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Question Number 185243 by Mastermind last updated on 19/Jan/23

$$\mathrm{Using}\epsilon -\delta \mathrm{approach}\mathrm{prove}\mathrm{that}$$$$\mathrm{li}\underset{\mathrm{z}\to \mathrm{i}}{\mathrm{m}}\frac{{3\mathrm{z}}^4-{2\mathrm{z}}^3+{8\mathrm{z}}^2-2\mathrm{z}+5}{\mathrm{z}-\mathrm{i}}=4+4\mathrm{i}$$$$$$$$\mathrm{Help}!$$

Commented by Frix last updated on 19/Jan/23

$$\mathrm{Different}\mathrm{approach},\mathrm{but}:$$$$3z^4-2z^3+8z^2-2z+5=(z^2+1)(3z^2-2z+5)$$$$\frac{1}{z-\mathrm{i}}=\frac{z}{z^2+1}+\frac{1}{z^2+1}\mathrm{i}$$$$\Rightarrow$$$$\underset{z\to \mathrm{i}}{\lim }\frac{3z^4-2z^3+8z^2-2z+5}{z-\mathrm{i}}=$$$$=\underset{x\to \mathrm{i}}{\lim }(z+\mathrm{i})(3z^2-2z+5)=4+4\mathrm{i}$$

Commented by Mastermind last updated on 19/Jan/23

$$\mathrm{Anyways},\mathrm{thank}\mathrm{you}\mathrm{but}\mathrm{this}\mathrm{is}\mathrm{not}$$$$\epsilon -\delta \mathrm{approach}$$

Commented by Frix last updated on 19/Jan/23

$$\mathrm{I}\mathrm{know}.$$$$\mathrm{Since}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{The}\mathrm{Omniscient}\mathrm{Answering}$$$$\mathrm{Machine}(\mathrm{if}\mathrm{I}\mathrm{were}\mathrm{the}\mathrm{answer}\mathrm{would}\mathrm{have}$$$$\mathrm{been}42\mathrm{of}\mathrm{course})\mathrm{I}\mathrm{simply}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}$$$$ϵ-\delta \mathrm{approach}.\mathrm{You}\mathrm{tell}\mathrm{me}\mathrm{please}.$$

Answered by 123564 last updated on 19/Jan/23

Commented by Mastermind last updated on 20/Jan/23

$$\mathrm{Please}\mathrm{try}\mathrm{to}\mathrm{be}\mathrm{type},\mathrm{so}\mathrm{itwill}\mathrm{easily}$$$$\mathrm{to}\mathrm{understand}\mathrm{thank}\mathrm{you}$$

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