Is my proof that $\lim\limits_{n\to +\infty}\frac{u_{n+1}}{u_n}=1$ correct? I'm doing an exercise where $(u_n)$ is a numerical sequence which is decreasing and strictly positive.While $(u_n)$ is a numerical sequence which is decreasing and strictly positive, then $(u_n)$ is convergent and its limit is positive which we symbolise by $l$. Assume that $l\ne 0$.
I have to prove that  $\lim\limits_{n\to +\infty}\dfrac{u_{n+1}}{u_n}=1$. I'm not sure if my proof is correct or not. Can you please check it? Thank you very much!
Please excuse my English. We don't study Maths in English.
Let $\varepsilon\in ]0;l[$.
So $\exists N\in\mathbb{N},\,\forall n\in\mathbb{N},\,n>N\Longrightarrow |u_n-l|<\varepsilon$
Let $n\in\mathbb{N}$ such as $n>N$. We also have $n+1>n>N$.
Then:
$|u_{n+1}-u_n|=|(u_{n+1}-l)-(u_n-l)|\le |u_{n+1}-l|+|u_n-l|<2\varepsilon$ $(1)$
And we have $|u_n-l|<\varepsilon$ so $0<l-\varepsilon<u_n<l+\varepsilon$ and so whe have $0<\dfrac{1}{u_n}<\dfrac{1}{l-\varepsilon}$ $(2)$
Then $(1)\times (2)$ gives:
$\left|\dfrac{u_{n+1}}{u_n}-1\right|<\dfrac{2\varepsilon}{l-\varepsilon}$
We put $\varepsilon '=\dfrac{2\varepsilon}{l-\varepsilon}>0$. Then $\varepsilon=\dfrac{l\varepsilon '}{2+\varepsilon '}>0$.
While $\varepsilon '>0$ then $\dfrac{\varepsilon '}{2+\varepsilon '}<1$ and because $l>0$ we have then $\varepsilon=\dfrac{l\varepsilon '}{2+\varepsilon '}<l$
And so $\forall\varepsilon '\in\mathbb{R}^{+*},\,\exists\varepsilon\in ]0,l[,\,\varepsilon=\dfrac{l\varepsilon '}{2+\varepsilon '}$ and so $\varepsilon '$ covers $\mathbb{R}^{+*}$ where $\mathbb{R}^{+*}$ is the set of strictly positive real numbers. As a result we have then:$$\forall\varepsilon '\in\mathbb{R}^{+*},\,\exists N\in\mathbb{N},\,\forall n\in\mathbb{N},\, n>N\Longrightarrow\left|\dfrac{u_{n+1}}{u_n}-1\right| <\varepsilon '$$
And so  $\lim\limits_{n\to +\infty}\dfrac{u_{n+1}}{u_n}=1$
Edit: $\mathbb{R}^{+*}$ is the set of strictly positive real numbers.
Edit2: Assume that $l\ne 0$.
 A: Not really an answer to your question, but an alternative approach.
If $u_n\to \ell_1$ and $v_n\to \ell_2$ with $l_1\neq 0$, then $\frac{v_n}{u_n}\to \frac{\ell_2}{\ell_1}$ as a general rule. So if $v_n=u_{n+1}$...
A: As an exercise, we give a detailed argument  directly from the definition. Suppose that the sequence $(u_n)$ has limit $a\gt 0$. We want to show that for any $\epsilon\gt 0$, there is an $N$ such that 
$$1-\epsilon\lt \frac{u_{n+1}}{u_n}\le 1\tag{1}$$
if $n\gt N$. Note that 
$$\frac{u_{n+1}}{u_n}\ge \frac{a}{u_n},$$ so it suffices to make $\frac{a}{u_n}\gt 1-\epsilon$. This will be the case automatically if $\epsilon\ge 1$, so we may suppose that $\epsilon\lt 1$.
For $0\lt \epsilon\lt 1$ we have 
$$\frac{a}{u_n}\gt 1-\epsilon \quad\text{iff}\quad u_n \lt \frac{a}{1-\epsilon}   \quad\text{iff}\quad u_n-a\lt \frac{a\epsilon}{1-\epsilon}.$$
Since the sequence $(u_n)$ converges to $a$, there is an $N$ such that if $n\gt N$, then $u_n-a\lt \frac{a\epsilon}{1-\epsilon}$. For any such $n$, Inequality (1) will hold.
Remark: Informally, this is simpler than it looks. We can scale the sequence $(u_n)$ so that it has limit $1$. That does not change ratios. 
