# Limits of convergent sequences, where one is strictly smaller

So I have been learning about sequences recently and I am struggling with the following problem:

Given two convergent sequences $$(a_n)_{n \geq 0}$$ and $$(b_n)_{n \geq 0}$$, where $$a_n < b_n$$ for all $$n$$, prove that $$\lim a_n \leq \lim b_n$$ follows.

It seems obvious that this would be true, but I can't just quite prove it. I thought maybe looking at the difference of the two sequences $$b_n - a_n$$ would be a good idea. Since $$b_n > a_n$$ for all $$n$$ this sequence would contain only positive numbers. If I could somehow show, that this always converges to some $$d \geq 0$$ than I thought this would prove it, however I haven't really been able to make some progress there so now I'm back to square one. I would really appreciate some help.

Let $$d_n\geq 0$$. If $$d=\lim d_n<0$$ then for $$\varepsilon =-d/2>0$$ there is $$N$$ such that for $$n>N$$ $$|d_n-d|<-d/2$$. But $$|d_n-d|=d_n-d\geq -d> -d/2$$. A contradiction.
• I think you made a small error at the beginning. Shouldn't it be $d = \lim d_n < 0$? – roblox99 Mar 7 at 18:13
• However I only have one question left. Would this proof still be valid if if $d_n$ didn't have a limit? Or is it guaranteed to have one since its the difference of two convergent sequences? – roblox99 Mar 7 at 18:19