{
  "index": "1974-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-2. Let \\( y(x) \\) be a continuously differentiable real-valued function of a real variable \\( x \\). Show that if \\( \\left(y^{\\prime}\\right)^{2}+y^{3} \\rightarrow 0 \\) as \\( x \\rightarrow+\\infty \\), then \\( y(x) \\) and \\( y^{\\prime}(x) \\rightarrow 0 \\) as \\( x \\rightarrow+\\infty \\).",
  "solution": "B-2.\nIf \\( y^{\\prime}\\left(x_{n}\\right)=0 \\) for a sequence \\( \\left\\{x_{n}\\right\\} \\) approaching \\( +\\infty \\), the hypothesis insures that \\( y\\left(x_{n}\\right) \\rightarrow 0 \\). Since these \\( x_{n} \\) may include any relative maxima and minima, this case must have \\( y(x) \\rightarrow 0 \\) as \\( x \\rightarrow+\\infty \\). Then one also has \\( y^{\\prime}(x) \\rightarrow 0 \\) as \\( x \\rightarrow+x \\).\n\nIn the remaining case, there is an \\( x_{0} \\) such that for \\( x>x_{0} \\) one has \\( y^{\\prime} \\neq 0 \\) and so \\( \\left(y^{\\prime}\\right)^{2}>0 \\). We restrict ourselves to the \\( x \\) 's with \\( x>x_{0} \\) and consider two subcases:\n(a) \\( y^{\\prime}>0 \\). If \\( y \\) is unbounded above, so are \\( y^{3} \\) and \\( \\left(y^{\\prime}\\right)^{2}+y^{3} \\). This contradicts the hypothesis \\( \\left(y^{\\prime}\\right)^{2}+y^{3} \\rightarrow 0 \\) as \\( x \\rightarrow+x \\). If \\( y \\) is bounded above, it approaches a finite limit. Then \\( y^{3},\\left(y^{\\prime}\\right)^{2} \\), and \\( y^{\\prime} \\) approach limits. Since \\( y \\) is bounded, the limit for \\( y^{\\prime} \\) must be 0 . Then \\( y \\) also has 0 as its limit.\n(b) \\( y^{\\prime}<0 \\). There is no problem unless \\( y \\) is unbounded below. Then we may assume that \\( y<0 \\) and compare \\( y \\) to a solution of the differential equation\n\\[\ny^{\\prime}=-(1 / 2)|y|^{3 / 2}, \\quad y<0 .\n\\]\n\nEvery solution diverges to \\( -x \\) in a finite interval, hence so does \\( y(x) \\); this contradicts the hypothesis that \\( y \\) is defined and smooth for all large \\( x \\).",
  "vars": [
    "y",
    "x",
    "x_n",
    "x_0"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "y": "realfunc",
        "x": "indepvar",
        "x_n": "seqpointn",
        "x_0": "initpoint"
      },
      "question": "Problem:\n<<<\nB-2. Let \\( realfunc(indepvar) \\) be a continuously differentiable real-valued function of a real variable \\( indepvar \\). Show that if \\( \\left(realfunc^{\\prime}\\right)^{2}+realfunc^{3} \\rightarrow 0 \\) as \\( indepvar \\rightarrow+\\infty \\), then \\( realfunc(indepvar) \\) and \\( realfunc^{\\prime}(indepvar) \\rightarrow 0 \\) as \\( indepvar \\rightarrow+\\infty \\).\n>>>\n",
      "solution": "B-2.\nIf \\( realfunc^{\\prime}\\left(seqpointn\\right)=0 \\) for a sequence \\( \\left\\{seqpointn\\right\\} \\) approaching \\( +\\infty \\), the hypothesis insures that \\( realfunc\\left(seqpointn\\right) \\rightarrow 0 \\). Since these \\( seqpointn \\) may include any relative maxima and minima, this case must have \\( realfunc(indepvar) \\rightarrow 0 \\) as \\( indepvar \\rightarrow+\\infty \\). Then one also has \\( realfunc^{\\prime}(indepvar) \\rightarrow 0 \\) as \\( indepvar \\rightarrow+indepvar \\).