{
  "index": "2021-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "For every positive real number $x$, let\n\\[\ng(x) = \\lim_{r \\to 0} ((x+1)^{r+1} - x^{r+1})^{\\frac{1}{r}}.\n\\]\nFind $\\lim_{x \\to \\infty} \\frac{g(x)}{x}$.",
  "solution": "The limit is $e$.\n\n\\noindent\n\\textbf{First solution.}\nBy l'H\\^opital's Rule, we have\n\\begin{align*}\n&\\lim_{r\\to 0} \\frac{\\log((x+1)^{r+1}-x^{r+1})}{r} \\\\\n&\\quad = \\lim_{r\\to 0} \\frac{d}{dr} \\log((x+1)^{r+1}-x^{r+1}) \\\\\n&\\quad = \\lim_{r\\to 0} \\frac{(x+1)^{r+1}\\log(x+1)-x^{r+1}\\log x}{(x+1)^{r+1}-x^{r+1}} \\\\\n&\\quad = (x+1)\\log(x+1)-x\\log x,\n\\end{align*}\nwhere $\\log$ denotes natural logarithm. It follows that $g(x) = e^{(x+1)\\log(x+1)-x\\log x} = \\frac{(x+1)^{x+1}}{x^x}$. Thus\n\\[\n\\lim_{x\\to\\infty} \\frac{g(x)}{x} = \\left(\\lim_{x\\to\\infty}\\frac{x+1}{x}\\right) \\cdot \\left(\\lim_{x\\to\\infty} \\left(1+\\frac{1}{x}\\right)^x\\right) = 1\\cdot e = e.\n\\]\n\n\\noindent\n\\textbf{Second solution.}\nWe first write \n\\begin{align*}\n\\lim_{x \\to \\infty} \\frac{g(x)}{x} &= \\lim_{x \\to \\infty} \\lim_{r \\to 0} \\frac{((x+1)^{r+1} - x^{r+1})^{1/r}}{x} \\\\\n&= \\lim_{x \\to \\infty} \\lim_{r \\to 0} \\frac{((r+1) x^r + O(x^{r-1}))^{1/r}}{x}.\n\\end{align*}\nWe would like to interchange the order of the limits, but this requires some justification.\nUsing Taylor's theorem with remainder, for $x \\geq 1$, $r \\leq 1$\nwe can bound the error term $O(x^{r-1})$ in absolute value by $(r+1) r x^{r-1}$. This\nmeans that if we continue to rewrite the orginial limit as\n\\[\n\\lim_{r\\to 0} \\lim_{x\\to\\infty} (r+1+O(x^{-1}))^{1/r},\n\\]\nthe error term $O(x^{-1})$ is bounded in absolute value by $(r+1) r/x$.\nFor $x \\geq 1$, $r \\leq 1$ this quantity is bounded in absolute value by $(r+1)r$, \\emph{independently of $x$}. This allows us to continue by interchanging the order of the limits,\nobtaining \n\\begin{align*}\n&\\lim_{r\\to 0} \\lim_{x\\to\\infty} (r+1+O(x^{-1}))^{1/r} \\\\\n&\\quad = \\lim_{r\\to 0} (r+1)^{1/r} \\\\\n&\\quad = \\lim_{s\\to \\infty} (1+1/s)^{s} = e,\n\\end{align*}\nwhere in the last step we take $s = 1/r$.\n\n\\noindent\n\\textbf{Third solution.} (by Clayton Lungstrum)\nWe first observe that\n\\begin{align*}\n((x+1)^{r+1} - x^{r+1})^{1/r}\n&= \\left( \\int_x^{x+1} (r+1)u^r\\,du \\right)^{1/r} \\\\\n&= (r+1)^{1/r} \\left( \\int_x^{x+1} u^r\\,du \\right)^{1/r}.\n\\end{align*}\nSince $\\lim_{r \\to 0} (r+1)^{1/r} = e$, we deduce that\n\\[\ng(x) = e \\lim_{r \\to 0} \\left( \\int_x^{x+1} u^r\\,du \\right)^{1/r}.\n\\]\nFor $r > 0$, $u^r$ is increasing for $x \\leq u \\leq x+1$, so\n\\[\nx^r \\leq \\int_x^{x+1} u^r\\,du \\leq (x+1)^r;\n\\]\nfor $r < 0$, $u^r$ is decreasing for $x \\leq u \\leq x+1$, so\n\\[\nx^r \\geq \\int_x^{x+1} u^r\\,du \\geq (x+1)^r.\n\\]\nIn both cases, we deduce that\n\\[\nx \\leq \\left( \\int_x^{x+1} u^r\\,du \\right)^{1/r} \\leq x+1;\n\\]\napplying the squeeze theorem to the resulting inequality\n $e \\leq \\frac{g(x)}{x} \\leq e\\left( 1 + \\frac{1}{x} \\right)$\n yields the claimed limit.",
  "vars": [
    "x",
    "r",
    "u",
    "s"
  ],
  "params": [
