{
  "index": "2011-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "\\[\n\\lim_{r\\to\\infty} \\frac{r^c \\int_0^{\\pi/2} x^r \\sin x \\,dx}{\\int_0^{\\pi/2} x^r \\cos x \\,dx} = L.\n\\]",
  "solution": "We claim that $(c,L) = (-1,2/\\pi)$ works.\nWrite $f(r) = \\int_0^{\\pi/2} x^r\\sin x\\,dx$. Then\n\\[\nf(r)  < \\int_0^{\\pi/2} x^r\\,dx = \\frac{(\\pi/2)^{r+1}}{r+1}\n\\]\nwhile since $\\sin x \\geq 2x/\\pi$ for $x \\leq \\pi/2$,\n\\[\nf(r) > \\int_0^{\\pi/2} \\frac{2x^{r+1}}{\\pi} \\,dx = \\frac{(\\pi/2)^{r+1}}{r+2}.\n\\]\nIt follows that\n\\[\n\\lim_{r\\to\\infty} r \\left(\\frac{2}{\\pi}\\right)^{r+1} f(r) = 1,\n\\]\nwhence\n\\[\n\\lim_{r\\to\\infty} \\frac{f(r)}{f(r+1)} = \\lim_{r\\to\\infty}\n\\frac{r(2/\\pi)^{r+1}f(r)}{(r+1)(2/\\pi)^{r+2}f(r+1)} \\cdot\n\\frac{2(r+1)}{\\pi r} = \\frac{2}{\\pi}.\n\\]\n\nNow by integration by parts, we have\n\\[\n\\int_0^{\\pi/2} x^r\\cos x\\,dx = \\frac{1}{r+1} \\int_0^{\\pi/2} x^{r+1} \\sin x\\,dx\n    = \\frac{f(r+1)}{r+1}.\n\\]\nThus setting $c = -1$ in the given limit yields\n\\[\n\\lim_{r\\to\\infty} \\frac{(r+1)f(r)}{r f(r+1)} =\n\\frac{2}{\\pi},\n\\]\nas desired.",
  "vars": [
    "r",
    "x",
    "f"
  ],
  "params": [
    "c",
    "L"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "expindex",
        "x": "integrationvariable",
        "f": "integralfunc",
        "c": "powerconstant",
        "L": "limitvalue"
      },
      "question": "\\[\n\\lim_{expindex\\to\\infty} \\frac{expindex^{powerconstant} \\int_0^{\\pi/2} integrationvariable^{expindex} \\sin integrationvariable \\,dintegrationvariable}{\\int_0^{\\pi/2} integrationvariable^{expindex} \\cos integrationvariable \\,dintegrationvariable} = limitvalue.\n\\]",
      "solution": "We claim that $(powerconstant,limitvalue) = (-1,2/\\pi)$ works.\nWrite $integralfunc(expindex) = \\int_0^{\\pi/2} integrationvariable^{expindex}\\sin integrationvariable\\,dintegrationvariable$. Then\n\\[\nintegralfunc(expindex)  < \\int_0^{\\pi/2} integrationvariable^{expindex}\\,dintegrationvariable = \\frac{(\\pi/2)^{expindex+1}}{expindex+1}\n\\]\nwhile since $\\sin integrationvariable \\geq 2integrationvariable/\\pi$ for $integrationvariable \\leq \\pi/2$,\n\\[\nintegralfunc(expindex) > \\int_0^{\\pi/2} \\frac{2integrationvariable^{expindex+1}}{\\pi} \\,dintegrationvariable = \\frac{(\\pi/2)^{expindex+1}}{expindex+2}.