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  "body": "\n## Proof\n\n1. Fix a real $a \u003e 0$, and let $b = a/2 \u003e 0$. For $x \u003e 0$, [[real powers|0TV07AQT62]] are $x^c = e^{c \\ln x}$, where $\\ln$ is the [[natural logarithm|0044CPHJBV]].\n2. Let $x \\ge 1$, and let $t = \\ln x$. Since [[$\\ln$ is strictly increasing|0RKW2X63YR]] and $\\ln 1 = 0$, we have $t \\ge 0$.\n3. By the inequality [[$1 + y \\le e^y$|012C7GFF20]] with $y = b t$, we get $b t \\le 1 + b t \\le e^{b t} = x^b$. So $\\ln x \\le \\frac{1}{b} x^b$.\n4. By the [[exponential of a sum|0E8HNVBCSP]], $x^a = e^{2 b t} = e^{b t} e^{b t} = x^b \\cdot x^b$. Also $x^b \u003e 0$ by [[positivity of the exponential|0P9BNBW2W3]].\n5. Dividing step 3 by $x^a$, and using $\\ln x \\ge 0$, gives $0 \\le \\frac{\\ln x}{x^a} \\le \\frac{1}{b \\, x^b}$ for every $x \\ge 1$.\n6. Let $\\varepsilon \u003e 0$. Let $K = \\frac{1}{b \\varepsilon} \u003e 0$, and let $M = \\max(1, e^{(\\ln K)/b})$.\n7. Let $x \u003e M$. Then $\\ln x \u003e \\ln(e^{(\\ln K)/b}) = (\\ln K)/b$, since $\\ln$ is strictly increasing and $\\ln(e^y) = y$.\n8. So $b \\ln x \u003e \\ln K$. Since [[$\\exp$ is strictly increasing|0KN2V7840E]], $x^b = e^{b \\ln x} \u003e e^{\\ln K} = K$.\n9. By step 5, for $x \u003e M$ we get $0 \\le \\frac{\\ln x}{x^a} \u003c \\frac{1}{b K} = \\varepsilon$.\n10. By the [[definition of a limit at infinity|0Y12QYTJK3]], $\\lim_{x \\to \\infty} \\frac{\\ln x}{x^a} = 0$. That is, $\\ln x = o(x^a)$ as $x \\to \\infty$.\n\n## Idea\n\nThe inequality $1 + y \\le e^y$ gives $\\ln x \\le \\frac{2}{a} x^{a/2}$, so the logarithm is at most a multiple of half the power. Dividing by $x^a$ leaves a multiple of $x^{-a/2}$, which tends to $0$. Using half the exponent leaves room to spare.\n",
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