Help please

My answer has been questioned so I am going to do it a little more main stream and show that the 2 answers are the same. :)

$i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ i^6=-1\\ i^7=-i\\ \text{and so the pattern continues}\\ i^{4n}=1\\ i^{4n+1}=i\\ i^{4n+2}=-1\\ i^{4n+3}=-i$

Hence       $i^3=-1 \qquad and \qquad i^9=i$

$\begin{align} \displaystyle\lim_{x\rightarrow i }\; \frac{{z}^{3}+2}{{z}^9+8}& =\frac{-i+2}{i+8}\\ & \text {Rationalize the denominator}\\ &=\frac{2-i}{8+i}\times \frac{8-i}{8-i}\\ &=\frac{16-2i-8i-1}{64--1}\\ &=\frac{15-10i}{65}\\ &=\frac{3-2i}{13}\\ &=\frac{3}{13}-\frac{2i}{13}\\ \end{align}$

See Majid, the answer is the same       

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Hi Majid,    

while z is a complex number

$\begin{align} \displaystyle\lim_{x\rightarrow i }\; \frac{{z}^{3}+2}{{z}^9+8}& =\frac{-i+2}{i+8}\\ &=\frac{-i+2}{i+8}\times \frac{i-8}{i-8}\\ &=\frac{1+8i+2i-16}{-1-64}\\ &=\frac{10i-15}{-65}\\ &=\frac{2i-3}{-13}\\ &=\frac{3}{13}-\frac{2i}{13}\\ \end{align}$