real analysis - Is my proof that $limlimits_{nto +infty}dfrac{u_{n+1}}{u_n}=1$ correct?

I'm doing an exercise where $(u_n)$ is a numerical sequence which is decreasing and strictily positive.While $(u_n)$ is a numerical sequence which is decreasing and strictily 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

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 '}

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 $l\ne 0$.

Answer

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.