z\rightarrow 0} \frac{\cos (z)}{24z} =\infty$$

mr fantastic

#### transgalactic

ok now i tried to determine the power of the pole
i did
g(x)=1/f(x)
where x=0
and i got that g'(0)=0/0 which means that its not a first order pole
actualyy i should get zero but i got 0/0
??

#### mr fantastic

MHF Hall of Fame
ok now i tried to determine the power of the pole
i did
g(x)=1/f(x)
where x=0
and i got that g'(0)=0/0 which means that its not a first order pole
actualyy i should get zero but i got 0/0
??
Post #2 has given you a big hint.

The singularity of $$\displaystyle \displaystyle f(z) = \frac{z - \sin z}{z^4}$$ at z = 0 is a simple pole. Do what post #2 said to see why.

#### transgalactic

i want to solve it by the derivative way
not by developing into a series

#### mr fantastic

MHF Hall of Fame
i want to solve it by the derivative way
not by developing into a series
Well then, where is your trouble in using l'Hopital's Rule a couple of times to find that $$\displaystyle \displaystyle {\lim_{z \to 0} \frac{z(z - \sin z)}{z^4} = {\lim_{z \to 0} \frac{z - \sin z}{z^3}}$$ exists and is finite?