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

lim_(n→∞) ((φ^(n+1) −(−φ)^(−n−1) )/(φ^n −(−φ)^(−n) )) =   (JS ⊛)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\phi^{{n}+\mathrm{1}} −\left(−\phi\right)^{−{n}−\mathrm{1}} }{\phi^{{n}} −\left(−\phi\right)^{−{n}} }\:=\: \\ $$$$\left({JS}\:\circledast\right) \\ $$

Answered by bobhans last updated on 05/Jul/20

φ = (((√5) +1)/2) > 1 ; lim_(n→∞)  (−φ)^(−n) = lim_(n→∞) (1/((−φ)^n )) = 0  Then lim_(n→∞) ((φ^(n+1) −(−φ)^(−n−1) )/(φ^n −(−φ)^(−n) )) = lim_(n→∞) (φ^(n+1) /φ^n ) = φ  = (((√5) +1)/2) (Bob− )

$$\phi\:=\:\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}\:>\:\mathrm{1}\:;\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(−\phi\right)^{−{n}} =\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left(−\phi\right)^{{n}} }\:=\:\mathrm{0} \\ $$$${Then}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\phi^{{n}+\mathrm{1}} −\left(−\phi\right)^{−{n}−\mathrm{1}} }{\phi^{{n}} −\left(−\phi\right)^{−{n}} }\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\phi^{{n}+\mathrm{1}} }{\phi^{{n}} }\:=\:\phi \\ $$$$=\:\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}\:\left({Bob}− \right) \\ $$

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