\n\\]\nIt follows that\n\\[\n\\lim_{expindex\\to\\infty} expindex \\left(\\frac{2}{\\pi}\\right)^{expindex+1} integralfunc(expindex) = 1,\n\\]\nwhence\n\\[\n\\lim_{expindex\\to\\infty} \\frac{integralfunc(expindex)}{integralfunc(expindex+1)} = \\lim_{expindex\\to\\infty}\n\\frac{expindex(2/\\pi)^{expindex+1}integralfunc(expindex)}{(expindex+1)(2/\\pi)^{expindex+2}integralfunc(expindex+1)} \\cdot\n\\frac{2(expindex+1)}{\\pi expindex} = \\frac{2}{\\pi}.\n\\]\n\nNow by integration by parts, we have\n\\[\n\\int_0^{\\pi/2} integrationvariable^{expindex}\\cos integrationvariable\\,dintegrationvariable = \\frac{1}{expindex+1} \\int_0^{\\pi/2} integrationvariable^{expindex+1} \\sin integrationvariable\\,dintegrationvariable\n    = \\frac{integralfunc(expindex+1)}{expindex+1}.\n\\]\nThus setting $powerconstant = -1$ in the given limit yields\n\\[\n\\lim_{expindex\\to\\infty} \\frac{(expindex+1)integralfunc(expindex)}{expindex integralfunc(expindex+1)} =\n\\frac{2}{\\pi},\n\\]\nas desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "pinecone",
        "x": "sandglass",
        "f": "teaspoon",
        "c": "chinchilla",
        "L": "watermelon"
      },
      "question": "\\[\\n\\lim_{pinecone\\to\\infty} \\frac{pinecone^{chinchilla} \\int_0^{\\pi/2} sandglass^{pinecone} \\sin sandglass \\,d sandglass}{\\int_0^{\\pi/2} sandglass^{pinecone} \\cos sandglass \\,d sandglass} = watermelon.\\n\\]",
      "solution": "We claim that $(chinchilla,watermelon) = (-1,2/\\pi)$ works.\\nWrite $teaspoon(pinecone) = \\int_0^{\\pi/2} sandglass^{pinecone}\\sin sandglass\\,d sandglass$. Then\\n\\[\\nteaspoon(pinecone)  < \\int_0^{\\pi/2} sandglass^{pinecone}\\,d sandglass = \\frac{(\\pi/2)^{pinecone+1}}{pinecone+1}\\n\\]\\nwhile since $\\sin sandglass \\geq 2sandglass/\\pi$ for $sandglass \\leq \\pi/2$,\\n\\[\\nteaspoon(pinecone) > \\int_0^{\\pi/2} \\frac{2sandglass^{pinecone+1}}{\\pi} \\,d sandglass = \\frac{(\\pi/2)^{pinecone+1}}{pinecone+2}.\\n\\]\\nIt follows that\\n\\[\\n\\lim_{pinecone\\to\\infty} pinecone \\left(\\frac{2}{\\pi}\\right)^{pinecone+1} teaspoon(pinecone) = 1,\\n\\]\\nwhence\\n\\[\\n\\lim_{pinecone\\to\\infty} \\frac{teaspoon(pinecone)}{teaspoon(pinecone+1)} = \\lim_{pinecone\\to\\infty}\\n\\frac{pinecone(2/\\pi)^{pinecone+1}teaspoon(pinecone)}{(pinecone+1)(2/\\pi)^{pinecone+2}teaspoon(pinecone+1)} \\cdot\\n\\frac{2(pinecone+1)}{\\pi pinecone} = \\frac{2}{\\pi}.