{
  "index": "2019-A-6",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Let $g$ be a real-valued function that is continuous on the closed interval $[0,1]$ and twice differentiable on \nthe open interval $(0,1)$. Suppose that for some real number $r>1$, \n\\[\n\\lim_{x \\to 0^+} \\frac{g(x)}{x^r} = 0.\n\\]\nProve that either\n\\[\n\\lim_{x \\to 0^+} g'(x) = 0 \\qquad \\mbox{or} \\qquad \\limsup_{x \\to 0^+} x^r |g''(x)| = \\infty.\n\\]",
  "solution": "\\textbf{Solution 1.}\n(by Harm Derksen)\nWe assume that $\\limsup_{x \\to 0^+} x^r |g''(x)| < \\infty$\nand deduce that $\\lim_{x \\to 0^+} g'(x) = 0$.\nNote that\n\\[\n\\limsup_{x \\to 0^+} x^r \\sup\\{| g''(\\xi)|: \\xi \\in [x/2, x]\\}\n< \\infty.\n\\]\nSuppose for the moment that there exists a function $h$ on $(0,1)$ which is positive, nondecreasing, and satisfies\n\\[\n\\lim_{x \\to 0^+} \\frac{g(x)}{h(x)} = \\lim_{x \\to 0^+} \\frac{h(x)}{x^r} = 0.\n\\]\nFor some $c>0$, $h(x) < x^r < x$ for $x \\in (0,c)$. By Taylor's theorem with remainder, we can find a function $\\xi$ on $(0,c)$ such that\n$\\xi(x) \\in [x-h(x),x]$ and\n\\[\ng(x-h(x)) = g(x) - g'(x) h(x) + \\frac{1}{2} g''(\\xi(x)) h(x)^2.\n\\]\nWe can thus express $g'(x)$ as\n\\[\n\\frac{g(x)}{h(x)} + \\frac{1}{2} x^r g''(\\xi(x)) \\frac{h(x)}{x^r}\n-  \\frac{g(x-h(x))}{h(x-h(x))} \\frac{h(x-h(x))}{h(x)}.\n\\]\nAs $x \\to 0^+$, $g(x)/h(x)$, $g(x-h(x))/h(x-h(x))$, and\n$h(x)/x^r$ tend to 0, while $x^r g''(\\xi(x))$ remains bounded\n(because $\\xi(x) \\geq x-h(x) \\geq x - x^r \\geq x/2$ for $x$ small)\nand $h(x-h(x))/h(x)$ is bounded in $(0,1]$.\nHence $\\lim_{x \\to 0^+} g'(x) = 0$ as desired.\n\nIt thus only remains to produce a function $h$ with the desired properties; this amounts to ``inserting'' a function between $g(x)$ and $x^r$ while taking care to ensure the positive and nondecreasing properties.\nOne of many options is $h(x) = x^r \\sqrt{f(x)}$ where\n\\[\nf(x) = \\sup\\{|z^{-r} g(z)|: z \\in (0,x)\\},\n\\]\nso that\n\\[\n\\frac{h(x)}{x^r} = \\sqrt{f(x)}, \\qquad \\frac{g(x)}{h(x)} = \\sqrt{f(x)} x^{-r} g(x).\n\\]\n\n\\noindent\n\\textbf{Solution 2.}\nWe argue by contradiction. Assume that $\\limsup_{x\\to 0^+} x^r|g''(x)|<\\infty$, so that there is an $M$ such that $|g''(x)| < M x^{-r}$ for all $x$; and that $\\lim_{x\\to 0^+} g'(x) \\neq 0$, so that there is an $\\epsilon_0>0$ and a sequence $x_n\\to 0$ with $|g'(x_n)| > \\epsilon_0$ for all $n$.\n\nNow let $\\epsilon>0$ be arbitrary. Since $\\lim_{x\\to 0^+} g(x) x^{-r} = 0$, there is a $\\delta>0$ for which $|g(x)|<\\epsilon x^r$ for all $x<\\delta$.\nChoose $n$ sufficiently large that $\\frac{\\epsilon_0 x_n^r}{2M}<x_n$ and $x_n < \\delta/2$; then $x_n+\\frac{\\epsilon_0 x_n^r}{2M} < 2 x_n < \\delta$. In addition,\nwe have $|g'(x)| > \\epsilon_0/2$ for all $x\\in [x_n,x_n+\\frac{\\epsilon_0 x_n^r}{2M}]$ since $|g'(x_n)| > \\epsilon_0$ and $|g''(x)| < Mx^{-r} \\leq M x_n^{-r}$ in this range. It follows that\n\\begin{align*}\n\\frac{\\epsilon_0^2}{2} \\frac{x_n^r}{2M} &<\n|g(x_n+\\frac{\\epsilon_0 x_n^r}{2M}) - g(x_n)| \\\\\n&\\leq |g(x_n+\\frac{\\epsilon_0 x_n^r}{2M})|+|g(x_n)| \\\\\n&< \\epsilon \\left((x_n+\\frac{\\epsilon_0 x_n^r}{2M})^r+x_n^r\\right) \\\\\n&< \\epsilon(1+2^r)x_n^r,\n\\end{align*}\nwhence $4M(1+2^r)\\epsilon > \\epsilon_0^2$. Since $\\epsilon>0$ is arbitrary and $M,r,\\epsilon_0$ are fixed, this gives the desired contradiction.\n\n\n\\noindent\n\\textbf{Remark.}\nHarm Derksen points out that the ``or'' in the problem need not be exclusive. For example, take\n\\[\ng(x) = \\begin{cases} x^5\\sin(x^{-3}) & x \\in (0,1] \\\\\n0 & x = 0.\n\\end{cases}\n\\]\nThen for $x \\in (0,1)$,\n\\begin{align*}\ng'(x) &= 5x^4\\sin(x^{-3})-3x\\cos(x^{-3}) \\\\\ng''(x) &=(20x^3-9x^{-3})\\sin(x^{-3})-18\\cos(x^{-3}).\n\\end{align*}\nFor $r=2$, $\\lim_{x\\to 0^+}x^{-r}g(x)=\\lim_{x\\to 0^+}x^3\\sin(x^{-3})=0$, $\\lim_{x\\to 0^+}g'(x)=0$ and\n$x^rg''(x)=(20x^5-9x^{-1})\\sin(x^{-3})-18x^2\\cos(x^{-3})$ is unbounded as $x\\to 0^+$.\n(Note that $g'(x)$ is not differentiable at $x=0$.)",
  "vars": [
    "g",
    "h",
    "x",
    "x_n",
    "z",
    "n",
    "\\\\xi"
  ],
  "params": [
    "r",
    "c",
    "M",
    "\\\\epsilon",
    "\\\\epsilon_0",
    "\\\\delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "g": "funcval",
        "h": "auxfunc",
        "x": "coordxx",
        "x_n": "seqcoord",
        "z": "varzeta",
        "n": "indexn",
        "\\xi": "xipoint",
        "r": "exppower",
