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  "body": "\n## Example\n\nLet $a_n = 1/n$ for each integer $n \\ge 1$.\nThen $a_n \\to 0$ as $n \\to \\infty$.\nLet $\\varepsilon \u003e 0$. By the Archimedean property of $\\mathbb{R}$, there is a natural number $N$ with $N \u003e 1/\\varepsilon$.\nFor each $n \\ge N$ we get $|a_n - 0| = 1/n \\le 1/N \u003c \\varepsilon$.\nBy the [[epsilon-N definition|09G6KHS6GY]], this shows $1/n \\to 0$.\n\n## Remarks\n\nThe proof uses the Archimedean property, which says that the natural numbers are not bounded above in $\\mathbb{R}$.\nWith the [[algebra of limits|07XBM3A9J5]], it gives many more limits, for example $1/n^2 \\to 0$ and $(n + 1)/n \\to 1$.\nMathlib states it for $1/n$ over $\\mathbb{R}$ as `tendsto_one_div_atTop_nhds_zero_nat`.\n\n## Explanations\n\n### L1 human\n\nThe numbers $1, 1/2, 1/3, 1/4, \\dots$ get smaller and smaller, and they approach $0$.\nTo get $1/n$ below $0.001$, it is enough to take $n \u003e 1000$.\n",
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