\n\nIn the remaining case, there is an \\( initpoint \\) such that for \\( indepvar>initpoint \\) one has \\( realfunc^{\\prime} \\neq 0 \\) and so \\( \\left(realfunc^{\\prime}\\right)^{2}>0 \\). We restrict ourselves to the \\( indepvar \\) 's with \\( indepvar>initpoint \\) and consider two subcases:\n(a) \\( realfunc^{\\prime}>0 \\). If \\( realfunc \\) is unbounded above, so are \\( realfunc^{3} \\) and \\( \\left(realfunc^{\\prime}\\right)^{2}+realfunc^{3} \\). This contradicts the hypothesis \\( \\left(realfunc^{\\prime}\\right)^{2}+realfunc^{3} \\rightarrow 0 \\) as \\( indepvar \\rightarrow+indepvar \\). If \\( realfunc \\) is bounded above, it approaches a finite limit. Then \\( realfunc^{3},\\left(realfunc^{\\prime}\\right)^{2} \\), and \\( realfunc^{\\prime} \\) approach limits. Since \\( realfunc \\) is bounded, the limit for \\( realfunc^{\\prime} \\) must be 0 . Then \\( realfunc \\) also has 0 as its limit.\n(b) \\( realfunc^{\\prime}<0 \\). There is no problem unless \\( realfunc \\) is unbounded below. Then we may assume that \\( realfunc<0 \\) and compare \\( realfunc \\) to a solution of the differential equation\n\\[\nrealfunc^{\\prime}=-(1 / 2)|realfunc|^{3 / 2}, \\quad realfunc<0 .\n\\]\n\nEvery solution diverges to \\( -indepvar \\) in a finite interval, hence so does \\( realfunc(indepvar) \\); this contradicts the hypothesis that \\( realfunc \\) is defined and smooth for all large \\( indepvar \\).\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "y": "rainstorm",
        "x": "buttercup",
        "x_n": "peppermint",
        "x_0": "lighthouse"
      },
      "question": "B-2. Let \\( rainstorm(buttercup) \\) be a continuously differentiable real-valued function of a real variable \\( buttercup \\). Show that if \\( \\left(rainstorm^{\\prime}\\right)^{2}+rainstorm^{3} \\rightarrow 0 \\) as \\( buttercup \\rightarrow+\\infty \\), then \\( rainstorm(buttercup) \\) and \\( rainstorm^{\\prime}(buttercup) \\rightarrow 0 \\) as \\( buttercup \\rightarrow+\\infty \\).",
      "solution": "B-2.\nIf \\( rainstorm^{\\prime}\\left(peppermint\\right)=0 \\) for a sequence \\( \\left\\{peppermint\\right\\} \\) approaching \\( +\\infty \\), the hypothesis insures that \\( rainstorm\\left(peppermint\\right) \\rightarrow 0 \\). Since these \\( peppermint \\) may include any relative maxima and minima, this case must have \\( rainstorm(buttercup) \\rightarrow 0 \\) as \\( buttercup \\rightarrow+\\infty \\). Then one also has \\( rainstorm^{\\prime}(buttercup) \\rightarrow 0 \\) as \\( buttercup \\rightarrow+buttercup \\).\n\nIn the remaining case, there is an \\( lighthouse \\) such that for \\( buttercup>lighthouse \\) one has \\( rainstorm^{\\prime} \\neq 0 \\) and so \\( \\left(rainstorm^{\\prime}\\right)^{2}>0 \\). We restrict ourselves to the \\( buttercup \\) 's with \\( buttercup>lighthouse \\) and consider two subcases:\n(a) \\( rainstorm^{\\prime}>0 \\). If \\( rainstorm \\) is unbounded above, so are \\( rainstorm^{3} \\) and \\( \\left(rainstorm^{\\prime}\\right)^{2}+rainstorm^{3} \\). This contradicts the hypothesis \\( \\left(rainstorm^{\\prime}\\right)^{2}+rainstorm^{3} \\rightarrow 0 \\) as \\( buttercup \\rightarrow+buttercup \\). If \\( rainstorm \\) is bounded above, it approaches a finite limit. Then \\( rainstorm^{3},\\left(rainstorm^{\\prime}\\right)^{2} \\), and \\( rainstorm^{\\prime} \\) approach limits. Since \\( rainstorm \\) is bounded, the limit for \\( rainstorm^{\\prime} \\) must be 0 . Then \\( rainstorm \\) also has 0 as its limit.