Let $(x_n)_{n \ge 1}$ with $x_1>4$ and $x_{n+1}=3+\frac{4}{x_n}, n \ge 1$.
Prove that $\lim_{n\to \infty} x_n=4$ and find $\lim_{n\to \infty}( \frac{x_{2n+1}+x_{2n}}{8})^n$.
I proved that $\lim_{n\to \infty} x_n=4$, and for the second limit I wrote $$\left(\frac{x_{2n}+x_{2n+1}}{8}\right)^n=\exp\left\{n\ln\left(\frac{x_{2n}+x_{2n+1}}{8}\right)\right\},$$ but I don't understand how to calculate this. Please help!
Sketch of proof : step 1: $ \ln(1-x)=-x-x^2-\cdots -x^n+ o(x^n)\ $, so you have to determine the limit of $$ n(x_{2n}+x_{2n+1}-8)$$
Step 2 : From a certain point $N$, $x_n\in[3,5]$ so (with $f(x):=3+\frac{4}{x}$) $$|x_{n+1}-x_{n}|=|f(x_{n})-f(x_{n-1})|\leq \sup_{c\in[3,5]}|f'(c)||x_{n}-x_{n-1}|< C |x_{n}-x_{n-1}|< 2 C^{n-N}$$ with $0<C<1$, for $n>N$. So, for $2n>N$, $$ |x_{2n+1}-x_{2n}|\leq 2 C^{2n-N} $$ sept 3: from the particular form of that equation, it is easy to prove that the subsequence $x_{2n}$ is decreasing to $4$ (since $x_0>4$) and $x_{2n+1}$ is increasing to $4$ (and $x_p>0$ or all $p$). So, $x_{2n+1}\leq 4\leq x_{2n}$ and $$ |x_{2n}+x_{2n+1}-8 |\leq |x_{2n+1}-4|+|x_{2n}-4|=|x_{2n+1}-x_{2n}|\leq 2 C^{2n-N}$$ so $$\lim_{n\rightarrow \infty}n(x_{2n}+x_{2n+1}-8) =0$$ You can conclude.