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  "body": "\n## Counterexample\n\nThe function $f \\colon \\mathbb{R} \\to \\mathbb{R}$ with $f(x) = x^2$ is [[continuous|0P27W9RW0G]] but not [[uniformly continuous|0Q4DR37DVE]].\nFor $n \\ge 1$, let $x_n = n$ and $y_n = n + 1/n$. Then $|x_n - y_n| = 1/n \\to 0$, but\n$$f(y_n) - f(x_n) = 2 + \\frac{1}{n^2} \u003e 2.$$\nSo no $\\delta$ works for $\\varepsilon = 2$.\n\n## Remarks\n\nOn each bounded interval $[-c, c]$, the function $x^2$ is Lipschitz with constant $2c$. So it is uniformly continuous there.\nThe problem is that the slope $2x$ is not bounded on $\\mathbb{R}$.\n\n## Explanations\n\n### L1 human\n\nThe graph of $x^2$ gets steeper and steeper as $x$ grows.\nSo a fixed small change in $x$ gives larger and larger changes in $x^2$.\n",
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  "name": "x^2 is not uniformly continuous on the real line",
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