\n(b) \\( rainstorm^{\\prime}<0 \\). There is no problem unless \\( rainstorm \\) is unbounded below. Then we may assume that \\( rainstorm<0 \\) and compare \\( rainstorm \\) to a solution of the differential equation\n\\[\nrainstorm^{\\prime}=-(1 / 2)|rainstorm|^{3 / 2}, \\quad rainstorm<0 .\n\\]\n\nEvery solution diverges to \\( -buttercup \\) in a finite interval, hence so does \\( rainstorm(buttercup) \\); this contradicts the hypothesis that \\( rainstorm \\) is defined and smooth for all large \\( buttercup \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "y": "stationary",
        "x": "invariant",
        "x_n": "singleton",
        "x_0": "finalpoint"
      },
      "question": "B-2. Let \\( stationary(invariant) \\) be a continuously differentiable real-valued function of a real variable \\( invariant \\). Show that if \\( \\left(stationary^{\\prime}\\right)^{2}+stationary^{3} \\rightarrow 0 \\) as \\( invariant \\rightarrow+\\infty \\), then \\( stationary(invariant) \\) and \\( stationary^{\\prime}(invariant) \\rightarrow 0 \\) as \\( invariant \\rightarrow+\\infty \\).",
      "solution": "B-2.\nIf \\( stationary^{\\prime}\\left(singleton\\right)=0 \\) for a sequence \\( \\left\\{singleton\\right\\} \\) approaching \\( +\\infty \\), the hypothesis insures that \\( stationary\\left(singleton\\right) \\rightarrow 0 \\). Since these \\( singleton \\) may include any relative maxima and minima, this case must have \\( stationary(invariant) \\rightarrow 0 \\) as \\( invariant \\rightarrow+\\infty \\). Then one also has \\( stationary^{\\prime}(invariant) \\rightarrow 0 \\) as \\( invariant \\rightarrow+invariant \\).\n\nIn the remaining case, there is an \\( finalpoint \\) such that for \\( invariant>finalpoint \\) one has \\( stationary^{\\prime} \\neq 0 \\) and so \\( \\left(stationary^{\\prime}\\right)^{2}>0 \\). We restrict ourselves to the \\( invariant \\) 's with \\( invariant>finalpoint \\) and consider two subcases:\n(a) \\( stationary^{\\prime}>0 \\). If \\( stationary \\) is unbounded above, so are \\( stationary^{3} \\) and \\( \\left(stationary^{\\prime}\\right)^{2}+stationary^{3} \\). This contradicts the hypothesis \\( \\left(stationary^{\\prime}\\right)^{2}+stationary^{3} \\rightarrow 0 \\) as \\( invariant \\rightarrow+invariant \\). If \\( stationary \\) is bounded above, it approaches a finite limit. Then \\( stationary^{3},\\left(stationary^{\\prime}\\right)^{2} \\), and \\( stationary^{\\prime} \\) approach limits. Since \\( stationary \\) is bounded, the limit for \\( stationary^{\\prime} \\) must be 0. Then \\( stationary \\) also has 0 as its limit.\n(b) \\( stationary^{\\prime}<0 \\). There is no problem unless \\( stationary \\) is unbounded below. Then we may assume that \\( stationary<0 \\) and compare \\( stationary \\) to a solution of the differential equation\n\\[\nstationary^{\\prime}=-(1 / 2)|stationary|^{3 / 2}, \\quad stationary<0 .\n\\]\n\nEvery solution diverges to \\( -invariant \\) in a finite interval, hence so does \\( stationary(invariant) \\); this contradicts the hypothesis that \\( stationary \\) is defined and smooth for all large \\( invariant \\)."