    "g",
    "O"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvar",
        "r": "rateparm",
        "u": "upperint",
        "s": "scalefac",
        "g": "gfunction",
        "O": "bigohsym"
      },
      "question": "For every positive real number $realvar$, let\n\\[\ngfunction(realvar) = \\lim_{rateparm \\to 0} ((realvar+1)^{rateparm+1} - realvar^{rateparm+1})^{\\frac{1}{rateparm}}.\n\\]\nFind $\\lim_{realvar \\to \\infty} \\frac{gfunction(realvar)}{realvar}$.",
      "solution": "The limit is $e$.\n\n\\noindent\n\\textbf{First solution.}\nBy l'H\\^opital's Rule, we have\n\\begin{align*}\n&\\lim_{rateparm\\to 0} \\frac{\\log\\!\\big((realvar+1)^{rateparm+1}-realvar^{rateparm+1}\\big)}{rateparm} \\\\\n&\\quad = \\lim_{rateparm\\to 0} \\frac{d}{d rateparm}\\, \\log\\!\\big((realvar+1)^{rateparm+1}-realvar^{rateparm+1}\\big) \\\\\n&\\quad = \\lim_{rateparm\\to 0} \\frac{(realvar+1)^{rateparm+1}\\log(realvar+1)-realvar^{rateparm+1}\\log realvar}{(realvar+1)^{rateparm+1}-realvar^{rateparm+1}} \\\\\n&\\quad = (realvar+1)\\log(realvar+1)-realvar\\log realvar,\n\\end{align*}\nwhere $\\log$ denotes the natural logarithm. It follows that $gfunction(realvar)=e^{(realvar+1)\\log(realvar+1)-realvar\\log realvar}= \\dfrac{(realvar+1)^{realvar+1}}{realvar^{realvar}}$. Thus\n\\[\n\\lim_{realvar\\to\\infty} \\frac{gfunction(realvar)}{realvar}= \\left(\\lim_{realvar\\to\\infty}\\frac{realvar+1}{realvar}\\right)\\!\\cdot\\!\\left(\\lim_{realvar\\to\\infty}\\left(1+\\frac{1}{realvar}\\right)^{realvar}\\right)=1\\cdot e = e.\n\\]\n\n\\noindent\n\\textbf{Second solution.}\nWe first write\n\\begin{align*}\n\\lim_{realvar \\to \\infty} \\frac{gfunction(realvar)}{realvar} &= \\lim_{realvar \\to \\infty} \\lim_{rateparm \\to 0} \\frac{((realvar+1)^{rateparm+1} - realvar^{rateparm+1})^{1/rateparm}}{realvar} \\\\\n&= \\lim_{realvar \\to \\infty} \\lim_{rateparm \\to 0} \\frac{((rateparm+1) realvar^{rateparm} + \\bigohsym(realvar^{rateparm-1}))^{1/rateparm}}{realvar}.\n\\end{align*}\nWe would like to interchange the order of the limits, but this requires some justification.\nUsing Taylor's theorem with remainder, for $realvar \\geq 1$, $rateparm \\leq 1$\nwe can bound the error term $\\bigohsym(realvar^{rateparm-1})$ in absolute value by $(rateparm+1)rateparm\\,realvar^{rateparm-1}$. This\nmeans that if we continue to rewrite the original limit as\n\\[\n\\lim_{rateparm\\to 0} \\lim_{realvar\\to\\infty} (rateparm+1+\\bigohsym(realvar^{-1}))^{1/rateparm},\n\\]\nthe error term $\\bigohsym(realvar^{-1})$ is bounded in absolute value by $(rateparm+1)rateparm/realvar$.\nFor $realvar \\geq 1$, $rateparm \\leq 1$ this quantity is bounded in absolute value by $(rateparm+1)rateparm$, independently of $realvar$. This allows us to continue by interchanging the order of the limits,\nobtaining\n\\begin{align*}\n&\\lim_{rateparm\\to 0} \\lim_{realvar\\to\\infty} (rateparm+1+\\bigohsym(realvar^{-1}))^{1/rateparm} \\\\\n&\\quad = \\lim_{rateparm\\to 0} (rateparm+1)^{1/rateparm} \\\\\n&\\quad = \\lim_{scalefac\\to \\infty} \\left(1+\\frac{1}{scalefac}\\right)^{scalefac} = e,\n\\end{align*}\nwhere in the last step we take $scalefac = 1/rateparm$.