        "c": "smallpos",
        "M": "boundup",
        "\\epsilon": "tolsmall",
        "\\epsilon_0": "tolspeca",
        "\\delta": "deltapos"
      },
      "question": "Let $funcval$ be a real-valued function that is continuous on the closed interval $[0,1]$ and twice differentiable on the open interval $(0,1)$. Suppose that for some real number $exppower>1$, \n\\[\n\\lim_{coordxx \\to 0^+} \\frac{funcval(coordxx)}{coordxx^{exppower}} = 0.\n\\]\nProve that either\n\\[\n\\lim_{coordxx \\to 0^+} funcval'(coordxx) = 0 \\qquad \\mbox{or} \\qquad \\limsup_{coordxx \\to 0^+} coordxx^{exppower} |funcval''(coordxx)| = \\infty.\n\\]",
      "solution": "\\textbf{Solution 1.}\n(by Harm Derksen)\nWe assume that $\\limsup_{coordxx \\to 0^+} coordxx^{exppower} |funcval''(coordxx)| < \\infty$\nand deduce that $\\lim_{coordxx \\to 0^+} funcval'(coordxx) = 0$.\nNote that\n\\[\n\\limsup_{coordxx \\to 0^+} coordxx^{exppower} \\sup\\{| funcval''(xipoint)|: xipoint \\in [coordxx/2, coordxx]\\}\n< \\infty.\n\\]\nSuppose for the moment that there exists a function $auxfunc$ on $(0,1)$ which is positive, nondecreasing, and satisfies\n\\[\n\\lim_{coordxx \\to 0^+} \\frac{funcval(coordxx)}{auxfunc(coordxx)} = \\lim_{coordxx \\to 0^+} \\frac{auxfunc(coordxx)}{coordxx^{exppower}} = 0.\n\\]\nFor some $smallpos>0$, $auxfunc(coordxx) < coordxx^{exppower} < coordxx$ for $coordxx \\in (0,smallpos)$. By Taylor's theorem with remainder, we can find a function $xipoint$ on $(0,smallpos)$ such that\n$xipoint(coordxx) \\in [coordxx-auxfunc(coordxx),coordxx]$ and\n\\[\nfuncval(coordxx-auxfunc(coordxx)) = funcval(coordxx) - funcval'(coordxx) \\, auxfunc(coordxx) + \\frac{1}{2} funcval''(xipoint(coordxx)) \\, auxfunc(coordxx)^2.\n\\]\nWe can thus express $funcval'(coordxx)$ as\n\\[\n\\frac{funcval(coordxx)}{auxfunc(coordxx)} + \\frac{1}{2} coordxx^{exppower} funcval''(xipoint(coordxx)) \\frac{auxfunc(coordxx)}{coordxx^{exppower}}\n-  \\frac{funcval(coordxx-auxfunc(coordxx))}{auxfunc(coordxx-auxfunc(coordxx))} \\frac{auxfunc(coordxx-auxfunc(coordxx))}{auxfunc(coordxx)}.\n\\]\nAs $coordxx \\to 0^+$, $funcval(coordxx)/auxfunc(coordxx)$, $funcval(coordxx-auxfunc(coordxx))/auxfunc(coordxx-auxfunc(coordxx))$, and\n$auxfunc(coordxx)/coordxx^{exppower}$ tend to $0$, while $coordxx^{exppower} funcval''(xipoint(coordxx))$ remains bounded\n(because $xipoint(coordxx) \\ge coordxx-auxfunc(coordxx) \\ge coordxx - coordxx^{exppower} \\ge coordxx/2$ for $coordxx$ small)\nand $auxfunc(coordxx-auxfunc(coordxx))/auxfunc(coordxx)$ is bounded in $(0,1]$.\nHence $\\lim_{coordxx \\to 0^+} funcval'(coordxx) = 0$ as desired.\n\nIt thus only remains to produce a function $auxfunc$ with the desired properties; this amounts to ``inserting'' a function between $funcval(coordxx)$ and $coordxx^{exppower}$ while taking care to ensure the positive and nondecreasing properties.\nOne of many options is $auxfunc(coordxx) = coordxx^{exppower} \\sqrt{f(coordxx)}$ where\n\\[\nf(coordxx) = \\sup\\{|varzeta^{-exppower} \\, funcval(varzeta)|: varzeta \\in (0,coordxx)\\},\n\\]\nso that\n\\[\n\\frac{auxfunc(coordxx)}{coordxx^{exppower}} = \\sqrt{f(coordxx)}, \\qquad \\frac{funcval(coordxx)}{auxfunc(coordxx)} = \\sqrt{f(coordxx)} \\, coordxx^{-exppower} funcval(coordxx).\n\\]\n\n\\noindent\n\\textbf{Solution 2.}\nWe argue by contradiction. Assume that $\\limsup_{coordxx\\to 0^+} coordxx^{exppower}|funcval''(coordxx)|<\\infty$, so that there is a $boundup$ such that $|funcval''(coordxx)| < boundup \\, coordxx^{-exppower}$ for all $coordxx$; and that $\\lim_{coordxx\\to 0^+} funcval'(coordxx) \\neq 0$, so that there is an $tolspeca>0$ and a sequence $seqcoord\\to 0$ with $|funcval'(seqcoord)| > tolspeca$ for all $indexn$.\n\nNow let $tolsmall>0$ be arbitrary. Since $\\lim_{coordxx\\to 0^+} funcval(coordxx) \\, coordxx^{-exppower} = 0$, there is a $deltapos>0$ for which $|funcval(coordxx)|<tolsmall \\, coordxx^{exppower}$ for all $coordxx<deltapos$.