\\n\\]\\n\\nNow by integration by parts, we have\\n\\[\\n\\int_0^{\\pi/2} sandglass^{pinecone}\\cos sandglass\\,d sandglass = \\frac{1}{pinecone+1} \\int_0^{\\pi/2} sandglass^{pinecone+1} \\sin sandglass\\,d sandglass\\n    = \\frac{teaspoon(pinecone+1)}{pinecone+1}.\\n\\]\\nThus setting $chinchilla = -1$ in the given limit yields\\n\\[\\n\\lim_{pinecone\\to\\infty} \\frac{(pinecone+1)teaspoon(pinecone)}{pinecone teaspoon(pinecone+1)} =\\n\\frac{2}{\\pi},\\n\\]\\nas desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "corepoint",
        "x": "knownvalue",
        "f": "constant",
        "c": "variable",
        "L": "beginning"
      },
      "question": "\\[\n\\lim_{corepoint\\to\\infty} \\frac{corepoint^{variable} \\int_0^{\\pi/2} knownvalue^{corepoint} \\sin knownvalue \\,dknownvalue}{\\int_0^{\\pi/2} knownvalue^{corepoint} \\cos knownvalue \\,dknownvalue} = beginning.\n\\]",
      "solution": "We claim that $(variable,beginning) = (-1,2/\\pi)$ works.\nWrite $constant(corepoint) = \\int_0^{\\pi/2} knownvalue^{corepoint}\\sin knownvalue\\,dknownvalue$. Then\n\\[\nconstant(corepoint)  < \\int_0^{\\pi/2} knownvalue^{corepoint}\\,dknownvalue = \\frac{(\\pi/2)^{corepoint+1}}{corepoint+1}\n\\]\nwhile since $\\sin knownvalue \\geq 2knownvalue/\\pi$ for $knownvalue \\leq \\pi/2$,\n\\[\nconstant(corepoint) > \\int_0^{\\pi/2} \\frac{2\\,knownvalue^{corepoint+1}}{\\pi} \\,dknownvalue = \\frac{(\\pi/2)^{corepoint+1}}{corepoint+2}.\n\\]\nIt follows that\n\\[\n\\lim_{corepoint\\to\\infty} corepoint \\left(\\frac{2}{\\pi}\\right)^{corepoint+1} constant(corepoint) = 1,\n\\]\nwhence\n\\[\n\\lim_{corepoint\\to\\infty} \\frac{constant(corepoint)}{constant(corepoint+1)} = \\lim_{corepoint\\to\\infty}\n\\frac{corepoint(2/\\pi)^{corepoint+1}constant(corepoint)}{(corepoint+1)(2/\\pi)^{corepoint+2}constant(corepoint+1)} \\cdot\n\\frac{2(corepoint+1)}{\\pi corepoint} = \\frac{2}{\\pi}.\n\\]\n\nNow by integration by parts, we have\n\\[\n\\int_0^{\\pi/2} knownvalue^{corepoint}\\cos knownvalue\\,dknownvalue = \\frac{1}{corepoint+1} \\int_0^{\\pi/2} knownvalue^{corepoint+1} \\sin knownvalue\\,dknownvalue\n    = \\frac{constant(corepoint+1)}{corepoint+1}.\n\\]\nThus setting $variable = -1$ in the given limit yields\n\\[\n\\lim_{corepoint\\to\\infty} \\frac{(corepoint+1)constant(corepoint)}{corepoint \\, constant(corepoint+1)} =\n\\frac{2}{\\pi},\n\\]\nas desired."
