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Question Number 185374 by Mastermind last updated on 20/Jan/23

$$\mathrm{li}\underset{\mathrm{z}\to \mathrm{\infty }}{\mathrm{m}}\frac{{\mathrm{iz}}^3+\mathrm{iz}-1}{(2\mathrm{z}+3\mathrm{i}){(\mathrm{z}-\mathrm{i})}^2}$$$$$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by MJS_new last updated on 20/Jan/23

$$\mathrm{if}z\to \mathrm{\infty }\mathrm{means}z\to \mathrm{\infty }+0\mathrm{i}\Rightarrow \mathrm{we}\mathrm{can}\mathrm{let}z\in \mathbb{R}$$$$\mathrm{and}\mathrm{i}\mathrm{is}\mathrm{constant}$$$$\frac{{cz}^3+cz-1}{(2z+3c){(z-c)}^2}=\frac{{cz}^3+cz-1}{2z^3-{cz}^2-4c^{2}z+3c^3}$$$$\Rightarrow \mathrm{the}\mathrm{limit}\mathrm{is}\frac{c}{2}=\frac{\mathrm{i}}{2}$$

Commented by Mastermind last updated on 21/Jan/23

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{boss}$$

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