    },
    "garbled_string": {
      "map": {
        "y": "qzxwvtnp",
        "x": "hjgrksla",
        "x_n": "vbnmzxcv",
        "x_0": "plmoknij"
      },
      "question": "B-2. Let \\( qzxwvtnp(hjgrksla) \\) be a continuously differentiable real-valued function of a real variable \\( hjgrksla \\). Show that if \\( \\left(qzxwvtnp^{\\prime}\\right)^{2}+qzxwvtnp^{3} \\rightarrow 0 \\) as \\( hjgrksla \\rightarrow+\\infty \\), then \\( qzxwvtnp(hjgrksla) \\) and \\( qzxwvtnp^{\\prime}(hjgrksla) \\rightarrow 0 \\) as \\( hjgrksla \\rightarrow+\\infty \\).",
      "solution": "B-2.\nIf \\( qzxwvtnp^{\\prime}\\left(vbnmzxcv_{n}\\right)=0 \\) for a sequence \\( \\left\\{vbnmzxcv_{n}\\right\\} \\) approaching \\( +\\infty \\), the hypothesis insures that \\( qzxwvtnp\\left(vbnmzxcv_{n}\\right) \\rightarrow 0 \\). Since these \\( vbnmzxcv_{n} \\) may include any relative maxima and minima, this case must have \\( qzxwvtnp(hjgrksla) \\rightarrow 0 \\) as \\( hjgrksla \\rightarrow+\\infty \\). Then one also has \\( qzxwvtnp^{\\prime}(hjgrksla) \\rightarrow 0 \\) as \\( hjgrksla \\rightarrow+hjgrksla \\).\n\nIn the remaining case, there is an \\( plmoknij_{0} \\) such that for \\( hjgrksla>plmoknij_{0} \\) one has \\( qzxwvtnp^{\\prime} \\neq 0 \\) and so \\( \\left(qzxwvtnp^{\\prime}\\right)^{2}>0 \\). We restrict ourselves to the \\( hjgrksla \\) 's with \\( hjgrksla>plmoknij_{0} \\) and consider two subcases:\n(a) \\( qzxwvtnp^{\\prime}>0 \\). If \\( qzxwvtnp \\) is unbounded above, so are \\( qzxwvtnp^{3} \\) and \\( \\left(qzxwvtnp^{\\prime}\\right)^{2}+qzxwvtnp^{3} \\). This contradicts the hypothesis \\( \\left(qzxwvtnp^{\\prime}\\right)^{2}+qzxwvtnp^{3} \\rightarrow 0 \\) as \\( hjgrksla \\rightarrow+hjgrksla \\). If \\( qzxwvtnp \\) is bounded above, it approaches a finite limit. Then \\( qzxwvtnp^{3},\\left(qzxwvtnp^{\\prime}\\right)^{2} \\), and \\( qzxwvtnp^{\\prime} \\) approach limits. Since \\( qzxwvtnp \\) is bounded, the limit for \\( qzxwvtnp^{\\prime} \\) must be 0 . Then \\( qzxwvtnp \\) also has 0 as its limit.\n(b) \\( qzxwvtnp^{\\prime}<0 \\). There is no problem unless \\( qzxwvtnp \\) is unbounded below. Then we may assume that \\( qzxwvtnp<0 \\) and compare \\( qzxwvtnp \\) to a solution of the differential equation\n\\[\nqzxwvtnp^{\\prime}=-(1 / 2)|qzxwvtnp|^{3 / 2}, \\quad qzxwvtnp<0 .\n\\]\n\nEvery solution diverges to \\( -hjgrksla \\) in a finite interval, hence so does \\( qzxwvtnp(hjgrksla) \\); this contradicts the hypothesis that \\( qzxwvtnp \\) is defined and smooth for all large \\( hjgrksla \\)."
