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Use Binet's formula for the Fibonacci numbers to show $$\lim_{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}} = \phi.$$

Could anyone give me a hint please?

Intuition
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    See the similar question: https://math.stackexchange.com/questions/728973/fibonacci-numbers-and-golden-ratio-phi-lim-sqrtnf-n?rq=1 – Robert Z Jun 25 '19 at 07:57

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Here is a solution if you do not know Binet's formula. Let $L = \lim_{n \to \infty}a_n = \lim_{n \to \infty}\frac{u_{n+1}}{u_n}$. Thus $$a_{n+1} = \frac{u_{n+2}}{u_{n+1}}$$ $$\implies a_{n+1} = \frac{u_{n+1}+u_n}{u_{n+1}}=1 + \frac{u_n}{u_{n+1}}.$$ Therefore $$L=\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left( 1 + \frac{u_n}{u_{n+1}}\right) = 1 + \lim_{n \to \infty}\frac{u_n}{u_{n+1}} = 1 + \frac{1}{L}.$$ Thus $$L = 1 + 1/L \implies L^2 - L - 1 = 0.$$ Taking the positive root, we have $L=(1+\sqrt{5})/2 = \phi.$

tmaj
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Hint

Binet formula write $$u_n=\frac{ \left(1+\sqrt{5}\right)^n-\left(1-\sqrt{5}\right)^n}{2^n\sqrt{5}}=\frac{ \left(1+\sqrt{5}\right)^n}{2^n\sqrt{5}}\Big(1-\left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n\Big)$$

Look how small is the factor $\frac{1-\sqrt{5}}{1+\sqrt{5}}\approx -0.382$.

Claude Leibovici
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$$\frac{u_{n+1}}{u_n}=\frac{\phi^{n+1}-\psi^{n+1}}{\phi^n-\psi^n}=\phi\frac{1-\left(\dfrac\psi\phi\right)^{n+1}}{1-\left(\dfrac\psi\phi\right)^n}.$$

As $$\left|\dfrac\psi\phi\right|<1,$$ the ratio (quickly) tends to $1$.