    },
    "garbled_string": {
      "map": {
        "r": "qzxwvtnp",
        "x": "hjgrksla",
        "f": "snvbdmwe",
        "c": "klpqrstu",
        "L": "wjfhxqzp"
      },
      "question": "\\[\\n\\lim_{qzxwvtnp\\to\\infty} \\frac{qzxwvtnp^{klpqrstu} \\int_0^{\\pi/2} hjgrksla^{qzxwvtnp} \\sin hjgrksla \\,dhjgrksla}{\\int_0^{\\pi/2} hjgrksla^{qzxwvtnp} \\cos hjgrksla \\,dhjgrksla} = wjfhxqzp.\\n\\]",
      "solution": "We claim that $(klpqrstu,wjfhxqzp) = (-1,2/\\pi)$ works.\\nWrite $snvbdmwe(qzxwvtnp) = \\int_0^{\\pi/2} hjgrksla^{qzxwvtnp}\\sin hjgrksla\\,dhjgrksla$. Then\\n\\[\\nsnvbdmwe(qzxwvtnp)  < \\int_0^{\\pi/2} hjgrksla^{qzxwvtnp}\\,dhjgrksla = \\frac{(\\pi/2)^{qzxwvtnp+1}}{qzxwvtnp+1}\\n\\]\\nwhile since $\\sin hjgrksla \\geq 2hjgrksla/\\pi$ for $hjgrksla \\leq \\pi/2$,\\n\\[\\nsnvbdmwe(qzxwvtnp) > \\int_0^{\\pi/2} \\frac{2hjgrksla^{qzxwvtnp+1}}{\\pi} \\,dhjgrksla = \\frac{(\\pi/2)^{qzxwvtnp+1}}{qzxwvtnp+2}.\\n\\]\\nIt follows that\\n\\[\\n\\lim_{qzxwvtnp\\to\\infty} qzxwvtnp \\left(\\frac{2}{\\pi}\\right)^{qzxwvtnp+1} snvbdmwe(qzxwvtnp) = 1,\\n\\]\\nwhence\\n\\[\\n\\lim_{qzxwvtnp\\to\\infty} \\frac{snvbdmwe(qzxwvtnp)}{snvbdmwe(qzxwvtnp+1)} = \\lim_{qzxwvtnp\\to\\infty}\\n\\frac{qzxwvtnp(2/\\pi)^{qzxwvtnp+1}snvbdmwe(qzxwvtnp)}{(qzxwvtnp+1)(2/\\pi)^{qzxwvtnp+2}snvbdmwe(qzxwvtnp+1)} \\cdot\\n\\frac{2(qzxwvtnp+1)}{\\pi qzxwvtnp} = \\frac{2}{\\pi}.\\n\\]\\nNow by integration by parts, we have\\n\\[\\n\\int_0^{\\pi/2} hjgrksla^{qzxwvtnp}\\cos hjgrksla\\,dhjgrksla = \\frac{1}{qzxwvtnp+1} \\int_0^{\\pi/2} hjgrksla^{qzxwvtnp+1} \\sin hjgrksla\\,dhjgrksla\\n    = \\frac{snvbdmwe(qzxwvtnp+1)}{qzxwvtnp+1}.\\n\\]\\nThus setting $klpqrstu = -1$ in the given limit yields\\n\\[\\n\\lim_{qzxwvtnp\\to\\infty} \\frac{(qzxwvtnp+1)snvbdmwe(qzxwvtnp)}{qzxwvtnp snvbdmwe(qzxwvtnp+1)} =\\n\\frac{2}{\\pi},\\n\\]\\nas desired."
    },
    "kernel_variant": {
      "question": "Fix θ = π⁄6.  \nFor every real r > −1 set  \n\n Fᵣ = ∬_{x≥0,\\,y≥0,\\,x+y≤θ} (x y)^{\\,r}\\,sin[3(x+y)] \\,dx\\,dy,\n\n Gᵣ = 1⁄3 ∬_{x≥0,\\,y≥0,\\,x+y≤θ} (x y)^{\\,r+½}\\,cos[3(x+y)] \\,dx\\,dy.\n\na)  Show that there is a unique real constant c such that  \n\n   lim_{r→∞}  r^{\\,c}\\,\\dfrac{Fᵣ}{Gᵣ}\n\nexists, is finite and positive.\n\nb)  Determine that constant c and evaluate the limit  \n   L = lim_{r→∞}  r^{\\,c}\\,Fᵣ/Gᵣ.",
