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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.

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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 $$\lim_{n\to+\infty}\frac{|x_{n+1}-x^{*}|}{|x_{n}-x^{*}|^p}|x_{n}-x^{*}|^{p-1}=M\cdot 0=0$$

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  • $\begingroup$ 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? $\endgroup$ Commented Dec 16, 2020 at 17:38
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    $\begingroup$ 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. $\endgroup$
    – Gauge_name
    Commented Dec 16, 2020 at 17:40

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