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  "body": "\n## Proof\n\nWe use the [[epsilon-delta form of the limit|0GBZDAFPX8]] and the [[epsilon-N form|09G6KHS6GY]] for sequences. The converse is proved by contraposition.\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. Assume first that $\\lim_{x \\to a} f(x) = L$, in the sense of the [[limit of a function|0TXTSEMHR5]]. Let $(x_n)$ be a sequence in $D \\setminus \\set{a}$ with $x_n \\to a$. Let $\\varepsilon \u003e 0$.\n3. The epsilon-delta form gives $\\delta \u003e 0$ with $\\abs{f(x) - L} \u003c \\varepsilon$ for all $x \\in D$ with $0 \u003c \\abs{x - a} \u003c \\delta$.\n4. Since $x_n \\to a$, the epsilon-N form gives $N$ with $\\abs{x_n - a} \u003c \\delta$ for all $n \\ge N$.\n5. For $n \\ge N$, also $x_n \\ne a$, so $0 \u003c \\abs{x_n - a} \u003c \\delta$. Step 3 gives $\\abs{f(x_n) - L} \u003c \\varepsilon$.\n6. So [[$f(x_n) \\to L$|0XKWSSY8P2]] by the epsilon-N form. This proves the first direction.\n7. Conversely, assume that $\\lim_{x \\to a} f(x) = L$ fails. We show that the sequence condition fails.\n8. By the epsilon-delta form, there is $\\varepsilon \u003e 0$ with this property: for each $\\delta \u003e 0$ there is $x \\in D$ with $0 \u003c \\abs{x - a} \u003c \\delta$ and $\\abs{f(x) - L} \\ge \\varepsilon$.\n9. For each natural number $n$, apply step 8 with $\\delta = 1/(n+1)$. This gives $x_n \\in D$ with $0 \u003c \\abs{x_n - a} \u003c 1/(n+1)$ and $\\abs{f(x_n) - L} \\ge \\varepsilon$.\n10. Each $x_n$ lies in $D \\setminus \\set{a}$, since $x_n \\ne a$.\n11. Let $\\eta \u003e 0$. There is $m \\ge 1$ with [[$1/m \u003c \\eta$|0WH916TJNY]]. For $n \\ge m$, $\\abs{x_n - a} \u003c 1/(n+1) \u003c 1/m \u003c \\eta$. So $x_n \\to a$.\n12. But $\\abs{f(x_n) - L} \\ge \\varepsilon$ for every $n$. So no $N$ works for this $\\varepsilon$ in the epsilon-N form, and $f(x_n) \\not\\to L$.\n13. So the sequence condition fails. By contraposition, the sequence condition implies $\\lim_{x \\to a} f(x) = L$.\n\n## Idea\n\nIf $f(x)$ is close to $L$ near $a$, then $f(x_n)$ is close to $L$ once $x_n$ is near $a$. If the limit fails, one $\\varepsilon$ is violated arbitrarily close to $a$. Picking one bad point within $1/(n+1)$ of $a$ for each $n$ gives a sequence that witnesses the failure.\n",
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