\n\n\\noindent\n\\textbf{Third solution.} (by Clayton Lungstrum)\nWe first observe that\n\\begin{align*}\n((realvar+1)^{rateparm+1} - realvar^{rateparm+1})^{1/rateparm}\n&= \\left( \\int_{realvar}^{realvar+1} (rateparm+1)\\,upperint^{rateparm}\\,d upperint \\right)^{1/rateparm} \\\\\n&= (rateparm+1)^{1/rateparm} \\left( \\int_{realvar}^{realvar+1} upperint^{rateparm}\\,d upperint \\right)^{1/rateparm}.\n\\end{align*}\nSince $\\lim_{rateparm \\to 0} (rateparm+1)^{1/rateparm} = e$, we deduce that\n\\[\ngfunction(realvar) = e \\lim_{rateparm \\to 0} \\left( \\int_{realvar}^{realvar+1} upperint^{rateparm}\\,d upperint \\right)^{1/rateparm}.\n\\]\nFor $rateparm > 0$, $upperint^{rateparm}$ is increasing for $realvar \\leq upperint \\leq realvar+1$, so\n\\[\nrealvar^{rateparm} \\leq \\int_{realvar}^{realvar+1} upperint^{rateparm}\\,d upperint \\leq (realvar+1)^{rateparm};\n\\]\nfor $rateparm < 0$, $upperint^{rateparm}$ is decreasing for $realvar \\leq upperint \\leq realvar+1$, so\n\\[\nrealvar^{rateparm} \\geq \\int_{realvar}^{realvar+1} upperint^{rateparm}\\,d upperint \\geq (realvar+1)^{rateparm}.\n\\]\nIn both cases, we deduce that\n\\[\nrealvar \\leq \\left( \\int_{realvar}^{realvar+1} upperint^{rateparm}\\,d upperint \\right)^{1/rateparm} \\leq realvar+1;\n\\]\napplying the squeeze theorem to the resulting inequality\n $e \\leq \\frac{gfunction(realvar)}{realvar} \\leq e\\left( 1 + \\frac{1}{realvar} \\right)$\n yields the claimed limit."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "paintbrush",
        "r": "belltower",
        "u": "moonlight",
        "s": "driftwood",
        "g": "pineapple",
        "O": "candlestick"
      },
      "question": "For every positive real number $paintbrush$, let\n\\[\n\\pineapple(paintbrush) = \\lim_{belltower \\to 0} ((paintbrush+1)^{belltower+1} - paintbrush^{belltower+1})^{\\frac{1}{belltower}}.\n\\]\nFind $\\lim_{paintbrush \\to \\infty} \\frac{\\pineapple(paintbrush)}{paintbrush}$.",
      "solution": "The limit is $e$.\n\n\\noindent\n\\textbf{First solution.}\nBy l'H\\^opital's Rule, we have\n\\begin{align*}\n&\\lim_{belltower\\to 0} \\frac{\\log((paintbrush+1)^{belltower+1}-paintbrush^{belltower+1})}{belltower} \\\\\n&\\quad = \\lim_{belltower\\to 0} \\frac{d}{d belltower} \\log((paintbrush+1)^{belltower+1}-paintbrush^{belltower+1}) \\\\\n&\\quad = \\lim_{belltower\\to 0} \\frac{(paintbrush+1)^{belltower+1}\\log(paintbrush+1)-paintbrush^{belltower+1}\\log paintbrush}{(paintbrush+1)^{belltower+1}-paintbrush^{belltower+1}} \\\\\n&\\quad = (paintbrush+1)\\log(paintbrush+1)-paintbrush\\log paintbrush,\n\\end{align*}\nwhere $\\log$ denotes natural logarithm. It follows that $\\pineapple(paintbrush) = e^{(paintbrush+1)\\log(paintbrush+1)-paintbrush\\log paintbrush} = \\frac{(paintbrush+1)^{paintbrush+1}}{paintbrush^{paintbrush}}$. Thus\n\\[\n\\lim_{paintbrush\\to\\infty} \\frac{\\pineapple(paintbrush)}{paintbrush} = \\left(\\lim_{paintbrush\\to\\infty}\\frac{paintbrush+1}{paintbrush}\\right) \\cdot \\left(\\lim_{paintbrush\\to\\infty} \\left(1+\\frac{1}{paintbrush}\\right)^{paintbrush}\\right) = 1\\cdot e = e.\n\\]\n\n\\noindent\n\\textbf{Second solution.