1. proof for sequences

Given that sequences a and b converges respectively, to the limits L1 and L2. Use def of limit of sequence to prove that if a>=b for all natural numbers n, then L1>=L2.

Suppose on the contrary that L1 < L2. Let epsilon = (L2-L1)/2
, which is positive.
Let n = max(N1;N2) + 1. Then n > N1 and n > N2. So we have
...............

a) why are they using epsilon = (L2-L1)/2
b) what does it mean by 'n=max(N1,N2) + 1?' why using this?

calculus is really giving me headache now.

2. Originally Posted by serenalee
Given that sequences a and b converges respectively, to the limits L1 and L2. Use def of limit of sequence to prove that if a>=b for all natural numbers n, then L1>=L2.

Suppose on the contrary that L1 < L2. Let epsilon = (L2-L1)/2
, which is positive.
Let n = max(N1;N2) + 1. Then n > N1 and n > N2. So we have
...............

a) why are they using epsilon = (L2-L1)/2
b) what does it mean by 'n=max(N1,N2) + 1?' why using this?

calculus is really giving me headache now.
The definition of the limit of sequence is
$\forall\epsilon>0,\exists N,\forall n>N,|a_n-a|<\epsilon$,right?
I guess you may have some trouble with understanding this definition. So you may define $\lim_{n\rightarrow \infty} a_n=a$as:
if $n$ is large enough, then $|a_n-a|$will be as small as possible.
Let's do this problem in this way.
Let $\lim_{n\rightarrow \infty} a_n=L_1,\lim_{n\rightarrow \infty}b_n=L_2$
Then we suppose $L_1.
Imagine a $N$ which is extremely large. This $N$ makes $|a_n-L_1|,|b_n-L_2|$ both extremely small from $n>N$. Since $L_1, $|L_1-L_2|$can not be "extremely small". If you write $a_n\approx L_1,b_n\approx L_2$ for the sake of convenience, then we will have $a_nsince $a_n\approx L_1. So we have found the contradiction.
If you understand this, then all the things such as $(L_2-L_1)/2,\max\{N_1,N_2\}+1$will become technical ,trivial and easy to understand!!