\nChoose $indexn$ sufficiently large that $\\frac{tolspeca \\, seqcoord^{exppower}}{2boundup}<seqcoord$ and $seqcoord < deltapos/2$; then $seqcoord+\\frac{tolspeca \\, seqcoord^{exppower}}{2boundup} < 2 \\, seqcoord < deltapos$. In addition,\nwe have $|funcval'(coordxx)| > tolspeca/2$ for all $coordxx\\in [seqcoord,seqcoord+\\frac{tolspeca \\, seqcoord^{exppower}}{2boundup}]$ since $|funcval'(seqcoord)| > tolspeca$ and $|funcval''(coordxx)| < boundup \\, coordxx^{-exppower} \\le boundup \\, seqcoord^{-exppower}$ in this range. It follows that\n\\begin{align*}\n\\frac{tolspeca^2}{2} \\frac{seqcoord^{exppower}}{2boundup} &<\n|funcval(seqcoord+\\tfrac{tolspeca \\, seqcoord^{exppower}}{2boundup}) - funcval(seqcoord)| \\\n&\\le |funcval(seqcoord+\\tfrac{tolspeca \\, seqcoord^{exppower}}{2boundup})|+|funcval(seqcoord)| \\\\\n&< tolsmall \\left((seqcoord+\\tfrac{tolspeca \\, seqcoord^{exppower}}{2boundup})^{exppower}+seqcoord^{exppower}\\right) \\\\\n&< tolsmall(1+2^{exppower}) seqcoord^{exppower},\n\\end{align*}\nwhence $4boundup(1+2^{exppower})tolsmall > tolspeca^2$. Since $tolsmall>0$ is arbitrary and $boundup,exppower,tolspeca$ are fixed, this gives the desired contradiction.\n\n\\noindent\n\\textbf{Remark.}\nHarm Derksen points out that the ``or'' in the problem need not be exclusive. For example, take\n\\[\nfuncval(coordxx) = \\begin{cases} coordxx^5\\sin(coordxx^{-3}) & coordxx \\in (0,1] \\\\ 0 & coordxx = 0.\\end{cases}\n\\]\nThen for $coordxx \\in (0,1)$,\n\\begin{align*}\nfuncval'(coordxx) &= 5\\,coordxx^4\\sin(coordxx^{-3})-3\\,coordxx\\cos(coordxx^{-3}) \\\\\nfuncval''(coordxx) &=(20\\,coordxx^3-9\\,coordxx^{-3})\\sin(coordxx^{-3})-18\\cos(coordxx^{-3}).\n\\end{align*}\nFor $exppower=2$, $\\lim_{coordxx\\to 0^+}coordxx^{-exppower}funcval(coordxx)=\\lim_{coordxx\\to 0^+}coordxx^3\\sin(coordxx^{-3})=0$, $\\lim_{coordxx\\to 0^+}funcval'(coordxx)=0$ and\n$coordxx^{exppower}funcval''(coordxx)=(20\\,coordxx^5-9\\,coordxx^{-1})\\sin(coordxx^{-3})-18\\,coordxx^2\\cos(coordxx^{-3})$ is unbounded as $coordxx\\to 0^+$.\n(Note that $funcval'(coordxx)$ is not differentiable at $coordxx = 0$.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "g": "shoreline",
        "h": "gallantry",
        "x": "pinecone",
        "x_n": "moonlight",
        "z": "evergreen",
        "n": "rainstorm",
        "\\\\xi": "lighthouse",
        "r": "hydrofoil",
        "c": "ridgepole",
        "M": "waterfall",
        "\\\\epsilon": "peppermint",
        "\\\\epsilon_0": "thunderbolt",
        "\\\\delta": "hummingbird"
      },
      "question": "Let $shoreline$ be a real-valued function that is continuous on the closed interval $[0,1]$ and twice differentiable on the open interval $(0,1)$. Suppose that for some real number $hydrofoil>1$, \n\\[\n\\lim_{pinecone \\to 0^+} \\frac{shoreline(pinecone)}{pinecone^{hydrofoil}} = 0.\n\\]\nProve that either\n\\[\n\\lim_{pinecone \\to 0^+} shoreline'(pinecone) = 0 \\qquad \\mbox{or} \\qquad \\limsup_{pinecone \\to 0^+} pinecone^{hydrofoil} |shoreline''(pinecone)| = \\infty.\n\\]",
      "solution": "\\textbf{Solution 1.}\n(by Harm Derksen)\nWe assume that $\\limsup_{pinecone \\to 0^+} pinecone^{hydrofoil} |shoreline''(pinecone)| < \\infty$\nand deduce that $\\lim_{pinecone \\to 0^+} shoreline'(pinecone) = 0$.\nNote that\n\\[\n\\limsup_{pinecone \\to 0^+} pinecone^{hydrofoil} \\sup\\{| shoreline''(lighthouse)|: lighthouse \\in [pinecone/2, pinecone]\\}\n< \\infty.\n\\]\nSuppose for the moment that there exists a function $gallantry$ on $(0,1)$ which is positive, nondecreasing, and satisfies\n\\[\n\\lim_{pinecone \\to 0^+} \\frac{shoreline(pinecone)}{gallantry(pinecone)} = \\lim_{pinecone \\to 0^+} \\frac{gallantry(pinecone)}{pinecone^{hydrofoil}} = 0.\n\\]\nFor some $ridgepole>0$, $gallantry(pinecone) < pinecone^{hydrofoil} < pinecone$ for $pinecone \\in (0,ridgepole)$. By Taylor's theorem with remainder, we can find a function $lighthouse$ on $(0,ridgepole)$ such that\n$lighthouse(pinecone) \\in [pinecone-gallantry(pinecone),pinecone]$ and\n\\[\nshoreline(pinecone-gallantry(pinecone)) = shoreline(pinecone) - shoreline'(pinecone) gallantry(pinecone) + \\frac{1}{2} shoreline''(lighthouse(pinecone)) gallantry(pinecone)^2.\n\\]\nWe can thus express $shoreline'(pinecone)$ as\n\\[\n\\frac{shoreline(pinecone)}{gallantry(pinecone)} + \\frac{1}{2} pinecone^{hydrofoil} shoreline''(lighthouse(pinecone)) \\frac{gallantry(pinecone)}{pinecone^{hydrofoil}}\n-  \\frac{shoreline(pinecone-gallantry(pinecone))}{gallantry(pinecone-gallantry(pinecone))} \\frac{gallantry(pinecone-gallantry(pinecone))}{gallantry(pinecone)}.\n\\]\nAs $pinecone \\to 0^+$, $shoreline(pinecone)/gallantry(pinecone)$, $shoreline(pinecone-gallantry(pinecone))/gallantry(pinecone-gallantry(pinecone))$, and\n$gallantry(pinecone)/pinecone^{hydrofoil}$ tend to 0, while $pinecone^{hydrofoil} shoreline''(lighthouse(pinecone))$ remains bounded\n(because $lighthouse(pinecone) \\geq pinecone-gallantry(pinecone) \\geq pinecone - pinecone^{hydrofoil} \\geq pinecone/2$ for $pinecone$ small)\nand $gallantry(pinecone-gallantry(pinecone))/gallantry(pinecone)$ is bounded in $(0,1]$.