      "solution": "Step 1.  Separating the two variables.  \nIntroduce the change of variables  \n\n s = x + y  (0 ≤ s ≤ θ),  t = x/(x + y)  (0 ≤ t ≤ 1),\n\nwhose Jacobian is |∂(x,y)/∂(s,t)| = s.  Thus  \n\n x = s t, y = s(1−t), dx dy = s\\,dt ds.\n\nHence  \n\n (xy)^{r} = s^{2r}\\,t^{r}(1−t)^{r}, (xy)^{r+½} = s^{2r+1}\\,t^{r+½}(1−t)^{r+½}.\n\nWith these and the fact that sin 3(x+y) = sin 3s, cos 3(x+y) = cos 3s we obtain  \n\n Fᵣ = B(r+1,r+1) · Iᵣ,  Gᵣ = 1⁄3 · B(r+3/2,r+3/2) · Jᵣ, (1)\n\nwhere  \n\n B(a,b) = Beta(a,b) = ∫_{0}^{1} t^{a−1}(1−t)^{b−1}dt,\n\n Iᵣ = ∫_{0}^{θ} s^{2r+1} sin 3s\\,ds,  Jᵣ = ∫_{0}^{θ} s^{2r+2} cos 3s\\,ds.\n\nStep 2.  Asymptotics of the Beta–quotient.  \nPut  \n\n Qᵣ := B(r+1,r+1) / B(r+3/2,r+3/2).\n\nUsing Stirling’s formula Γ(z+a) ~ √{2π}\\,z^{z+a-½}e^{-z}\\,(1+O(1/z)) we have\n\n Γ(r+1)  ~ √{2π}\\,r^{\\,r+½}e^{-r},  \n Γ(r+3/2)~ √{2π}\\,r^{\\,r+1 }e^{-r},  \n Γ(2r+2)/Γ(2r+3) = 1/(2r+2).\n\nConsequently  \n\n Qᵣ = [Γ(r+1)/Γ(r+3/2)]² · Γ(2r+3)/Γ(2r+2)  \n   ~ r^{−1} · (2r+2) → 2   (r→∞).   (2)\n\nStep 3.  Edge–dominated asymptotics of Iᵣ and Jᵣ.  \nSet n = 2r+1 and m = n+1 = 2r+2.  Writing s = θt in Iᵣ and Jᵣ one gets  \n\n Iᵣ = θ^{n+1} ∫_{0}^{1} t^{n} sin(3θt) dt,  \n Jᵣ = θ^{m+1} ∫_{0}^{1} t^{m} cos(3θt) dt. (3)\n\nBecause n, m → ∞ the integrals are controlled by a narrow layer near t=1.  \nPut t = 1 − u/n in the first integral, t = 1 − u/m in the second; then\n\n t^{n} ≈ e^{−u},  t^{m} ≈ e^{−u}.\n\nSince 3θ = π/2, we have near t = 1  \n\n sin(3θt) = sin[π/2 − (πu)/(2n)] = cos[(πu)/(2n)] = 1 + O(1/n²),  \n cos(3θt) = sin[(πu)/(2m)] ≈ (πu)/(2m) + O(1/m³).\n\nTherefore\n\n Iᵣ = θ^{n+1}·(1/n) ∫_{0}^{∞} e^{−u}du + O(θ^{n+1}/n²)\n   = θ^{2r+2}/(2r+1) + O(θ^{2r}/r²),   (4)\n\n Jᵣ = θ^{m+1}·(π/2m²) ∫_{0}^{∞} u e^{−u}du + O(θ^{m+1}/m³)\n   = π θ^{2r+3}/[2(2r+2)²] + O(θ^{2r+1}/r³).   (5)\n\nStep 4.  The quotient Iᵣ/Jᵣ.  \nCombining (4) and (5):\n\n Iᵣ/Jᵣ\n  = [θ^{2r+2}/(2r+1)] · [2(2r+2)²/(π θ^{2r+3})] · [1 + o(1)]\n  = (4/πθ) · r · [1 + o(1)].   (6)\n\nStep 5.  Completing the asymptotics of Fᵣ / Gᵣ.  \nFrom (1), (2) and (6):\n\n Fᵣ / Gᵣ\n  = 3 · Qᵣ · (Iᵣ/Jᵣ)\n  = 3 · [2 + o(1)] · [(4/πθ) r (1 + o(1))]\n  = (24/πθ)·r · [1 + o(1)].   (7)\n\nStep 6.  The correct scaling and the limit.  \nBecause Fᵣ/Gᵣ grows linearly with r, multiplying by r^{−1} yields a finite non–zero limit.  \nThus the unique constant is  \n\n c = −1. (8)\n\nUsing θ = π/6 in (7) we finally get  \n\n lim_{r→∞} r^{−1} Fᵣ/Gᵣ = (24/π) · (6/π) = 144/π². (9)\n\nHence\n\n L = 144 ⁄ π².\n\nAnswer: c = −1 and L = 144/π².",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.818514",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional setting: the problem involves a genuine 2-dimensional integral over a triangular domain, unlike the 1-dimensional originals.  \n2. Additional parameters: two separate exponents (r and r+½) and a non–trivial trigonometric frequency introduce asymmetric growth rates.  \n3. Multiple advanced techniques: the solution uses a change of variables, the Beta function, Stirling’s formula, edge-type Laplace/steepest-descent analysis, and delicate cancellation of leading asymptotic terms.  \n4. Interaction of concepts: the r-dependence coming from the Beta quotient and from the edge-dominated integrals balance only after careful second-order estimates; missing any of these details gives the wrong scale.  \n5. Depth: proving the existence of the limit and computing it require several pages of precise asymptotics, far beyond the single integration-by-parts argument that solves the original kernel variant.\n\nThese added layers make the enhanced problem substantially harder than both the initial problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix θ = π⁄6.  \nFor every real r > −1 set  \n\n Fᵣ = ∬_{x≥0,\\,y≥0,\\,x+y≤θ} (x y)^{\\,r}\\,sin[3(x+y)] \\,dx\\,dy,\n\n Gᵣ = 1⁄3 ∬_{x≥0,\\,y≥0,\\,x+y≤θ} (x y)^{\\,r+½}\\,cos[3(x+y)] \\,dx\\,dy.\n\na)  Show that there is a unique real constant c such that  \n\n   lim_{r→∞}  r^{\\,c}\\,\\dfrac{Fᵣ}{Gᵣ}\n\nexists, is finite and positive.\n\nb)  Determine that constant c and evaluate the limit  \n   L = lim_{r→∞}  r^{\\,c}\\,Fᵣ/Gᵣ.",
      "solution": "Step 1.  Separating the two variables.  \nIntroduce the change of variables  \n\n s = x + y  (0 ≤ s ≤ θ),  t = x/(x + y)  (0 ≤ t ≤ 1),\n\nwhose Jacobian is |∂(x,y)/∂(s,t)| = s.  Thus  \n\n x = s t, y = s(1−t), dx dy = s\\,dt ds.\n\nHence  \n\n (xy)^{r} = s^{2r}\\,t^{r}(1−t)^{r}, (xy)^{r+½} = s^{2r+1}\\,t^{r+½}(1−t)^{r+½}.\n\nWith these and the fact that sin 3(x+y) = sin 3s, cos 3(x+y) = cos 3s we obtain  \n\n Fᵣ = B(r+1,r+1) · Iᵣ,  Gᵣ = 1⁄3 · B(r+3/2,r+3/2) · Jᵣ, (1)\n\nwhere  \n\n B(a,b) = Beta(a,b) = ∫_{0}^{1} t^{a−1}(1−t)^{b−1}dt,\n\n Iᵣ = ∫_{0}^{θ} s^{2r+1} sin 3s\\,ds,  Jᵣ = ∫_{0}^{θ} s^{2r+2} cos 3s\\,ds.\n\nStep 2.  Asymptotics of the Beta–quotient.  \nPut  \n\n Qᵣ := B(r+1,r+1) / B(r+3/2,r+3/2).\n\nUsing Stirling’s formula Γ(z+a) ~ √{2π}\\,z^{z+a-½}e^{-z}\\,(1+O(1/z)) we have\n\n Γ(r+1)  ~ √{2π}\\,r^{\\,r+½}e^{-r},  \n Γ(r+3/2)~ √{2π}\\,r^{\\,r+1 }e^{-r},  \n Γ(2r+2)/Γ(2r+3) = 1/(2r+2).\n\nConsequently  \n\n Qᵣ = [Γ(r+1)/Γ(r+3/2)]² · Γ(2r+3)/Γ(2r+2)  \n   ~ r^{−1} · (2r+2) → 2   (r→∞).   (2)\n\nStep 3.  Edge–dominated asymptotics of Iᵣ and Jᵣ.  \nSet n = 2r+1 and m = n+1 = 2r+2.  