}\nWe first write \n\\begin{align*}\n\\lim_{paintbrush \\to \\infty} \\frac{\\pineapple(paintbrush)}{paintbrush} &= \\lim_{paintbrush \\to \\infty} \\lim_{belltower \\to 0} \\frac{((paintbrush+1)^{belltower+1} - paintbrush^{belltower+1})^{1/belltower}}{paintbrush} \\\\\n&= \\lim_{paintbrush \\to \\infty} \\lim_{belltower \\to 0} \\frac{((belltower+1) paintbrush^{belltower} + \\candlestick(paintbrush^{belltower-1}))^{1/belltower}}{paintbrush}.\n\\end{align*}\nWe would like to interchange the order of the limits, but this requires some justification.\nUsing Taylor's theorem with remainder, for $paintbrush \\ge 1$, $belltower \\le 1$ we can bound the error term $\\candlestick(paintbrush^{belltower-1})$ in absolute value by $(belltower+1)belltower paintbrush^{belltower-1}$. This means that if we continue to rewrite the original limit as\n\\[\n\\lim_{belltower\\to 0} \\lim_{paintbrush\\to\\infty} (belltower+1+\\candlestick(paintbrush^{-1}))^{1/belltower},\n\\]\nthe error term $\\candlestick(paintbrush^{-1})$ is bounded in absolute value by $(belltower+1)belltower/paintbrush$.\nFor $paintbrush \\ge 1$, $belltower \\le 1$ this quantity is bounded in absolute value by $(belltower+1)belltower$, \\emph{independently of $paintbrush$}. This allows us to continue by interchanging the order of the limits, obtaining\n\\begin{align*}\n&\\lim_{belltower\\to 0} \\lim_{paintbrush\\to\\infty} (belltower+1+\\candlestick(paintbrush^{-1}))^{1/belltower} \\\\\n&\\quad = \\lim_{belltower\\to 0} (belltower+1)^{1/belltower} \\\\\n&\\quad = \\lim_{driftwood\\to \\infty} (1+1/driftwood)^{driftwood} = e,\n\\end{align*}\nwhere in the last step we take $driftwood = 1/belltower$.\n\n\\noindent\n\\textbf{Third solution.} (by Clayton Lungstrum)\nWe first observe that\n\\begin{align*}\n((paintbrush+1)^{belltower+1} - paintbrush^{belltower+1})^{1/belltower} &= \\left( \\int_{paintbrush}^{paintbrush+1} (belltower+1)moonlight^{belltower}\\,d moonlight \\right)^{1/belltower} \\\\\n&= (belltower+1)^{1/belltower} \\left( \\int_{paintbrush}^{paintbrush+1} moonlight^{belltower}\\,d moonlight \\right)^{1/belltower}.\n\\end{align*}\nSince $\\lim_{belltower \\to 0} (belltower+1)^{1/belltower} = e$, we deduce that\n\\[\n\\pineapple(paintbrush) = e \\lim_{belltower \\to 0} \\left( \\int_{paintbrush}^{paintbrush+1} moonlight^{belltower}\\,d moonlight \\right)^{1/belltower}.\n\\]\nFor $belltower > 0$, $moonlight^{belltower}$ is increasing for $paintbrush \\le moonlight \\le paintbrush+1$, so\n\\[\npaintbrush^{belltower} \\le \\int_{paintbrush}^{paintbrush+1} moonlight^{belltower}\\,d moonlight \\le (paintbrush+1)^{belltower};\n\\]\nfor $belltower < 0$, $moonlight^{belltower}$ is decreasing for $paintbrush \\le moonlight \\le paintbrush+1$, so\n\\[\npaintbrush^{belltower} \\ge \\int_{paintbrush}^{paintbrush+1} moonlight^{belltower}\\,d moonlight \\ge (paintbrush+1)^{belltower}.\n\\]\nIn both cases, we deduce that\n\\[\npaintbrush \\le \\left( \\int_{paintbrush}^{paintbrush+1} moonlight^{belltower}\\,d moonlight \\right)^{1/belltower} \\le paintbrush+1;\n\\]\napplying the squeeze theorem to the resulting inequality $e \\le \\frac{\\pineapple(paintbrush)}{paintbrush} \\le e\\left(1+\\frac{1}{paintbrush}\\right)$ yields the claimed limit."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "tinyvalue",