\nHence $\\lim_{pinecone \\to 0^+} shoreline'(pinecone) = 0$ as desired.\n\nIt thus only remains to produce a function $gallantry$ with the desired properties; this amounts to ``inserting'' a function between $shoreline(pinecone)$ and $pinecone^{hydrofoil}$ while taking care to ensure the positive and nondecreasing properties.\nOne of many options is $gallantry(pinecone) = pinecone^{hydrofoil} \\sqrt{f(pinecone)}$ where\n\\[\nf(pinecone) = \\sup\\{|evergreen^{-hydrofoil} shoreline(evergreen)|: evergreen \\in (0,pinecone)\\},\n\\]\nso that\n\\[\n\\frac{gallantry(pinecone)}{pinecone^{hydrofoil}} = \\sqrt{f(pinecone)}, \\qquad \\frac{shoreline(pinecone)}{gallantry(pinecone)} = \\sqrt{f(pinecone)} pinecone^{-hydrofoil} shoreline(pinecone).\n\\]\n\n\\noindent\n\\textbf{Solution 2.}\nWe argue by contradiction. Assume that $\\limsup_{pinecone\\to 0^+} pinecone^{hydrofoil}|shoreline''(pinecone)|<\\infty$, so that there is an $waterfall$ such that $|shoreline''(pinecone)| < waterfall pinecone^{-hydrofoil}$ for all $pinecone$; and that $\\lim_{pinecone\\to 0^+} shoreline'(pinecone) \\neq 0$, so that there is an $thunderbolt>0$ and a sequence $moonlight\\to 0$ with $|shoreline'(moonlight)| > thunderbolt$ for all $rainstorm$.\n\nNow let $peppermint>0$ be arbitrary. Since $\\lim_{pinecone\\to 0^+} shoreline(pinecone) pinecone^{-hydrofoil} = 0$, there is a $hummingbird>0$ for which $|shoreline(pinecone)|<peppermint pinecone^{hydrofoil}$ for all $pinecone<hummingbird$.\nChoose $rainstorm$ sufficiently large that $\\frac{thunderbolt moonlight^{hydrofoil}}{2waterfall}<moonlight$ and $moonlight < hummingbird/2$; then $moonlight+\\frac{thunderbolt moonlight^{hydrofoil}}{2waterfall} < 2 moonlight < hummingbird$. In addition,\nwe have $|shoreline'(pinecone)| > thunderbolt/2$ for all $pinecone\\in [moonlight,moonlight+\\frac{thunderbolt moonlight^{hydrofoil}}{2waterfall}]$ since $|shoreline'(moonlight)| > thunderbolt$ and $|shoreline''(pinecone)| < waterfall pinecone^{-hydrofoil} \\leq waterfall moonlight^{-hydrofoil}$ in this range. It follows that\n\\begin{align*}\n\\frac{thunderbolt^2}{2} \\frac{moonlight^{hydrofoil}}{2waterfall} &<\n|shoreline(moonlight+\\tfrac{thunderbolt moonlight^{hydrofoil}}{2waterfall}) - shoreline(moonlight)| \\\\\n&\\leq |shoreline(moonlight+\\tfrac{thunderbolt moonlight^{hydrofoil}}{2waterfall})|+|shoreline(moonlight)| \\\\\n&< peppermint \\left((moonlight+\\tfrac{thunderbolt moonlight^{hydrofoil}}{2waterfall})^{hydrofoil}+moonlight^{hydrofoil}\\right) \\\\\n&< peppermint(1+2^{hydrofoil})moonlight^{hydrofoil},\n\\end{align*}\nwhence $4waterfall(1+2^{hydrofoil})peppermint > thunderbolt^2$. Since $peppermint>0$ is arbitrary and $waterfall,hydrofoil,thunderbolt$ are fixed, this gives the desired contradiction.\n\n\\noindent\n\\textbf{Remark.}\nHarm Derksen points out that the ``or'' in the problem need not be exclusive. For example, take\n\\[\nshoreline(pinecone) = \\begin{cases} pinecone^5\\sin(pinecone^{-3}) & pinecone \\in (0,1] \\\\\n0 & pinecone = 0.\n\\end{cases}\n\\]\nThen for $pinecone \\in (0,1)$,\n\\begin{align*}\nshoreline'(pinecone) &= 5pinecone^4\\sin(pinecone^{-3})-3pinecone\\cos(pinecone^{-3}) \\\\\nshoreline''(pinecone) &=(20pinecone^3-9pinecone^{-3})\\sin(pinecone^{-3})-18\\cos(pinecone^{-3}).\n\\end{align*}\nFor $hydrofoil=2$, $\\lim_{pinecone\\to 0^+}pinecone^{-hydrofoil}shoreline(pinecone)=\\lim_{pinecone\\to 0^+}pinecone^3\\sin(pinecone^{-3})=0$, $\\lim_{pinecone\\to 0^+}shoreline'(pinecone)=0$ and\n$pinecone^{hydrofoil}shoreline''(pinecone)=(20pinecone^5-9pinecone^{-1})\\sin(pinecone^{-3})-18pinecone^2\\cos(pinecone^{-3})$ is unbounded as $pinecone\\to 0^+$.\n(Note that $shoreline'(pinecone)$ is not differentiable at $pinecone=0$.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "g": "dysfunction",
        "h": "declinefn",
        "x": "staticpoint",
        "x_n": "constantseq",
        "z": "dullpoint",
        "n": "solitude",
        "\\\\xi": "steadystate",
        "r": "basevalue",
        "c": "variance",
        "M": "minimizer",
        "\\\\epsilon": "bulkvalue",
        "\\\\epsilon_0": "massivezero",
        "\\\\delta": "largeshift"
      },
