Math Help - limit of an increasing sequence

1. limit of an increasing sequence

2. Originally Posted by john_n82

Since x is the supremum of { $x_{n}:n\in n$},

GIVEN ε>0 there exists kεN AND SUCH that:

$x-\epsilon< x_{k}\leq x

But since { $x_{n}$} is increasing and bounded by x:

for all n $x_{n}\leq x$ and

for $n\geq k$ $x_{n}\geq x_{k}$

hence :

$x-\epsilon

THUS $\lim_{n\rightarrow\infty}{x_{n}} = x$

3. Thank you for responding. There is a line that I don't understand. Could you please explain further more?

Originally Posted by xalk
for $n\geq k$ $x_{n}\geq x_{k}$

4. Originally Posted by john_n82
Thank you for responding. There is a line that I don't understand. Could you please explain further more?
You know that the sequence is increasing.

Therefore, $x_n < x_{n + 1} < x_{n + 2} < \cdots < x_{n + K}$.
So if $j>n$ then $x_n < x_j$