It is well-known that the solution to the Fibonacci sequence is
$$F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)$$
Show that
$$\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\frac{1+\sqrt{5}}{2}$$
Let
$$\varphi=\frac{1+\sqrt{5}}{2}\implies -\varphi^{-1}=\frac{1-\sqrt{5}}{2}$$
Hence $$ \lim_{n\to\infty}\frac{F_{n+1}}{F_n} =\frac{\varphi^{n+1}-(-\varphi)^{-(n+1)}}{\varphi^n - (-\varphi)^{-n}}=\lim_{n\to\infty}\frac{\varphi + \frac{(-1)^{n+1}}{\varphi^{2n+1}}}{1+\frac{(-1)^n}{\varphi^{2n}}}$$
Now both $\frac{(-1)^{n+1}}{\varphi^{2n+1}}$ and $\frac{(-1)^n}{\varphi^{2n}}$ are alternating sequences. By the Leibniz test, both of them approach $0$ because $\displaystyle\lim_{n\to\infty}\frac{1}{\varphi^n}=0$. Setting this result back to the previous equation leads to the desired results immediately.