      "question": "Let $dysfunction$ be a real-valued function that is continuous on the closed interval $[0,1]$ and twice differentiable on the open interval $(0,1)$. Suppose that for some real number $basevalue>1$, \n\\[\n\\lim_{staticpoint \\to 0^+} \\frac{dysfunction(staticpoint)}{staticpoint^{basevalue}} = 0.\n\\]\nProve that either\n\\[\n\\lim_{staticpoint \\to 0^+} dysfunction'(staticpoint) = 0 \\qquad \\mbox{or} \\qquad \\limsup_{staticpoint \\to 0^+} staticpoint^{basevalue} |dysfunction''(staticpoint)| = \\infty.\n\\]",
      "solution": "\\textbf{Solution 1.}\n(by Harm Derksen)\nWe assume that $\\limsup_{staticpoint \\to 0^+} staticpoint^{basevalue} |dysfunction''(staticpoint)| < \\infty$\nand deduce that $\\lim_{staticpoint \\to 0^+} dysfunction'(staticpoint) = 0$.\nNote that\n\\[\n\\limsup_{staticpoint \\to 0^+} staticpoint^{basevalue} \\sup\\{| dysfunction''(steadystate)|: steadystate \\in [staticpoint/2, staticpoint]\\}\n< \\infty.\n\\]\nSuppose for the moment that there exists a function $declinefn$ on $(0,1)$ which is positive, nondecreasing, and satisfies\n\\[\n\\lim_{staticpoint \\to 0^+} \\frac{dysfunction(staticpoint)}{declinefn(staticpoint)} = \\lim_{staticpoint \\to 0^+} \\frac{declinefn(staticpoint)}{staticpoint^{basevalue}} = 0.\n\\]\nFor some $variance>0$, $declinefn(staticpoint) < staticpoint^{basevalue} < staticpoint$ for $staticpoint \\in (0,variance)$. By Taylor's theorem with remainder, we can find a function $steadystate$ on $(0,variance)$ such that\n$steadystate(staticpoint) \\in [staticpoint-declinefn(staticpoint),staticpoint]$ and\n\\[\ndysfunction(staticpoint-declinefn(staticpoint)) = dysfunction(staticpoint) - dysfunction'(staticpoint) declinefn(staticpoint) + \\frac{1}{2} dysfunction''(steadystate(staticpoint)) declinefn(staticpoint)^2.\n\\]\nWe can thus express $dysfunction'(staticpoint)$ as\n\\[\n\\frac{dysfunction(staticpoint)}{declinefn(staticpoint)} + \\frac{1}{2} staticpoint^{basevalue} dysfunction''(steadystate(staticpoint)) \\frac{declinefn(staticpoint)}{staticpoint^{basevalue}}\n-  \\frac{dysfunction(staticpoint-declinefn(staticpoint))}{declinefn(staticpoint-declinefn(staticpoint))} \\frac{declinefn(staticpoint-declinefn(staticpoint))}{declinefn(staticpoint)}.\n\\]\nAs $staticpoint \\to 0^+$, $dysfunction(staticpoint)/declinefn(staticpoint)$, $dysfunction(staticpoint-declinefn(staticpoint))/declinefn(staticpoint-declinefn(staticpoint))$, and\n$declinefn(staticpoint)/staticpoint^{basevalue}$ tend to 0, while $staticpoint^{basevalue} dysfunction''(steadystate(staticpoint))$ remains bounded\n(because $steadystate(staticpoint) \\geq staticpoint-declinefn(staticpoint) \\geq staticpoint - staticpoint^{basevalue} \\geq staticpoint/2$ for $staticpoint$ small)\nand $declinefn(staticpoint-declinefn(staticpoint))/declinefn(staticpoint)$ is bounded in $(0,1]$.\nHence $\\lim_{staticpoint \\to 0^+} dysfunction'(staticpoint) = 0$ as desired.\n\nIt thus only remains to produce a function $declinefn$ with the desired properties; this amounts to ``inserting'' a function between $dysfunction(staticpoint)$ and $staticpoint^{basevalue}$ while taking care to ensure the positive and nondecreasing properties.\nOne of many options is $declinefn(staticpoint) = staticpoint^{basevalue} \\sqrt{f(staticpoint)}$ where\n\\[\nf(staticpoint) = \\sup\\{|dullpoint^{-basevalue} dysfunction(dullpoint)|: dullpoint \\in (0,staticpoint)\\},\n\\]\nso that\n\\[\n\\frac{declinefn(staticpoint)}{staticpoint^{basevalue}} = \\sqrt{f(staticpoint)}, \\qquad \\frac{dysfunction(staticpoint)}{declinefn(staticpoint)} = \\sqrt{f(staticpoint)} staticpoint^{-basevalue} dysfunction(staticpoint).\n\\]\n\n\\noindent\n\\textbf{Solution 2.}\nWe argue by contradiction. Assume that $\\limsup_{staticpoint\\to 0^+} staticpoint^{basevalue}|dysfunction''(staticpoint)|<\\infty$, so that there is a $minimizer$ such that $|dysfunction''(staticpoint)| < minimizer\\, staticpoint^{-basevalue}$ for all $staticpoint$; and that $\\lim_{staticpoint\\to 0^+} dysfunction'(staticpoint) \\neq 0$, so that there is a $massivezero>0$ and a sequence $constantseq\\to 0$ with $|dysfunction'(constantseq)| > massivezero$ for all $solitude$.\n\nNow let $bulkvalue>0$ be arbitrary. Since $\\lim_{staticpoint\\to 0^+} dysfunction(staticpoint) staticpoint^{-basevalue} = 0$, there is a $largeshift>0$ for which $|dysfunction(staticpoint)|<bulkvalue\\, staticpoint^{basevalue}$ for all $staticpoint<largeshift$.