If $x_n$ is an monotonically increasing sequence that converges to $x$ then all terms of $x_n$ is less than for equal to $x$

Question

I have come across that if a sequence of real numbers, say $x_n$ which is monotonically increasing and converges to a real number $x$ then $x_n \leq x$ for all $n\geq 1$. Is this statement is correct? If correct then why , according to me suppose if $x_1>x$ then since X-n is monotonically increasing so, $x_2\geq x_1$ so, distance between $x_2$ and $x$ is more than $x_1$ and $x$, similarly all terms will go further and further from x , and if we choose $\epsilon$ to be $d(x_1,x)/2$ then no terms of the sequence will be in this $\epsilon$- neighborhood of x which is a contradiction/ AM I right ?? Is there any rigorous proof of this??

Answer 1

Yes, you are right. You took $\varepsilon=\frac12\lvert x_1-x\rvert$, but I would even drop that $\frac12$ factor, since it is not needed.

Answer 2

You can also use the fact that limits preserve inequalities weakly. In symbols if $a_n\leq b_n$ then $\lim_{n\to \infty} a_n\leq \lim_{n\to\infty} b_n$.

Now let $m$ be a fixed positive integer. Then since $x_n$ is increasing we have $x_m\leq x_n$ for all $n>m$. Taking limits as $n\to \infty $ we see that $x_m\leq x$. Since $m$ is arbitrary we are done.

Note that if the sequence $x_n$ is strictly increasing then we have $x_n<x$ for all positive integers $n$. From the above argument we have $x_n\leq x$ and replacing $n$ by $(n+1)$ we get $x_{n+1}\leq x$. Since $x_n$ is strictly increasing we have $x_n<x_{n+1}$ and thus we get $x_n<x$.