(Solved):
8. (Bonus) Recall that the sequence \( \left\{x_{n}\right\}_{n=0}^{\infty} \) converges to \( \bar ...
8. (Bonus) Recall that the sequence \( \left\{x_{n}\right\}_{n=0}^{\infty} \) converges to \( \bar{x} \) of order \( \alpha \) if \( \lim _{n \rightarrow \infty} \frac{\left|x_{n+1}-\bar{x}\right|}{\left|x_{n}-\bar{x}\right|^{\alpha}}=k \) as \( x_{n} \rightarrow \bar{x} \) and converges superlinearly to \( \bar{x} \) if \( \lim _{n \rightarrow \infty} \frac{\left|x_{n+1}-\bar{x}\right|}{\left|x_{n}-\bar{x}\right|}=0 \) as \( x_{n} \rightarrow \bar{x} \). (a) Show that if \( x_{n} \rightarrow \bar{x} \) of order \( \alpha \) for \( \alpha>1 \), then \( \left\{x_{n}\right\} \) is superlinearly convergent to \( \bar{x} \). (b) Suppose that \( \left\{x_{n}\right\} \) is superlinearly convergent to \( \bar{x} \). Show that \( \lim _{n \rightarrow \infty} \frac{\left|x_{n+1}-x_{n}\right|}{\left|x_{n}-\bar{x}\right|}=1 \).