\nChoose $solitude$ sufficiently large that $\\frac{massivezero\\, constantseq^{basevalue}}{2\\,minimizer}<constantseq$ and $constantseq < largeshift/2$; then $constantseq+\\frac{massivezero\\, constantseq^{basevalue}}{2\\,minimizer} < 2\\, constantseq < largeshift$. In addition,\nwe have $|dysfunction'(staticpoint)| > massivezero/2$ for all $staticpoint\\in [constantseq,constantseq+\\frac{massivezero\\, constantseq^{basevalue}}{2\\,minimizer}]$ since $|dysfunction'(constantseq)| > massivezero$ and $|dysfunction''(staticpoint)| < minimizer\\,staticpoint^{-basevalue} \\leq minimizer\\, constantseq^{-basevalue}$ in this range. It follows that\n\\begin{align*}\n\\frac{massivezero^{2}}{2} \\frac{constantseq^{basevalue}}{2\\,minimizer} &<\n|dysfunction(constantseq+\\tfrac{massivezero\\, constantseq^{basevalue}}{2\\,minimizer}) - dysfunction(constantseq)| \\\\\n&\\leq |dysfunction(constantseq+\\tfrac{massivezero\\, constantseq^{basevalue}}{2\\,minimizer})|+|dysfunction(constantseq)| \\\\\n&< bulkvalue \\left((constantseq+\\tfrac{massivezero\\, constantseq^{basevalue}}{2\\,minimizer})^{basevalue}+constantseq^{basevalue}\\right) \\\\\n&< bulkvalue(1+2^{basevalue})constantseq^{basevalue},\n\\end{align*}\nwhence $4\\,minimizer(1+2^{basevalue})\\,bulkvalue > massivezero^{2}$. Since $bulkvalue>0$ is arbitrary and $minimizer,basevalue,massivezero$ are fixed, this gives the desired contradiction.\n\n\\noindent\n\\textbf{Remark.}\nHarm Derksen points out that the ``or'' in the problem need not be exclusive. For example, take\n\\[\ndysfunction(staticpoint) = \\begin{cases} staticpoint^{5}\\sin(staticpoint^{-3}) & staticpoint \\in (0,1] \\\\ 0 & staticpoint = 0. \\end{cases}\n\\]\nThen for $staticpoint \\in (0,1)$,\n\\begin{align*}\ndysfunction'(staticpoint) &= 5staticpoint^{4}\\sin(staticpoint^{-3})-3staticpoint\\cos(staticpoint^{-3}) \\\\\ndysfunction''(staticpoint) &=(20staticpoint^{3}-9staticpoint^{-3})\\sin(staticpoint^{-3})-18\\cos(staticpoint^{-3}).\n\\end{align*}\nFor $basevalue=2$, $\\lim_{staticpoint\\to 0^+}staticpoint^{-basevalue}dysfunction(staticpoint)=\\lim_{staticpoint\\to 0^+}staticpoint^{3}\\sin(staticpoint^{-3})=0$, $\\lim_{staticpoint\\to 0^+}dysfunction'(staticpoint)=0$ and\n$staticpoint^{basevalue}dysfunction''(staticpoint)=(20staticpoint^{5}-9staticpoint^{-1})\\sin(staticpoint^{-3})-18staticpoint^{2}\\cos(staticpoint^{-3})$ is unbounded as $staticpoint\\to 0^+$. (Note that $dysfunction'(staticpoint)$ is not differentiable at $staticpoint=0$.)"
    },
    "garbled_string": {
      "map": {
        "g": "qzxwvtnp",
        "h": "hjgrksla",
        "x": "mdlkeprb",
        "x_n": "fanhqcts",
        "z": "kqspvemd",
        "n": "sqytrdvb",
        "\\xi": "wopajrks",
        "r": "blmqzstc",
        "c": "oytdjkea",
        "M": "vczpiqrm",
        "\\epsilon": "dufgxwse",
        "\\epsilon_0": "atnfrpql",
        "\\delta": "irmchvsa"
      },
      "question": "Let $qzxwvtnp$ be a real-valued function that is continuous on the closed interval $[0,1]$ and twice differentiable on the open interval $(0,1)$. Suppose that for some real number $blmqzstc>1$, \n\\[\n\\lim_{mdlkeprb \\to 0^+} \\frac{qzxwvtnp(mdlkeprb)}{mdlkeprb^{blmqzstc}} = 0.\n\\]\nProve that either\n\\[\n\\lim_{mdlkeprb \\to 0^+} qzxwvtnp'(mdlkeprb) = 0 \\qquad \\mbox{or} \\qquad \\limsup_{mdlkeprb \\to 0^+} mdlkeprb^{blmqzstc} |qzxwvtnp''(mdlkeprb)| = \\infty.\n\\]",
      "solution": "\\textbf{Solution 1.}\n(by Harm Derksen)\nWe assume that $\\limsup_{mdlkeprb \\to 0^+} mdlkeprb^{blmqzstc} |qzxwvtnp''(mdlkeprb)| < \\infty$\nand deduce that $\\lim_{mdlkeprb \\to 0^+} qzxwvtnp'(mdlkeprb) = 0$.\nNote that\n\\[\n\\limsup_{mdlkeprb \\to 0^+} mdlkeprb^{blmqzstc} \\sup\\{| qzxwvtnp''(wopajrks)|: wopajrks \\in [mdlkeprb/2, mdlkeprb]\\}\n< \\infty.\n\\]\nSuppose for the moment that there exists a function $hjgrksla$ on $(0,1)$ which is positive, nondecreasing, and satisfies\n\\[\n\\lim_{mdlkeprb \\to 0^+} \\frac{qzxwvtnp(mdlkeprb)}{hjgrksla(mdlkeprb)} = \\lim_{mdlkeprb \\to 0^+} \\frac{hjgrksla(mdlkeprb)}{mdlkeprb^{blmqzstc}} = 0.\n\\]\nFor some $oytdjkea>0$, $hjgrksla(mdlkeprb) < mdlkeprb^{blmqzstc} < mdlkeprb$ for $mdlkeprb \\in (0,oytdjkea)$. By Taylor's theorem with remainder, we can find a function $wopajrks$ on $(0,oytdjkea)$ such that\n$wopajrks(mdlkeprb) \\in [mdlkeprb-hjgrksla(mdlkeprb),mdlkeprb]$ and\n\\[\nqzxwvtnp(mdlkeprb-hjgrksla(mdlkeprb)) = qzxwvtnp(mdlkeprb) - qzxwvtnp'(mdlkeprb) hjgrksla(mdlkeprb) + \\frac{1}{2} qzxwvtnp''(wopajrks(mdlkeprb)) hjgrksla(mdlkeprb)^2.