\(\displaystyle \frac{z-\sin z}{z^4}\)
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

\(\displaystyle \frac{z-\sin z}{z^4}\)
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

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
??

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
??

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?