# Inequality between two sequences preserved in the limit? [duplicate]

Let $(a_n)_{n\in \mathbb{N}}$ and $(b_n)_{n\in \mathbb{N}}$ be two real sequences that satisfy $a_n\geq b_n, \forall n \in \mathbb{N}$ and converge to some $a,b$, respectively.

Is it always true that $a \geq b$?

## marked as duplicate by Martin Sleziak, R_D, JMP, M. Vinay, Davide Giraudo real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 7 '16 at 9:43

• Yes. What can you say about the limit of $(a_n-b_n)$? – David Mitra Jun 29 '13 at 12:55
• Suppose you have $a_n \geqslant b_n$, $a_n \to a$, $b_n \to b$ and $a < b$. Try to find a contradiction. – Daniel Fischer Jun 29 '13 at 12:55
Yes. If you assume that $a < b$, you can take $\varepsilon = (b-a)/2$ and use the definition of the limit of a sequence to come up with a contradiction, i.e. find an $n$ such that $a_n < b_n$.