I have used approximation : $$e^{h_{1}}=1+h_{1}+h^2_{1}+o(h^3_{1}) $$ $$\cos(h_{2})=1-\frac{h^2_{2}}{2}+o(h^4_{2})$$ $$\sin(h_{2})=h_{2}-o(h^3_{2})$$ Using this as a definition :$ e^h = e^{h_{1}}[cos(h_{2})+i\sin(h_{2})]$ not $e^h =1+h+\frac{h^2}{2}+...$ to be more specific I want to calculate this limit : $$\lim_{h\to 0}\frac{e^{h_{1}}\cos(h_{2})}{h}=1\ ,\ \lim_{h\to 0}\frac{e^{h_{1}}\sin(h_{2})}{h}=0$$ without a lot of calculation
and after a lot of tedious calculations I got the result ,but I wonder if there is a quick way to prove that this limit is equal to 1 without a lot of calculations?
Thank you ...
Well you can immediately write : $$e^h=1+h+\frac{h^2}{2}+o(h^2)$$ Hence : $$\frac{e^h - 1}{h}=1+\frac{h}{2}+o(h)$$ And you have your result.
There is no need to separate imaginary and real part.