# Simple limit I need to prove by limit definitions

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.

• ${a_{n+1}\over a_n}-1={1\over a_n}\bigl[ (a_{n+1} - a) -( a_n - a)\bigr]$. – David Mitra Nov 16 '19 at 16:34
• By using the triangle inequality I get $\frac{1}{a_n} \cdot 2 \cdot \epsilon$ how do I continue from there? – Daniel Segal Nov 16 '19 at 16:41
• There is an $N>0$ and a $\delta>0$ with $|a_m|>\delta$ for all $m\ge N$ (since your limit is non-zero). – David Mitra Nov 16 '19 at 16:44

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 $$a$$.