Order of Convergence Proof with p>1 and M>0 So I am struggling with a problem on my homework, the problem statement is Assuming $x_{n}\rightarrow x^*$, show that any sequence that satisfies $$\lim\limits_{n \to \infty} \frac{|x_{n+1}-x^*|}{|x_{n}-x^*|^p}= M$$ with $p>1$ for some $M>0$ also satisfies $$\lim\limits_{n \to \infty} \frac{|x_{n+1}-x^*|}{|x_{n}-x^*|}= 0$$.
I looked at this problem for a bit and think it is as simple as seeing that $x_{n+1}$ goes to $x^*$. This would make the numerator of the limit $= 0$, but if I use this methodology, then I would have the bottom of the limit $=0$ as well which would cause an error in my methodology. I would appreciate any hints or nudges in the correct direction. This is my first experience with this so a lighter explanation would help.

First of all note that $\lim_{n\to+\infty}|x_{n}-x^{*}|=0$ since the sequence is convergent. Then $\forall p>1$ the following holds:
$$\lim_{n\to+\infty}|x_{n}-x^{*}|^{p-1}=0$$
Now multiply and divide by $|x_n-x^{*}|^p$ the argument of the limit you want to show to hold to get
$$\lim_{n\to+\infty}\frac{|x_{n+1}-x^{*}|}{|x_{n}-x^{*}|^p}|x_{n}-x^{*}|^{p-1}=M\cdot 0=0$$
