# 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 $$\lim_{n\to+\infty}\frac{|x_{n+1}-x^{*}|}{|x_{n}-x^{*}|^p}|x_{n}-x^{*}|^{p-1}=M\cdot 0=0$$
• I understand that $\lim_{n\to+\infty}|x_{n}-x^{*}|=0$ But regarding the power that we can take it to, can we just make it p instead of p-1 as long as p is positive? Commented Dec 16, 2020 at 17:38
• Yes, of course. This fact holds because of the fact that the function $|y|\mapsto|y|^p$ is continuous for every $p>0$. Therefore you can bring the limit inside. Commented Dec 16, 2020 at 17:40