\n\\]\nWe can thus express $qzxwvtnp'(mdlkeprb)$ as\n\\[\n\\frac{qzxwvtnp(mdlkeprb)}{hjgrksla(mdlkeprb)} + \\frac{1}{2} mdlkeprb^{blmqzstc} qzxwvtnp''(wopajrks(mdlkeprb)) \\frac{hjgrksla(mdlkeprb)}{mdlkeprb^{blmqzstc}}\n-  \\frac{qzxwvtnp(mdlkeprb-hjgrksla(mdlkeprb))}{hjgrksla(mdlkeprb-hjgrksla(mdlkeprb))} \\frac{hjgrksla(mdlkeprb-hjgrksla(mdlkeprb))}{hjgrksla(mdlkeprb)}.\n\\]\nAs $mdlkeprb \\to 0^+$, $qzxwvtnp(mdlkeprb)/hjgrksla(mdlkeprb)$, $qzxwvtnp(mdlkeprb-hjgrksla(mdlkeprb))/hjgrksla(mdlkeprb-hjgrksla(mdlkeprb))$, and\n$hjgrksla(mdlkeprb)/mdlkeprb^{blmqzstc}$ tend to $0$, while $mdlkeprb^{blmqzstc} qzxwvtnp''(wopajrks(mdlkeprb))$ remains bounded\n(because $wopajrks(mdlkeprb) \\ge mdlkeprb-hjgrksla(mdlkeprb) \\ge mdlkeprb - mdlkeprb^{blmqzstc} \\ge mdlkeprb/2$ for $mdlkeprb$ small)\nand $hjgrksla(mdlkeprb-hjgrksla(mdlkeprb))/hjgrksla(mdlkeprb)$ is bounded in $(0,1]$.\nHence $\\lim_{mdlkeprb \\to 0^+} qzxwvtnp'(mdlkeprb) = 0$ as desired.\n\nIt thus only remains to produce a function $hjgrksla$ with the desired properties; this amounts to ``inserting'' a function between $qzxwvtnp(mdlkeprb)$ and $mdlkeprb^{blmqzstc}$ while taking care to ensure the positive and nondecreasing properties.\nOne of many options is $hjgrksla(mdlkeprb) = mdlkeprb^{blmqzstc} \\sqrt{f(mdlkeprb)}$ where\n\\[\nf(mdlkeprb) = \\sup\\{|kqspvemd^{-blmqzstc} qzxwvtnp(kqspvemd)|: kqspvemd \\in (0,mdlkeprb)\\},\n\\]\nso that\n\\[\n\\frac{hjgrksla(mdlkeprb)}{mdlkeprb^{blmqzstc}} = \\sqrt{f(mdlkeprb)}, \\qquad \\frac{qzxwvtnp(mdlkeprb)}{hjgrksla(mdlkeprb)} = \\sqrt{f(mdlkeprb)} mdlkeprb^{-blmqzstc} qzxwvtnp(mdlkeprb).\n\\]\n\n\\noindent\n\\textbf{Solution 2.}\nWe argue by contradiction. Assume that $\\limsup_{mdlkeprb\\to 0^+} mdlkeprb^{blmqzstc}|qzxwvtnp''(mdlkeprb)|<\\infty$, so that there is an $vczpiqrm$ such that $|qzxwvtnp''(mdlkeprb)| < vczpiqrm mdlkeprb^{-blmqzstc}$ for all $mdlkeprb$; and that $\\lim_{mdlkeprb\\to 0^+} qzxwvtnp'(mdlkeprb) \\neq 0$, so that there is an $atnfrpql>0$ and a sequence $fanhqcts\\to 0$ with $|qzxwvtnp'(fanhqcts)| > atnfrpql$ for all $sqytrdvb$.\n\nNow let $dufgxwse>0$ be arbitrary. Since $\\lim_{mdlkeprb\\to 0^+} qzxwvtnp(mdlkeprb) mdlkeprb^{-blmqzstc} = 0$, there is a $irmchvsa>0$ for which $|qzxwvtnp(mdlkeprb)|<dufgxwse mdlkeprb^{blmqzstc}$ for all $mdlkeprb<irmchvsa$.\nChoose $sqytrdvb$ sufficiently large that $\\frac{atnfrpql fanhqcts^{blmqzstc}}{2vczpiqrm}<fanhqcts$ and $fanhqcts < irmchvsa/2$; then $fanhqcts+\\frac{atnfrpql fanhqcts^{blmqzstc}}{2vczpiqrm} < 2 fanhqcts < irmchvsa$. In addition,\nwe have $|qzxwvtnp'(mdlkeprb)| > atnfrpql/2$ for all $mdlkeprb\\in [fanhqcts,fanhqcts+\\frac{atnfrpql fanhqcts^{blmqzstc}}{2vczpiqrm}]$ since $|qzxwvtnp'(fanhqcts)| > atnfrpql$ and $|qzxwvtnp''(mdlkeprb)| < vczpiqrm mdlkeprb^{-blmqzstc} \\leq vczpiqrm fanhqcts^{-blmqzstc}$ in this range. It follows that\n\\begin{align*}\n\\frac{atnfrpql^2}{2} \\frac{fanhqcts^{blmqzstc}}{2vczpiqrm} &<\n|qzxwvtnp(fanhqcts+\\frac{atnfrpql fanhqcts^{blmqzstc}}{2vczpiqrm}) - qzxwvtnp(fanhqcts)| \\\\\n&\\leq |qzxwvtnp(fanhqcts+\\frac{atnfrpql fanhqcts^{blmqzstc}}{2vczpiqrm})|+|qzxwvtnp(fanhqcts)| \\\\\n&< dufgxwse \\left((fanhqcts+\\frac{atnfrpql fanhqcts^{blmqzstc}}{2vczpiqrm})^{blmqzstc}+fanhqcts^{blmqzstc}\\right) \\\\\n&< dufgxwse(1+2^{blmqzstc})fanhqcts^{blmqzstc},\n\\end{align*}\nwhence $4vczpiqrm(1+2^{blmqzstc})dufgxwse > atnfrpql^2$. Since $dufgxwse>0$ is arbitrary and $vczpiqrm,blmqzstc,atnfrpql$ are fixed, this gives the desired contradiction.\n\n\\noindent\n\\textbf{Remark.}\nHarm Derksen points out that the ``or'' in the problem need not be exclusive. For example, take\n\\[\nqzxwvtnp(mdlkeprb) = \\begin{cases} mdlkeprb^5\\sin(mdlkeprb^{-3}) & mdlkeprb \\in (0,1] \\\\\n0 & mdlkeprb = 0.\n\\end{cases}\n\\]\nThen for $mdlkeprb \\in (0,1)$,\n\\begin{align*}\nqzxwvtnp'(mdlkeprb) &= 5mdlkeprb^4\\sin(mdlkeprb^{-3})-3mdlkeprb\\cos(mdlkeprb^{-3}) \\\\\nqzxwvtnp''(mdlkeprb) &=(20mdlkeprb^3-9mdlkeprb^{-3})\\sin(mdlkeprb^{-3})-18\\cos(mdlkeprb^{-3}).\n\\end{align*}\nFor $blmqzstc=2$, $\\lim_{mdlkeprb\\to 0^+}mdlkeprb^{-blmqzstc}qzxwvtnp(mdlkeprb)=\\lim_{mdlkeprb\\to 0^+}mdlkeprb^3\\sin(mdlkeprb^{-3})=0$, $\\lim_{mdlkeprb\\to 0^+}qzxwvtnp'(mdlkeprb)=0$ and\n$mdlkeprb^{blmqzstc}qzxwvtnp''(mdlkeprb)=(20mdlkeprb^5-9mdlkeprb^{-1})\\sin(mdlkeprb^{-3})-18mdlkeprb^2\\cos(mdlkeprb^{-3})$ is unbounded as $mdlkeprb\\to 0^+$.\n(Note that $qzxwvtnp'(mdlkeprb)$ is not differentiable at $mdlkeprb=0$.)"
