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The sequence of $\begin{Bmatrix} {x}_{n} \end{Bmatrix}$ is strictly decreasing,$\quad \lim_{n\to\infty }{x}_{n}=0 \quad $,and$\quad \lim_{n\to\infty }{y}_{n}=0 .$

From the above conditions,whether we can get the following conclusion:

If $\quad \lim_{n\to\infty }\frac{{y}_{n+1}-{y}_{n}}{{x}_{n+1}-{x}_{n}}=\infty. then \quad \lim_{n\to\infty }\frac{{y}_{n}}{{x}_{n}}=\infty $.

If the conclusion is correct, please give the proof; or some ounterexamples.

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I think the question meant to ask to find a counterexample to the claim that the implication hold. –  Gina Jun 24 at 6:43
    
I'm not sure what you meant by $\infty$ is not $+\infty$? Can you clarify exactly what needed to changed by editing your question please? –  Gina Jun 24 at 8:47

1 Answer 1

(The question is a bit confusing. I am assuming that you are looking for counterexample to the claim where the last implication hold, ie. an example to the claim you put in the question)

No counterexample exist, since the implication indeed hold. For any $M>0$ there exist an $N$ such that for all $n>N$ then $\frac{y_{n}-y_{n+1}}{x_{n}-x_{n+1}}>M$. Consider any fixed $n>N$. Then we have $y_{k}-y_{k+1}>M(x_{k}-x_{k+1})$ for all $k\geq n$ so by performing the telescoping sum we get $y_{n}-y_{k}>M(x_{n}-x_{k})$. Taking the limit as $k\rightarrow\infty$ give us $y_{n}-0\geq M(x_{n}-0)$ so $\frac{y_{n}}{x_{n}}\geq M$. Hence for any $M>0$ there exist an $N$ such that for all $n>N$ then $\frac{y_{n}}{x_{n}}\geq M$. In other word, $\lim\limits_{n\rightarrow\infty}\frac{y_{n}}{x_{n}}=\infty$

EDIT: In light of the OP's comment: if the conclusion were meant to be:

$\lim\limits_{n\rightarrow\infty}|\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}|=\infty$ then $\lim\limits_{n\rightarrow\infty}|\frac{y_{n}}{x_{n}}|=\infty$

(ie. add in the absolute value) then this one is incorrect and a counterexample can be given:

$y_{n}=\frac{(-1)^{n}}{n};x_{n}=\frac{1}{n}$

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