$\left(x_n\right) \in \mathbb{R}$ which converges to $l$ where $\alpha
Given a sequence $\left(x_n\right) \in \mathbb{R}$  which converges to $l \in \mathbb{R} $ where $$ \alpha<x_n \text{      for all  } n \in \mathbb{N}.$$ show that $\alpha \leq l$.
I tried to prove this. I got an idea but I cannot express myself mathematically which is 100% correct. Here is my proof, Suggest a Correct Mathematical Proof for pure maths students. Thank You.
Attempt
Given $\left(x_n\right)$ converges to $l$
$\implies$ for any $\epsilon >0$ there exists an $n_0\left(\epsilon\right)$ such that for all $ n\geq n_0\left(\epsilon\right)$ we have $l-\epsilon < x_n < l+ \epsilon$ because  $\alpha < x_n$, this we have $\alpha< x_n<l-\epsilon$ for all $ n\geq n_0\left(\epsilon\right)$.
Hence we have $\alpha- l<\epsilon$ hence $\alpha- l$ cannot take positive values hence $\alpha- l \leq 0$ hence the conclusion.
But I am not sure if I had provided every required details, I need someone to make it Logically and Mathematically correct.
 A: Suppose for a contradiction that $\alpha > l$. Then $\varepsilon = \alpha - l > 0$. Since $x_n \rightarrow l$, therefore $|x_n - l| < \varepsilon = \alpha - l$ for sufficiently large $n$. Then for sufficiently large $n$
$x_n - l \leq |x_n - l| < \alpha - l \Longrightarrow x_n < \alpha$
This contradicts the fact that $x_n > \alpha$ for all $n$. 
A: If $\alpha >l$ take $\epsilon =\alpha -l$ to get a  contradiction.
A: You know that for each $\epsilon>0$, there is an $n_{\epsilon}$ such that for $n\geq n_{\epsilon}$, $\alpha<x_{n}<l+\epsilon$. Thus, $\alpha<l+\epsilon$. But this holds for all $\epsilon$, so it must be that $\alpha\leq l$. 
A: $\alpha \lt x_n \forall n .$
Now,$(x_n)\to l\implies \exists$ finitely many n such that  $ x_n \not\in(l-\varepsilon,l+\varepsilon)\therefore\alpha\lt x_n(\forall n) \le l-\varepsilon$(for finitely many n)$ \lt l, $.That is $\alpha\lt l$.
The case,$\alpha=l$ may occur when,$l \lt x_n \forall n$,i.e. $(x_n)$ is decreasing and  doesn't attain its limit($x_n \ne l $,for any n).For example,$\left(\dfrac{1}{n}\right)\to 0$,but any term of the sequence never $=0$.In such case,we may have $\alpha=l$