    },
    "kernel_variant": {
      "question": "Let f be a real-valued function that is continuous on the closed interval [0,3] and twice differentiable on the open interval (0,3).  Fix a real number α>1 and assume\n\n   lim_{x→0⁺} f(x)/x^{α} = 0.\n\nProve that at least one of the following two statements holds:\n\n   (i)  lim_{x→0⁺} f'(x) = 0,\n   (ii) limsup_{x→0⁺} x^{α} |f''(x)| = ∞.",
      "solution": "Assume, with the goal of obtaining a contradiction, that neither (i) nor (ii) is true; that is\n\n   (1)   lim_{x→0⁺} f'(x) ≠ 0,  and\n   (2)   limsup_{x→0⁺} x^{α}|f''(x)| < ∞.\n\nFrom (2) there exist numbers M>0 and δ>0 such that\n\n   |f''(x)| ≤ M x^{−α},            0 < x < δ.           (3)\n\nBecause (1) fails, the non-zero limit of f′ does not exist; in particular\n\n   L := limsup_{x→0⁺} |f'(x)| > 0 (possibly L = ∞).\n\nChoose any finite constant c satisfying 0 < c < L.  (This is possible whether L is finite or infinite.)  By the definition of limsup there is a sequence (x_k)_{k≥1} with\n\n   x_k ↓ 0   and   |f'(x_k)| ≥ c     for every k,            (4)\n\nand with x_k < δ/2 once k is large enough.\n\nSet\n\n   Δ_k := (c/(2M)) x_k^{α}.                                   (5)\n\nBecause α>1, we have Δ_k/x_k = (c/(2M)) x_k^{α−1} → 0; hence, for all sufficiently large k,\n\n   0 < Δ_k < x_k   and   x_k + Δ_k < δ.                      (6)\n\n———————————————————————————\nStep 1: f′ is bounded away from 0 on [x_k , x_k+Δ_k].\n———————————————————————————\n\nFix such a large k and any t ∈ [x_k , x_k+Δ_k].  By the mean-value theorem there exists ξ between x_k and t for which\n\n   f′(t) − f′(x_k) = f''(ξ) (t − x_k).\n\nUsing (3), (5) and (6),\n\n   |f′(t) − f′(x_k)| ≤ M x_k^{−α} Δ_k = M x_k^{−α} (c/(2M)) x_k^{α} = c/2.\n\nConsequently,\n\n   |f′(t)| ≥ |f′(x_k)| − |f′(t) − f′(x_k)| ≥ c − c/2 = c/2,  ∀ t ∈ [x_k , x_k+Δ_k].    (7)\n\nTherefore f′ keeps the same sign on that interval and is bounded in magnitude below by c/2.\n\n———————————————————————————\nStep 2: A lower estimate for |f(x_k+Δ_k) − f(x_k)|.\n———————————————————————————\n\nIntegrating f′ over [x_k , x_k+Δ_k] and using (7),\n\n   |f(x_k+Δ_k) − f(x_k)| = ∫_{x_k}^{x_k+Δ_k} |f′(t)| dt ≥ (c/2) Δ_k = (c^2/(4M)) x_k^{α}.      (8)\n\n———————————————————————————\nStep 3: An upper estimate coming from the hypothesis on f.\n———————————————————————————\n\nBecause lim_{x→0⁺} f(x)/x^{α} = 0, for every ε>0 there exists δ_1>0 such that\n\n   |f(x)| < ε x^{α},    0 < x < δ_1.                        (9)\n\nFor large k we have x_k < min{δ/2, δ_1/2}.  Using (6) we then have x_k + Δ_k < 2x_k < δ_1, so that (9) applies to both x_k and x_k+Δ_k.  Hence\n\n   |f(x_k+Δ_k) − f(x_k)| ≤ |f(x_k+Δ_k)| + |f(x_k)|\n                          < ε[(x_k+Δ_k)^{α} + x_k^{α}] \n                          ≤ ε[(2x_k)^{α} + x_k^{α}] = ε(2^{α}+1) x_k^{α}.            (10)\n\n———————————————————————————\nStep 4: Contradiction.\n———————————————————————————\n\nCombine (8) and (10):\n\n   (c^2/(4M)) x_k^{α} ≤ |f(x_k+Δ_k) − f(x_k)| < ε(2^{α}+1) x_k^{α}.\n\nSince ε>0 is arbitrary, this is impossible.  Therefore our initial assumption that both (1) and (2) hold cannot be true.  At least one of them must fail, i.e. (i) or (ii) in the statement of the theorem must hold, completing the proof. ∎",
      "_meta": {
        "core_steps": [
          "Assume x^r g''(x) is bounded; pick sequence where |g'(x)| stays > const >0",
          "Use boundedness of x^r g'' to show g' varies little on a short x-scale proportionate to x^r",
          "Integrate g' over that short interval to force a change |g(x+Δ)-g(x)| ≥ c·x^r",
          "Use hypothesis g(x)/x^r → 0 to bound |g| by ε·x^r and get |g(x+Δ)-g(x)| ≤ 2ε·x^r",
          "Let ε→0 to obtain contradiction, hence lim_{x→0^+} g'(x)=0"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Length-1 interval; any positive upper end works because argument is local near 0",
            "original": "[0,1]"
          },
          "slot2": {
            "description": "Exponent in condition; needs to exceed 1 but precise value immaterial",
            "original": "r>1"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}