    },
    "kernel_variant": {
      "question": "Let y: ℝ → ℝ be a C¹–function that satisfies\n      7\\,[y'(x)]^{2} + y(x)^{5}\\;\\longrightarrow\\;0 \\quad (x\\to -\\infty).\nProve that the two limits\n      \\displaystyle\\lim_{x\\to-\\infty} y(x)\\quad\\text{and}\\quad\\lim_{x\\to-\\infty} y'(x)\nexist and are equal to 0.",
      "solution": "Put\n   F(x):=7\\,[y'(x)]^{2}+y(x)^{5}\\qquad(x\\in\\mathbb R),\nso that F(x)→0 as x→−∞.\nBecause 7[y'(x)]^{2}≥0 we have\n   (1)\\;\\; y(x)^{5} \\le F(x) \\qquad(\\text{all }x).\n\n-----------------------------------------------------------------\nA.  y is bounded above for all sufficiently negative x\n-----------------------------------------------------------------\nSince F(x)→0 there is a point X such that |F(x)|<1 whenever x<X.\nTake some x<X and assume, with the aim of obtaining a contradiction,\nthat y(x)>2.  Then y(x)^{5}>32, hence by (1)\n        F(x) \\ge y(x)^{5} > 32,\nwhich contradicts |F(x)|<1.  Consequently\n        y(x) \\le 2  \\quad (x<X).             (2)\nThus y is bounded above on (−∞,X], a fact that suffices for every\nsubsequent step of the proof.\n\n-----------------------------------------------------------------\nB.  First case – y′ has infinitely many zeros x_{n}→−∞\n-----------------------------------------------------------------\nAt such points F(x_{n}) = y(x_{n})^{5}→0, hence y(x_{n})→0.  Between\nconsecutive zeros the sign of y′ is constant, so y is monotone on each\ninterval [x_{n+1},x_{n}].  It follows that for x in this interval\n |y(x)|≤max\\{|y(x_{n})|,|y(x_{n+1})|\\}→0; hence y(x)→0 as x→−∞.\nUsing F(x)=7[y'(x)]^{2}+y(x)^{5} we immediately deduce y′(x)→0.\nTherefore the desired conclusion holds in this first case.\n\n-----------------------------------------------------------------\nC.  Second case – eventually y′ keeps one sign\n-----------------------------------------------------------------\nAssume now that there exists x_{0} with y′(x)≠0 for all x<x_{0}; then\ny′ is either strictly positive or strictly negative on (−∞,x_{0}).\n\nC1.  Sub-case y′>0 on (−∞,x_{0}).\nHere y decreases as we move to the left, so the limit\n        L:=\\lim_{x\\to-\\infty} y(x)\nexists (possibly infinite).  Three possibilities are considered.\n\nC1(a)  L>0.  By (1) we would have F(x)≥y(x)^{5}→L^{5}>0, contradicting\nF(x)→0.\n\nC1(b)  −∞<L<0.  Then 7[y′(x)]^{2}=−y(x)^{5}+F(x)→−L^{5}>0, so\nlim_{x\\to-\\infty}y′(x)=c:=\\sqrt{−L^{5}/7}>0.  Choose t<x_{0}.  For\nx<t we get\n   y(x)=y(t)−\\int_{x}^{t}y′(s)ds ≤ y(t)−(c/2)(t−x) \\to −∞\nas x→−∞, contradicting the finiteness of L.  Hence this alternative\nis impossible.\n\nC1(c)  L=−∞.  Because F(x)→0 we can pick X_{1}<x_{0} so that |F(x)|≤1\nfor all x<X_{1}.  Whenever x<X_{1} and y(x)≤−2 we have\n −y(x)^{5}−|F(x)|≥\\tfrac12(−y(x))^{5}, and therefore\n      |y′(x)|≥(1/\\sqrt{14})(−y(x))^{5/2}.            (3)\nSince y′>0 we obtain for such x\n  \\frac{d}{dx}[\\,(−y)^{−3/2}\\,]=\\tfrac32(−y)^{−5/2}y′(x)\n                                 ≥ \\tfrac32\\cdot\\tfrac1{\\sqrt{14}}=:k>0.