{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe compare open intervals with symmetric intervals $(c - r, c + r)$.\n\n1. Let $D \\subseteq \\R$, let $f \\colon D \\to \\R$, let $a$ be a limit point of $D$, and let $L \\in \\R$.\n2. Claim: if $V = (p, q)$ is an [[open interval|0TK8Y3J5RW]] containing a point $c$, then $r = \\min(c - p, q - c)$ is positive and $(c - r, c + r) \\subseteq V$.\n3. Proof of the claim: $p \u003c c \u003c q$ gives $r \u003e 0$. If $\\abs{y - c} \u003c r$, then $y \u003e c - r \\ge p$ and $y \u003c c + r \\le q$, so $y \\in V$.\n4. Assume first that $\\lim_{x \\to a} f(x) = L$ in the sense of the [[definition|0TXTSEMHR5]]. Let $\\varepsilon \u003e 0$.\n5. The set $V = (L - \\varepsilon, L + \\varepsilon)$ is an open interval containing $L$. The definition gives an open interval $U$ containing $a$ with $f(x) \\in V$ for all $x \\in U \\cap D$ with $x \\ne a$.\n6. By the claim, there is $\\delta \u003e 0$ with $(a - \\delta, a + \\delta) \\subseteq U$.\n7. Let $x \\in D$ with $0 \u003c \\abs{x - a} \u003c \\delta$. Then $x \\in U \\cap D$ and $x \\ne a$. So $f(x) \\in V$, that is, $\\abs{f(x) - L} \u003c \\varepsilon$.\n8. So the epsilon-delta condition holds.\n9. Conversely, assume the epsilon-delta condition. Let $V$ be an open interval containing $L$.\n10. By the claim, there is $\\varepsilon \u003e 0$ with $(L - \\varepsilon, L + \\varepsilon) \\subseteq V$. The condition gives $\\delta \u003e 0$ for this $\\varepsilon$.\n11. Let $U = (a - \\delta, a + \\delta)$. This is an open interval containing $a$.\n12. Let $x \\in U \\cap D$ with $x \\ne a$. Then $0 \u003c \\abs{x - a} \u003c \\delta$, so $\\abs{f(x) - L} \u003c \\varepsilon$. Hence $f(x) \\in (L - \\varepsilon, L + \\varepsilon) \\subseteq V$.\n13. So $\\lim_{x \\to a} f(x) = L$ in the sense of the definition. This proves the equivalence.\n\n## Idea\n\nThe definition speaks of arbitrary open intervals, the condition of symmetric ones of radius $\\varepsilon$ or $\\delta$. Every symmetric interval is an open interval, and every open interval around a point contains a symmetric one. So the two conditions say the same thing.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "26A03"
    ],
    "name": "Proof of the epsilon-delta form of the limit of a function",
    "proves": "0GBZDAFPX8",
    "sources": [
      {
        "locator": "(ε, δ)-definition of limit",
        "url": "https://en.wikipedia.org/wiki/Limit_of_a_function"
      }
    ],
    "tag": "QFP0HEWJDZ",
    "uses": [
      "0TXTSEMHR5",
      "0TK8Y3J5RW"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/QFP0HEWJDZ/",
  "links": [
    {
      "field": "proves",
      "target": "0GBZDAFPX8"
    },
    {
      "field": "uses",
      "target": "0TXTSEMHR5"
    },
    {
      "field": "uses",
      "target": "0TK8Y3J5RW"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/QFP0HEWJDZ.md",
  "msc": [
    "26A03"
  ],
  "name": "Proof of the epsilon-delta form of the limit of a function",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/QFP/QFP0HEWJDZ.md",
  "status": null,
  "tag": "QFP0HEWJDZ",
  "uri": "m2://QFP0HEWJDZ",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-epsilon-delta-form-of-the-limit-of-a-function/",
  "writing": {
    "tag": "QFP0HEWJDZ",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}