let $$\lim_{n \to \infty}\frac{f_{n+1}}{f_n} = \phi$$ by definition, where $$f_n$$ is the nth fibonacci number
then $$\implies \lim_{n\to \infty}1+\frac{f_{n-1}}{f_n}= \phi $$ $$\implies 1+\lim_{n\to \infty}\frac{f_{n-1}}{f_n}= \phi $$ $$\implies 1+\lim_{n\to \infty}\frac{f_{n}}{f_{n+1}}= \phi $$ $$\implies 1+(\lim_{n\to \infty}\frac{f_{n+1}}{f_{n}})^{-1} = \phi $$ $$\implies 1+ \frac{1}{\phi} = \phi $$
I am not sure if the reciprocal of a limit of ratios is the limit of the reciprocal of the ratios. Is my reasoning correct here?
Since $f_n>0$, and $x\mapsto x^{-1}$ is continuous in $(0, \infty) $, then is correct to say that $$\lim_{n\to\infty}\frac{f_{n-1}}{f_n}=\left(\lim_{n\to\infty}\frac{f_n}{f_{n-1}}\right) ^{-1}.$$