        "r": "colossal",
        "u": "constant",
        "s": "finiteval",
        "g": "fixedfunc",
        "O": "precision"
      },
      "question": "For every positive real number $tinyvalue$, let\n\\[\nfixedfunc(tinyvalue) = \\lim_{colossal \\to 0} ((tinyvalue+1)^{colossal+1} - tinyvalue^{colossal+1})^{\\frac{1}{colossal}}.\n\\]\nFind $\\lim_{tinyvalue \\to \\infty} \\frac{fixedfunc(tinyvalue)}{tinyvalue}$.",
      "solution": "The limit is $e$.\n\n\\noindent\n\\textbf{First solution.}\nBy l'H\\^opital's Rule, we have\n\\begin{align*}\n&\\lim_{colossal\\to 0} \\frac{\\log((tinyvalue+1)^{colossal+1}-tinyvalue^{colossal+1})}{colossal} \\\\\n&\\quad = \\lim_{colossal\\to 0} \\frac{d}{dcolossal} \\log((tinyvalue+1)^{colossal+1}-tinyvalue^{colossal+1}) \\\\\n&\\quad = \\lim_{colossal\\to 0} \\frac{(tinyvalue+1)^{colossal+1}\\log(tinyvalue+1)-tinyvalue^{colossal+1}\\log tinyvalue}{(tinyvalue+1)^{colossal+1}-tinyvalue^{colossal+1}} \\\\\n&\\quad = (tinyvalue+1)\\log(tinyvalue+1)-tinyvalue\\log tinyvalue,\n\\end{align*}\nwhere $\\log$ denotes natural logarithm. It follows that $fixedfunc(tinyvalue) = e^{(tinyvalue+1)\\log(tinyvalue+1)-tinyvalue\\log tinyvalue} = \\frac{(tinyvalue+1)^{tinyvalue+1}}{tinyvalue^{tinyvalue}}$. Thus\n\\[\n\\lim_{tinyvalue\\to\\infty} \\frac{fixedfunc(tinyvalue)}{tinyvalue} = \\left(\\lim_{tinyvalue\\to\\infty}\\frac{tinyvalue+1}{tinyvalue}\\right) \\cdot \\left(\\lim_{tinyvalue\\to\\infty} \\left(1+\\frac{1}{tinyvalue}\\right)^{tinyvalue}\\right) = 1\\cdot e = e.\n\\]\n\n\\noindent\n\\textbf{Second solution.}\nWe first write \n\\begin{align*}\n\\lim_{tinyvalue \\to \\infty} \\frac{fixedfunc(tinyvalue)}{tinyvalue} &= \\lim_{tinyvalue \\to \\infty} \\lim_{colossal \\to 0} \\frac{((tinyvalue+1)^{colossal+1} - tinyvalue^{colossal+1})^{1/colossal}}{tinyvalue} \\\\\n&= \\lim_{tinyvalue \\to \\infty} \\lim_{colossal \\to 0} \\frac{((colossal+1) tinyvalue^{colossal} + precision(tinyvalue^{colossal-1}))^{1/colossal}}{tinyvalue}.\n\\end{align*}\nWe would like to interchange the order of the limits, but this requires some justification.\nUsing Taylor's theorem with remainder, for $tinyvalue \\geq 1$, $colossal \\leq 1$\nwe can bound the error term $precision(tinyvalue^{colossal-1})$ in absolute value by $(colossal+1) colossal tinyvalue^{colossal-1}$. This\nmeans that if we continue to rewrite the orginial limit as\n\\[\n\\lim_{colossal\\to 0} \\lim_{tinyvalue\\to\\infty} (colossal+1+precision(tinyvalue^{-1}))^{1/colossal},\n\\]\nthe error term $precision(tinyvalue^{-1})$ is bounded in absolute value by $(colossal+1) colossal/tinyvalue$.\nFor $tinyvalue \\geq 1$, $colossal \\leq 1$ this quantity is bounded in absolute value by $(colossal+1)colossal$, \\emph{independently of $tinyvalue$}. This allows us to continue by interchanging the order of the limits,\nobtaining \n\\begin{align*}\n&\\lim_{colossal\\to 0} \\lim_{tinyvalue\\to\\infty} (colossal+1+precision(tinyvalue^{-1}))^{1/colossal} \\\\\n&\\quad = \\lim_{colossal\\to 0} (colossal+1)^{1/colossal} \\\\\n&\\quad = \\lim_{finiteval\\to \\infty} (1+1/finiteval)^{finiteval} = e,\n\\end{align*}\nwhere in the last step we take $finiteval = 1/colossal$.