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Question Number 101822 by john santu last updated on 04/Jul/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{{\varphi }^{n+1}-{(-\varphi )}^{-n-1}}{{\varphi }^n-{(-\varphi )}^{-n}}=$$$$(JS⊛)$$

Answered by bobhans last updated on 05/Jul/20

$$\varphi =\frac{\sqrt{5}+1}{2}>1;\underset{n\to \mathrm{\infty }}{\lim }{(-\varphi )}^{-n}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{{(-\varphi )}^n}=0$$$$Then\underset{n\to \mathrm{\infty }}{\lim }\frac{{\varphi }^{n+1}-{(-\varphi )}^{-n-1}}{{\varphi }^n-{(-\varphi )}^{-n}}=\underset{n\to \mathrm{\infty }}{\lim }\frac{{\varphi }^{n+1}}{{\varphi }^n}=\varphi$$$$=\frac{\sqrt{5}+1}{2}(Bob-)$$

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