Proof about a limit property clarification

[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| ≤ Mfor all n. Since lim t_n = t there exists N_1 such that n &gt; N1 implies |t_n − t| &lt;\epsilon/(2M) Also, since lim s_n = s there exists N_2 such that n &gt; N_2implies |s_n − s| &lt; \epsilon/(2(|t| + 1)) Then |s_nt_n − st| ≤ |s_n| · |t_n − t| + |t| · |s_n − s|<br /> ≤ M · (\epsilon/2M)+ |t| · (\epsilon/(2(|t| + 1))&lt;\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?

 

[bonfire09]

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

 

[Office_Shredder]

If your confusion is just in that you expect a &lt; and the book uses a \leq, then it's a simple observation that if a &lt; 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

 

[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))&lt;ϵ /(2)+ϵ /(2)?. It seems that should require another proof on its own or is it just possible just by observation?

 

[bonfire09]

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