\n\n\\noindent\n\\textbf{Third solution.} (by Clayton Lungstrum)\nWe first observe that\n\\begin{align*}\n((tinyvalue+1)^{colossal+1} - tinyvalue^{colossal+1})^{1/colossal}\n&= \\left( \\int_{tinyvalue}^{tinyvalue+1} (colossal+1)constant^{colossal}\\,dconstant \\right)^{1/colossal} \\\\\n&= (colossal+1)^{1/colossal} \\left( \\int_{tinyvalue}^{tinyvalue+1} constant^{colossal}\\,dconstant \\right)^{1/colossal}.\n\\end{align*}\nSince $\\lim_{colossal \\to 0} (colossal+1)^{1/colossal} = e$, we deduce that\n\\[\nfixedfunc(tinyvalue) = e \\lim_{colossal \\to 0} \\left( \\int_{tinyvalue}^{tinyvalue+1} constant^{colossal}\\,dconstant \\right)^{1/colossal}.\n\\]\nFor $colossal > 0$, $constant^{colossal}$ is increasing for $tinyvalue \\leq constant \\leq tinyvalue+1$, so\n\\[\n tinyvalue^{colossal} \\leq \\int_{tinyvalue}^{tinyvalue+1} constant^{colossal}\\,dconstant \\leq (tinyvalue+1)^{colossal};\n\\]\nfor $colossal < 0$, $constant^{colossal}$ is decreasing for $tinyvalue \\leq constant \\leq tinyvalue+1$, so\n\\[\n tinyvalue^{colossal} \\geq \\int_{tinyvalue}^{tinyvalue+1} constant^{colossal}\\,dconstant \\geq (tinyvalue+1)^{colossal}.\n\\]\nIn both cases, we deduce that\n\\[\n tinyvalue \\leq \\left( \\int_{tinyvalue}^{tinyvalue+1} constant^{colossal}\\,dconstant \\right)^{1/colossal} \\leq tinyvalue+1;\n\\]\napplying the squeeze theorem to the resulting inequality\n $e \\leq \\frac{fixedfunc(tinyvalue)}{tinyvalue} \\leq e\\left( 1 + \\frac{1}{tinyvalue} \\right)$\n yields the claimed limit."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "r": "hjgrksla",
        "u": "pmncblet",
        "s": "zplmhrxo",
        "g": "vbkyserq",
        "O": "trdhxamc"
      },
      "question": "For every positive real number $qzxwvtnp$, let\n\\[\nvbkyserq(qzxwvtnp) = \\lim_{hjgrksla \\to 0} ((qzxwvtnp+1)^{hjgrksla+1} - qzxwvtnp^{hjgrksla+1})^{\\frac{1}{hjgrksla}}.\n\\]\nFind $\\lim_{qzxwvtnp \\to \\infty} \\frac{vbkyserq(qzxwvtnp)}{qzxwvtnp}$.",
      "solution": "The limit is $e$.\n\n\\noindent\n\\textbf{First solution.}\nBy l'H\\^opital's Rule, we have\n\\begin{align*}\n&\\lim_{hjgrksla\\to 0} \\frac{\\log((qzxwvtnp+1)^{hjgrksla+1}-qzxwvtnp^{hjgrksla+1})}{hjgrksla} \\\\\n&\\quad = \\lim_{hjgrksla\\to 0} \\frac{d}{d hjgrksla} \\log((qzxwvtnp+1)^{hjgrksla+1}-qzxwvtnp^{hjgrksla+1}) \\\\\n&\\quad = \\lim_{hjgrksla\\to 0} \\frac{(qzxwvtnp+1)^{hjgrksla+1}\\log(qzxwvtnp+1)-qzxwvtnp^{hjgrksla+1}\\log qzxwvtnp}{(qzxwvtnp+1)^{hjgrksla+1}-qzxwvtnp^{hjgrksla+1}} \\\\\n&\\quad = (qzxwvtnp+1)\\log(qzxwvtnp+1)-qzxwvtnp\\log qzxwvtnp,\n\\end{align*}\nwhere $\\log$ denotes natural logarithm. It follows that $vbkyserq(qzxwvtnp) = e^{(qzxwvtnp+1)\\log(qzxwvtnp+1)-qzxwvtnp\\log qzxwvtnp} = \\frac{(qzxwvtnp+1)^{qzxwvtnp+1}}{qzxwvtnp^{qzxwvtnp}}$. Thus\n\\[\n\\lim_{qzxwvtnp\\to\\infty} \\frac{vbkyserq(qzxwvtnp)}{qzxwvtnp} = \\left(\\lim_{qzxwvtnp\\to\\infty}\\frac{qzxwvtnp+1}{qzxwvtnp}\\right) \\cdot \\left(\\lim_{qzxwvtnp\\to\\infty} \\left(1+\\frac{1}{qzxwvtnp}\\right)^{qzxwvtnp}\\right) = 1\\cdot e = e.