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Given that $\lim_{n\to \infty}a_n=a\neq0$, prove that: $$\lim_{n\to \infty}\frac{a_{n+1}}{a_{n}}=1$$

I have to prove it by the definition of limit ($\forall$ $\epsilon$ > 0 $\exists$ N $\in \mathbb{N}$ such that $\forall$ n > N : $|\frac{a_{n+1}}{a_n} - 1| < \epsilon$)

I've struggled to come up with an idea.

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    $\begingroup$ ${a_{n+1}\over a_n}-1={1\over a_n}\bigl[ (a_{n+1} - a) -( a_n - a)\bigr]$. $\endgroup$ – David Mitra Nov 16 '19 at 16:34
  • $\begingroup$ By using the triangle inequality I get $\frac{1}{a_n} \cdot 2 \cdot \epsilon$ how do I continue from there? $\endgroup$ – Daniel Segal Nov 16 '19 at 16:41
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    $\begingroup$ There is an $N>0$ and a $\delta>0$ with $|a_m|>\delta$ for all $m\ge N$ (since your limit is non-zero). $\endgroup$ – David Mitra Nov 16 '19 at 16:44
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Hint: Since $\lim_{n \to \infty} a_n = a$, we have that for any $\delta>0$, eventually $a_n,a_{n+1} \in (a-\delta,a+\delta)$. Thus $$\frac{a_{n+1}}{a_n} < \frac{a+\delta}{a-\delta} =1+\frac{2\delta}{a-\delta}$$ and $$\frac{a_{n+1}}{a_n}> \frac{a-\delta}{a+\delta} = 1-\frac{2\delta}{a+\delta}.$$ Now for a given $\epsilon>0$, you need to show you can choose $\delta$ so that both $\frac{2\delta}{a-\delta}<\epsilon$ and $\frac{2\delta}{a+\delta}<\epsilon$. Your choice of $\delta$ might depend on $a$.

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