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  "body": "\n## Proof\n\nWe extend $f$ and $g$ by $0$ at $c$ and compare $f(x)/g(x)$ with a quotient of derivatives through the Cauchy mean value theorem. Here $L$ is a real number.\n\n1. Define $F, G \\colon I \\to \\R$ by $F(x) = f(x)$ and $G(x) = g(x)$ for $x \\ne c$, and $F(c) = G(c) = 0$.\n2. For $x \\ne c$, the functions $F$ and $G$ agree with $f$ and $g$ near $x$. So they are differentiable at $x$, and hence [[continuous|07ZMX4ANJS]] there.\n3. At $c$, we have $F(x) \\to 0 = F(c)$ and $G(x) \\to 0 = G(c)$ as $x \\to c$. So $F$ and $G$ are [[continuous at|0EEE2CZB2A]] $c$.\n4. Fix $x \\in I$ with $x \\ne c$. Let $J$ be the closed interval with endpoints $c$ and $x$. Since $I$ is an [[interval|0TK8Y3J5RW]], $J \\subseteq I$.\n5. On $J$, $F$ and $G$ are continuous. On the interior of $J$, they are differentiable with derivatives $f'$ and $g'$.\n6. By the [[mean value theorem|0BARQY52N5]] on $J$, there is $s$ strictly between $c$ and $x$ with $G(x) - G(c) = g'(s)(x - c)$. This form holds whichever endpoint is smaller.\n7. So $g(x) = g'(s)(x - c)$. Both factors are nonzero, so $g(x) \\ne 0$.\n8. Let $\\alpha \u003c \\beta$ be the endpoints of $J$. The [[Cauchy mean value theorem|02SYZAWM2K]] gives $t$ strictly between them with $(F(\\beta) - F(\\alpha)) G'(t) = (G(\\beta) - G(\\alpha)) F'(t)$.\n9. Swapping $\\alpha$ and $\\beta$ multiplies both sides by $-1$. So $(F(x) - F(c)) g'(t) = (G(x) - G(c)) f'(t)$, that is, $f(x) g'(t) = g(x) f'(t)$.\n10. Dividing by $g(x) g'(t) \\ne 0$ gives $\\dfrac{f(x)}{g(x)} = \\dfrac{f'(t)}{g'(t)}$, where $t$ lies strictly between $c$ and $x$.\n11. Now let $V$ be an open interval containing $L$. By the definition of the [[limit|0TXTSEMHR5]], there is an open interval $U$ containing $c$ with $f'(t)/g'(t) \\in V$ for all $t \\in U \\cap I$ with $t \\ne c$.\n12. The set $W = U \\cap I$ is an open interval containing $c$. Let $x \\in W$ with $x \\ne c$.\n13. The point $t$ of step 10 lies between $c$ and $x$, so $t \\in W$, and $t \\ne c$. Hence $f(x)/g(x) = f'(t)/g'(t) \\in V$.\n14. So for every open interval $V$ containing $L$ there is an open interval $W$ containing $c$ with $f(x)/g(x) \\in V$ for $x \\in W$, $x \\ne c$.\n15. By the definition of the limit, $f(x)/g(x) \\to L$ as $x \\to c$.\n\n## Idea\n\nOnce $f$ and $g$ are set to $0$ at $c$, the quotient $f(x)/g(x)$ is a quotient of increments over the interval from $c$ to $x$. The Cauchy mean value theorem says this equals $f'(t)/g'(t)$ at some $t$ closer to $c$ than $x$. So whatever $f'/g'$ does near $c$, $f/g$ does too.\n",
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