\n\\]\n\n\\noindent\n\\textbf{Second solution.}\nWe first write \n\\begin{align*}\n\\lim_{qzxwvtnp \\to \\infty} \\frac{vbkyserq(qzxwvtnp)}{qzxwvtnp} &= \\lim_{qzxwvtnp \\to \\infty} \\lim_{hjgrksla \\to 0} \\frac{((qzxwvtnp+1)^{hjgrksla+1} - qzxwvtnp^{hjgrksla+1})^{1/hjgrksla}}{qzxwvtnp} \\\\\n&= \\lim_{qzxwvtnp \\to \\infty} \\lim_{hjgrksla \\to 0} \\frac{((hjgrksla+1) qzxwvtnp^{hjgrksla} + trdhxamc(qzxwvtnp^{hjgrksla-1}))^{1/hjgrksla}}{qzxwvtnp}.\n\\end{align*}\nWe would like to interchange the order of the limits, but this requires some justification.\nUsing Taylor's theorem with remainder, for $qzxwvtnp \\ge 1$, $hjgrksla \\le 1$\nwe can bound the error term $trdhxamc(qzxwvtnp^{hjgrksla-1})$ in absolute value by $(hjgrksla+1) hjgrksla qzxwvtnp^{hjgrksla-1}$. This\nmeans that if we continue to rewrite the orginial limit as\n\\[\n\\lim_{hjgrksla\\to 0} \\lim_{qzxwvtnp\\to\\infty} (hjgrksla+1+trdhxamc(qzxwvtnp^{-1}))^{1/hjgrksla},\n\\]\nthe error term $trdhxamc(qzxwvtnp^{-1})$ is bounded in absolute value by $(hjgrksla+1) hjgrksla/qzxwvtnp$.\nFor $qzxwvtnp \\ge 1$, $hjgrksla \\le 1$ this quantity is bounded in absolute value by $(hjgrksla+1)hjgrksla$, \\emph{independently of $qzxwvtnp$}. This allows us to continue by interchanging the order of the limits,\nobtaining \n\\begin{align*}\n&\\lim_{hjgrksla\\to 0} \\lim_{qzxwvtnp\\to\\infty} (hjgrksla+1+trdhxamc(qzxwvtnp^{-1}))^{1/hjgrksla} \\\\\n&\\quad = \\lim_{hjgrksla\\to 0} (hjgrksla+1)^{1/hjgrksla} \\\\\n&\\quad = \\lim_{zplmhrxo\\to \\infty} (1+1/zplmhrxo)^{zplmhrxo} = e,\n\\end{align*}\nwhere in the last step we take $zplmhrxo = 1/hjgrksla$.\n\n\\noindent\n\\textbf{Third solution.} (by Clayton Lungstrum)\nWe first observe that\n\\begin{align*}\n((qzxwvtnp+1)^{hjgrksla+1} - qzxwvtnp^{hjgrksla+1})^{1/hjgrksla}\n&= \\left( \\int_{qzxwvtnp}^{qzxwvtnp+1} (hjgrksla+1)pmncblet^{hjgrksla}\\,dpmncblet \\right)^{1/hjgrksla} \\\\\n&= (hjgrksla+1)^{1/hjgrksla} \\left( \\int_{qzxwvtnp}^{qzxwvtnp+1} pmncblet^{hjgrksla}\\,dpmncblet \\right)^{1/hjgrksla}.\n\\end{align*}\nSince $\\lim_{hjgrksla \\to 0} (hjgrksla+1)^{1/hjgrksla} = e$, we deduce that\n\\[\nvbkyserq(qzxwvtnp) = e \\lim_{hjgrksla \\to 0} \\left( \\int_{qzxwvtnp}^{qzxwvtnp+1} pmncblet^{hjgrksla}\\,dpmncblet \\right)^{1/hjgrksla}.\n\\]\nFor $hjgrksla > 0$, $pmncblet^{hjgrksla}$ is increasing for $qzxwvtnp \\le pmncblet \\le qzxwvtnp+1$, so\n\\[\nqzxwvtnp^{hjgrksla} \\le \\int_{qzxwvtnp}^{qzxwvtnp+1} pmncblet^{hjgrksla}\\,dpmncblet \\le (qzxwvtnp+1)^{hjgrksla};\n\\]\nfor $hjgrksla < 0$, $pmncblet^{hjgrksla}$ is decreasing for $qzxwvtnp \\le pmncblet \\le qzxwvtnp+1$, so\n\\[\nqzxwvtnp^{hjgrksla} \\ge \\int_{qzxwvtnp}^{qzxwvtnp+1} pmncblet^{hjgrksla}\\,dpmncblet \\ge (qzxwvtnp+1)^{hjgrksla}.\n\\]\nIn both cases, we deduce that\n\\[\nqzxwvtnp \\le \\left( \\int_{qzxwvtnp}^{qzxwvtnp+1} pmncblet^{hjgrksla}\\,dpmncblet \\right)^{1/hjgrksla} \\le qzxwvtnp+1;\n\\]\napplying the squeeze theorem to the resulting inequality\n $e \\le \\frac{vbkyserq(qzxwvtnp)}{qzxwvtnp} \\le e\\left( 1 + \\frac{1}{qzxwvtnp} \\right)$\n yields the claimed limit."
