1. ## limes proof

hello,

I am given a being the limes of a sequence an and s being a real number

(a) let all elements of the seqence be >= 0, the it follows that a >= 0 as well.
(b) if an >= s for all n, then it follows that a >= s.
(c) if an <= s for all n, then it follows that a <= s.

How can I arrive at (b) and (c) coming from a? I first thought about epsilon criteria, but that didn't get me anywhere. Greatly appreciating your help...

oz.

2. Originally Posted by ozfingwe
hello,

I am given a being the limes of a sequence an and s being a real number

(a) let all elements of the seqence be >= 0, the it follows that a >= 0 as well.
(b) if an >= s for all n, then it follows that a >= s.
(c) if an <= s for all n, then it follows that a <= s.

How can I arrive at (b) and (c) coming from a? I first thought about epsilon criteria, but that didn't get me anywhere. Greatly appreciating your help...

oz.

Oops! Misread. So given (a) consider $\displaystyle \tilde{a}_n=a_n-s$ and $\displaystyle b_n=s-a_n$

3. Originally Posted by ozfingwe
I am given a being the limits of a sequence an and s being a real number
(b) if an >= s for all n, then it follows that a >= s.
Here is another way to do part (b).
Suppose that $\displaystyle a<s$, then $\displaystyle s-a>0$.
So $\displaystyle \left( {\exists N} \right)\left[ {\left| {a_N - a} \right| < s - a} \right] \Rightarrow \quad a_N < s$.