Writing s = θt in Iᵣ and Jᵣ one gets  \n\n Iᵣ = θ^{n+1} ∫_{0}^{1} t^{n} sin(3θt) dt,  \n Jᵣ = θ^{m+1} ∫_{0}^{1} t^{m} cos(3θt) dt. (3)\n\nBecause n, m → ∞ the integrals are controlled by a narrow layer near t=1.  \nPut t = 1 − u/n in the first integral, t = 1 − u/m in the second; then\n\n t^{n} ≈ e^{−u},  t^{m} ≈ e^{−u}.\n\nSince 3θ = π/2, we have near t = 1  \n\n sin(3θt) = sin[π/2 − (πu)/(2n)] = cos[(πu)/(2n)] = 1 + O(1/n²),  \n cos(3θt) = sin[(πu)/(2m)] ≈ (πu)/(2m) + O(1/m³).\n\nTherefore\n\n Iᵣ = θ^{n+1}·(1/n) ∫_{0}^{∞} e^{−u}du + O(θ^{n+1}/n²)\n   = θ^{2r+2}/(2r+1) + O(θ^{2r}/r²),   (4)\n\n Jᵣ = θ^{m+1}·(π/2m²) ∫_{0}^{∞} u e^{−u}du + O(θ^{m+1}/m³)\n   = π θ^{2r+3}/[2(2r+2)²] + O(θ^{2r+1}/r³).   (5)\n\nStep 4.  The quotient Iᵣ/Jᵣ.  \nCombining (4) and (5):\n\n Iᵣ/Jᵣ\n  = [θ^{2r+2}/(2r+1)] · [2(2r+2)²/(π θ^{2r+3})] · [1 + o(1)]\n  = (4/πθ) · r · [1 + o(1)].   (6)\n\nStep 5.  Completing the asymptotics of Fᵣ / Gᵣ.  \nFrom (1), (2) and (6):\n\n Fᵣ / Gᵣ\n  = 3 · Qᵣ · (Iᵣ/Jᵣ)\n  = 3 · [2 + o(1)] · [(4/πθ) r (1 + o(1))]\n  = (24/πθ)·r · [1 + o(1)].   (7)\n\nStep 6.  The correct scaling and the limit.  \nBecause Fᵣ/Gᵣ grows linearly with r, multiplying by r^{−1} yields a finite non–zero limit.  \nThus the unique constant is  \n\n c = −1. (8)\n\nUsing θ = π/6 in (7) we finally get  \n\n lim_{r→∞} r^{−1} Fᵣ/Gᵣ = (24/π) · (6/π) = 144/π². (9)\n\nHence\n\n L = 144 ⁄ π².\n\nAnswer: c = −1 and L = 144/π².",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.626421",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional setting: the problem involves a genuine 2-dimensional integral over a triangular domain, unlike the 1-dimensional originals.  \n2. Additional parameters: two separate exponents (r and r+½) and a non–trivial trigonometric frequency introduce asymmetric growth rates.  \n3. Multiple advanced techniques: the solution uses a change of variables, the Beta function, Stirling’s formula, edge-type Laplace/steepest-descent analysis, and delicate cancellation of leading asymptotic terms.  \n4. Interaction of concepts: the r-dependence coming from the Beta quotient and from the edge-dominated integrals balance only after careful second-order estimates; missing any of these details gives the wrong scale.  \n5. Depth: proving the existence of the limit and computing it require several pages of precise asymptotics, far beyond the single integration-by-parts argument that solves the original kernel variant.\n\nThese added layers make the enhanced problem substantially harder than both the initial problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}