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  "body": "\n## Statement\n\nLet $(a_n)$ and $(b_n)$ be real sequences that converge to real numbers $a$ and $b$.\nAssume that there is a natural number $N$ such that $a_n \\le b_n$ for all $n \\ge N$.\nThen $a \\le b$.\n\n## Remarks\n\nStrict inequalities do not pass to the limit. For example, $0 \u003c 1/n$ for all $n \\ge 1$, but $\\lim 1/n = 0$.\nWith a constant sequence $b_n = c$, it says: if $a_n \\le c$ for large $n$, then $\\lim a_n \\le c$.\nThe proof is by contradiction. If $a \u003e b$, take $\\varepsilon = (a - b)/2$. Then for large $n$ we get $b_n \u003c a_n$.\n\n## Explanations\n\n### L1 human\n\nIf one sequence is always below another, its limit cannot be above the limit of the other.\nThe limits can be equal, even when the terms are never equal.\n",
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