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  "body": "\n## Proof\n\nWe use [[sequential continuity|0PAQM3H57X]] and the quotient rule for limits of sequences.\n\n1. Let $D \\subseteq \\R$ and let $f, g \\colon D \\to \\R$ be [[continuous|0P27W9RW0G]] with $g(x) \\ne 0$ for all $x \\in D$. Then $f/g$ is defined on all of $D$.\n2. By the equivalence of continuity and sequential continuity, $f$ and $g$ are [[sequentially continuous|0MVDQEAB60]].\n3. Let $(x_n)$ be a sequence in $D$ that converges to a point $a \\in D$. By step 2, $f(x_n) \\to f(a)$ and $g(x_n) \\to g(a)$.\n4. We have $g(a) \\ne 0$. By the [[quotient rule for limits|03S6GWWKEZ]], there is $N$ such that $(f(x_n)/g(x_n))_{n \\ge N}$ converges to $f(a)/g(a)$.\n5. Let $\\varepsilon \u003e 0$. By the [[epsilon-N form|09G6KHS6GY]] for the sequence of step 4, there is $N' \\ge N$ with $\\abs{f(x_n)/g(x_n) - f(a)/g(a)} \u003c \\varepsilon$ for all $n \\ge N'$.\n6. The same $N'$ works for the whole sequence $(f(x_n)/g(x_n))_{n \\ge 0}$, which is defined by step 1. So $(f/g)(x_n) \\to (f/g)(a)$.\n7. Since $(x_n)$ and $a$ were arbitrary, $f/g$ is sequentially continuous. By the equivalence again, $f/g$ is continuous.\n\n## Idea\n\nSequential continuity reduces the question to the quotient rule for limits of sequences. The hypothesis $g \\ne 0$ on $D$ makes the quotient defined everywhere and makes the limit $g(a)$ nonzero.\n",
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