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Question Number 83710 by john santu last updated on 05/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{(1+\sqrt{5})}^x-{(1-\sqrt{5})}^x}{{(1+\sqrt{5})}^{x-1}-{(1-\sqrt{5})}^{x-1}}=?$$

Commented by john santu last updated on 05/Mar/20

$$\mathrm{let}1+\sqrt{5}=\mathrm{u}\mathrm{\&}1-\sqrt{5}=\mathrm{v}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{u}}^x-v^x}{u^{x-1}-v^{x-1}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{u}}^{-x}(u^x-v^x)}{u^{-x}(u^{x-1}-v^{x-1})}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1-{\mathrm{v}}^x{u}^{-x}}{u^{-1}-v^{x-1}{u}^{-x}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1-{\left(\frac{\mathrm{v}}{\mathrm{u}}\right)}^x}{u^{-1}(1-{\left(\frac{v}{u}\right)}^{x-1})}=$$$$sett=\frac{v}{u}\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }u\left(\frac{1-t^x}{1-t^{x-1}}\right)$$$$since\mid \mathrm{t}\mid =<1\Rightarrow u\times \underset{x\to \mathrm{\infty }}{\lim }\frac{1-0}{1-0}$$$$\Rightarrow u=\sqrt{5}+1\Leftarrow \mathrm{the}\mathrm{answer}$$

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