---
tag: 0N32V0E3XJ
kind: statement
role: proposition
name: Limits preserve non-strict inequalities
msc: [40A05]
uses: [0XKWSSY8P2, 09G6KHS6GY]
sources:
  - url: https://en.wikipedia.org/wiki/Limit_of_a_sequence
    locator: Properties
---

## Statement

Let $(a_n)$ and $(b_n)$ be real sequences that converge to real numbers $a$ and $b$.
Assume that there is a natural number $N$ such that $a_n \le b_n$ for all $n \ge N$.
Then $a \le b$.

## Remarks

Strict inequalities do not pass to the limit. For example, $0 < 1/n$ for all $n \ge 1$, but $\lim 1/n = 0$.
With a constant sequence $b_n = c$, it says: if $a_n \le c$ for large $n$, then $\lim a_n \le c$.
The proof is by contradiction. If $a > b$, take $\varepsilon = (a - b)/2$. Then for large $n$ we get $b_n < a_n$.

## Explanations

### L1 human

If one sequence is always below another, its limit cannot be above the limit of the other.
The limits can be equal, even when the terms are never equal.