Real Analysis: Proof of convergence

[uva123]

Homework Statement

Prove if {b_n} converges to B and B ≠ 0 and b_n ≠ 0 for all n, then there is M>0 such that |b_n|≥M for for all n.

Homework Equations

What I have so far:
I know that if {b_n} converges to B and B ≠ 0 then their is a positive real number M and a positive integer N such that if n≥N, then |b_n|≥M . (by lemma)
PROOF (of lemma)- (Note: let E be epsilon)
since B ≠ 0, (|B|)/2=E>0. There is N such that if n≥N, then
|b_n-B|<E. Let M=[(|B|)/2]. thus for n≥N,
|b_n|=|b_n-B+B|≥|B|-|b_n-B|≥|B|-[(|B|)/2]=[(|B|)/2]=M

The Attempt at a Solution

i know that this is not the entire proof i need but i don't know what changes need to be made. what variations do i need to make in the proof for the lemma? please help point me in the right direction!

 

[dickdick]

I don't see anything you need to change or amplify. Why do you think there is a weakness?

 

[silvermane]

Homework Statement

Prove if {b_n} converges to B and B ≠ 0 and b_n ≠ 0 for all n, then there is M>0 such that |b_n|≥M for for all n.

Homework Equations

What I have so far:
I know that if {b_n} converges to B and B ≠ 0 then their is a positive real number M and a positive integer N such that if n≥N, then |b_n|≥M . (by lemma)
PROOF (of lemma)- (Note: let E be epsilon)
since B ≠ 0, (|B|)/2=E>0. There is N such that if n≥N, then
|b_n-B|<E. Let M=[(|B|)/2]. thus for n≥N,
|b_n|=|b_n-B+B|≥|B|-|b_n-B|≥|B|-[(|B|)/2]=[(|B|)/2]=M

The Attempt at a Solution

i know that this is not the entire proof i need but i don't know what changes need to be made. what variations do i need to make in the proof for the lemma? please help point me in the right direction!

I think it looks fine. :)