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  "body": "\n## Proof\n\n1. From the [[defining series|0WB57D9488]], $\\sin 0 = 0$, since every term vanishes at $0$. Also $\\cos 0 = 1$, since only the first term of the cosine series is nonzero.\n2. For $x \\ne 0$, $\\frac{\\sin x}{x} = \\frac{\\sin(0 + x) - \\sin 0}{x}$. This is the difference quotient of $\\sin$ at $0$ with increment $x$.\n3. By the [[derivative of sine|00R4MWW532]], $\\sin$ is differentiable at $0$ with $\\sin'(0) = \\cos 0 = 1$.\n4. By the [[definition of the derivative|0F1WR9C83K]], the difference quotient in step 2 tends to $\\sin'(0)$ as $x \\to 0$.\n5. So $\\lim_{x \\to 0} \\frac{\\sin x}{x} = 1$.\n\n## Idea\n\nSince $\\sin 0 = 0$, the quotient $\\frac{\\sin x}{x}$ is exactly the slope of the secant of $\\sin$ from $0$ to $x$. As $x \\to 0$ this slope tends to the slope of the tangent at $0$, which is $\\cos 0 = 1$.\n",
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