Cauchy sequence in real numbers

I am struggling with proving this question

Show that if $a_n \leq x$ for all $n \geq 1$, then (the formal limit when n goes to infinity) $\lim_{n\rightarrow \infty} a_n \leq x$.

I am assuming by contradiction that $\lim_{n \rightarrow \infty}a_n > x$.

Then there exists a rational $q$ between them s.t $x < q < \lim_{n \rightarrow \infty}a_n$. Then $x = \lim x < q = \lim q < \lim a_n$, but I could not go to a contradiction from this. On the other hand, I am thinking to prove it by saying $x = \lim b_n$, and then if $a_n > b_n$ then we will have $x = \lim b_n < \lim a_n \leq \lim x = x$ which is a contradiction. So what is the better way to prove it? Thanks for any help.

• s.t x<q< LIM(a_n). Then x=LIM(x)<q=LIM(q)<LIM(a_n), but I could not go to a contradiction from this. On the other hand, I am thinking to prove it by saying x=LIM(b_n), and then if a_n>b_n then we will have x=LIM(b_n)<LIM(a_n)<=LIM(x)=x which is a contradiction. So what is the better way to prove it? Thanks for any help. – Ahmed Sep 14 '17 at 18:02
• Sorry this is the completeness for the question – Ahmed Sep 14 '17 at 18:03

Assume that $L:=\lim_{n\to +\infty} a_n>x$. By definition of limit, there is $N$ such that $L-\epsilon<a_n<L+\epsilon$ for all $n\geq N$. Try to find a contradiction by letting $\epsilon=(L-x)/2>0.$
Suppose $\lim_{n\to\infty}a_n=L>x$, and let $\epsilon=\frac{L-x}{2}$. Use the definition of the limit to show that for $n$ sufficiently large, $\lvert a_n-L\rvert<\epsilon$. Show that $a_n$ being this close to $L$ is a contradiction.