# Proof about a limit property clarification

1. Jun 3, 2013

### bonfire09

The book proves this limit and I am a bit confused how all the pieces fit together.
So the book proves "If $(s_n)$ converges to $s$ and $(t_n)$ converges to $t$, then $(s_nt_n)$ converges to $st$. That is, $lim(s_nt_n) = (lim s_n)(lim t_n)$.

The proof goes like this
Let $\epsilon> 0$ . By Theorem 9.1 there is a constant $M > 0$ such that
$|s_n| ≤ M$for all $n$. Since $lim t_n = t$ there exists $N_1$ such that $n > N1$ implies $|t_n − t| <\epsilon/(2M)$ Also, since $lim s_n = s$ there exists $N_2$ such that $n > N_2$implies $|s_n − s| < \epsilon/(2(|t| + 1))$ Then $|s_nt_n − st| ≤ |s_n| · |t_n − t| + |t| · |s_n − s| ≤ M · (\epsilon/2M)+ |t| · (\epsilon/(2(|t| + 1))<\epsilon/2+\epsilon/2=\epsilon$.

The part I do not understand about the proof is this jump in the inequality in the last step that is how is $|s_n| · |t_n − t| + |t| · |s_n − s|≤ M · (\epsilon/(2M))+ |t| · (\epsilon/(2(|t| + 1))$ instead of just less than?

Last edited: Jun 3, 2013
2. Jun 3, 2013

### bonfire09

Sorry I had accidentally posted this post before I had completed it. So I made several changes to it but now its completed.

3. Jun 3, 2013

### Office_Shredder

Staff Emeritus
If your confusion is just in that you expect a $<$ and the book uses a $\leq$, then it's a simple observation that if $a < b$, it is certainly true that $a\leq b$, but maybe it would have been more consistent on the part of the author to use a strict inequality there

4. Jun 3, 2013

### bonfire09

Oh yes I did forget to ask I also noticed that the author selects two different epsilon values that each limit that is the $\epsilon/2M$ and $\epsilon/(2|t|+1).$ that seem to be completely arbitrary. If I redid the proof and could I chose different values than the author? And also how does the author verify that $M⋅\epsilon/2M+|t|⋅(ϵ/(2(|t|+1))<ϵ /(2)+ϵ /(2)$?. It seems that should require another proof on its own or is it just possible just by observation?

Last edited: Jun 3, 2013
5. Jun 3, 2013

### bonfire09

Well anyways I think I figured out a simpler way of doing this proof. By definition of convergence let $\epsilon>0$. Since all convergent sequences are bounded there exists a real number call it $M$ such that $M>0 \text{ and }\ |s_n| < M \text{ for all n }$. Since
$lim t_n=t \text{ there exists }\ N_1 \text{ such that }\ n>N_1 \text{ implies } |t_n-t|< \frac{\epsilon}{2M}$. Similarly for $lim s_n=s \text{ there exists} \ N_2 \text{ such that }\ n>N_2 \text{ implies } |s_n-s|< \frac{\epsilon}{2M}$. Thus we get $|s_nt_n- st|≤|s_n|*|t_n-t|+|t|*|s_n-s|< M*\frac{\epsilon}{2M} + M*\frac{\epsilon}{2M}=\epsilon$. Hence $lim s_nt_n=(lim s_n)(lim t_n)$.

Last edited: Jun 3, 2013
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