\nIntegrating (from x with y(x)≤−2 up to X_{1}) yields\n (−y(x))^{−3/2} ≤ (−y(X_{1}))^{−3/2} − k(X_{1}−x),\nwhose right–hand side becomes negative for sufficiently negative x, an\nimpossibility.  Hence L=−∞ cannot occur.\n\nThe only option left is L=0.  Returning to F(x)=7[y′(x)]^{2}+y(x)^{5}\nwe conclude y(x)→0 and y′(x)→0 as x→−∞.\n\nC2.  Sub-case y′<0 on (−∞,x_{0}).\nNow y increases as x→−∞, so\n        S:=\\lim_{x\\to-\\infty} y(x)\nexists.\n\nC2(a)  S=+∞ contradicts the upper bound (2).\n\nC2(b)  S>0.  Then y(x)^{5}→S^{5}>0 so F(x)→+∞, contradiction.\n\nC2(c)  S<0.  Exactly as in C1(b) (but with y′<0) we find\nlim_{x\\to-\\infty}y′(x)=−\\sqrt{−S^{5}/7}<0, which again forces\ny(x)→−∞, contradicting the finiteness of S.\n\nThus S=0 and, using F, also y′(x)→0.\n\n-----------------------------------------------------------------\nD.  Conclusion\n-----------------------------------------------------------------\nAll possible cases lead to\n   \\boxed{\\displaystyle\\lim_{x\\to-\\infty}y(x)=0\\quad\\text{and}\\quad\\lim_{x\\to-\\infty}y'(x)=0}.",
      "_meta": {
        "core_steps": [
          "Locate a sequence of critical points; if it exists, (y')^2+y^3=0 there implies y→0, so whole function and derivative tend to 0.",
          "If no critical points appear after some X₀, then y′ keeps one sign, making y monotone on (X₀,∞).",
          "Monotone-increasing case: unbounded growth would make the positive y^3 term keep the sum away from 0, so y must be bounded ⇒ lim y exists and (y')→0 ⇒ y→0.",
          "Monotone-decreasing case: if y were unbounded below, inequality |y′|≳|y|^{3/2} would force blow-up in finite x, impossible; hence y bounded, limits exist, and hypothesis gives y→0 and y′→0.",
          "Thus in every scenario y(x)→0 and y′(x)→0 as x→+∞."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Exponent of the non-derivative term in the hypothesis expression (currently y^3). Any odd (or generally positive) power keeps the argument intact.",
            "original": "3"
          },
          "slot2": {
            "description": "Direction of the limit for x (currently x→+∞); the same proof works if we send x→−∞ instead.",
            "original": "+∞"
          },
          "slot3": {
            "description": "Implicit coefficient in front of (y′)^2 in the hypothesis (presently 1). Any positive constant would work.",
            "original": "1"
          },
          "slot4": {
            "description": "Constant chosen in the comparison ODE y′ = −(1/2)|y|^{3/2}; any positive constant serves the same purpose.",
            "original": "1/2"
          },
          "slot5": {
            "description": "Exponent 3/2 in the comparison inequality (half of the power in slot1); it changes consistently if slot1 changes.",
            "original": "3/2"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}