    },
    "kernel_variant": {
      "question": "For every positive real number x define the two–parameter expression  \n\\[\nK_x(r,s)=\\frac{(x+1)^{\\,r+1}-x^{\\,r+1}}{(x+1)^{\\,s+1}-x^{\\,s+1}},\\qquad r\\neq s ,\n\\]\nand set  \n\\[\nk(x)=\\lim_{(r,s)\\to(0,0)}\\Bigl(K_x(r,s)\\Bigr)^{\\frac1{\\,r-s\\,}},\n\\]\nwhere the double‐limit is taken with \\((r,s)\\) approaching \\((0,0)\\) arbitrarily in the open first quadrant.  \nDetermine the value of  \n\\[\n\\boxed{\\displaystyle\\lim_{x\\to\\infty}\\frac{k(x)}{x}}.\n\\]\n\n────────────────────────────────────────",
      "solution": "Step 1.  Re–interpret the two-variable limit.  \nPut  \n\\[\nF_x(t)=\\log\\!\\bigl((x+1)^{\\,t+1}-x^{\\,t+1}\\bigr)\\quad(t>-1).\n\\]\nThen  \n\\[\n\\log K_x(r,s)=F_x(r)-F_x(s),\\qquad \n\\Bigl(K_x(r,s)\\Bigr)^{\\frac1{r-s}}\n       =\\exp\\!\\Bigl(\\tfrac{F_x(r)-F_x(s)}{\\,r-s}\\Bigr).\n\\]\nHence  \n\\[\nk(x)=\\exp\\!\\Bigl(\\lim_{(r,s)\\to(0,0)}\n                 \\frac{F_x(r)-F_x(s)}{\\,r-s}\\Bigr)=\\exp\\!\\bigl(F_x'(0)\\bigr),\n\\]\nprovided the directional derivative \\(F_x'(0)\\) exists and the limit defining \\(k(x)\\) is path–independent.  All that remains is to show:\n\n(1) \\(F_x\\) is differentiable at \\(t=0\\);\n\n(2) the two-variable limit really equals \\(F_x'(0)\\);\n\n(3) evaluate \\(F_x'(0)\\) explicitly and take \\(x\\to\\infty\\).\n\nStep 2.  Differentiability of \\(F_x\\) at the origin.  \nBecause both summands in \\((x+1)^{\\,t+1}-x^{\\,t+1}\\) are strictly positive for \\(t>-1\\), \\(F_x\\) is well–defined and \\(C^{\\infty}\\) on \\((-1,\\infty)\\).  In particular\n\\[\nF_x'(t)=\\frac{(x+1)^{\\,t+1}\\log(x+1)-x^{\\,t+1}\\log x}\n              {(x+1)^{\\,t+1}-x^{\\,t+1}},\n\\]\nso \\(F_x\\) is differentiable at \\(t=0\\).\n\nStep 3.  Path–independence of the two-variable limit.  \nFor any \\(C^{1}\\)-function \\(G\\) one has\n\\[\n\\lim_{(u,v)\\to(a,a)}\\frac{G(u)-G(v)}{u-v}=G'(a),\n\\]\nbecause the numerator factors as \\((u-v)\\int_{0}^{1}G'(v+\\theta(u-v))\\,d\\theta\\).\nApplying this with \\(G=F_x\\) and \\(a=0\\) proves that\n\\[\n\\lim_{(r,s)\\to(0,0)}\\frac{F_x(r)-F_x(s)}{r-s}=F_x'(0),\n\\]\nindependently of the path, so the limit defining \\(k(x)\\) exists and equals\n\\[\nk(x)=\\exp\\!\\bigl(F_x'(0)\\bigr).\n\\]\n\nStep 4.  Closed form for \\(k(x)\\).  \nUsing the formula for \\(F_x'(t)\\) and putting \\(t=0\\),\n\\[\nF_x'(0)=\n\\frac{(x+1)\\log(x+1)-x\\log x}{(x+1)-x}\n=(x+1)\\log(x+1)-x\\log x.\n\\]\nHence\n\\[\nk(x)=\\exp\\!\\bigl((x+1)\\log(x+1)-x\\log x\\bigr)\n     =\\frac{(x+1)^{x+1}}{x^{\\,x}}\n\\]\n—the same closed expression that arose in the original problem, but now obtained from a non-trivial two-parameter limit.\n\nStep 5.  Asymptotics for large x.  \nWrite\n\\[\n\\frac{k(x)}{x}\n     =\\frac{x+1}{x}\\Bigl(1+\\frac1x\\Bigr)^{x}\n     =\\Bigl(1+\\frac1x\\Bigr)\\Bigl(1+\\frac1x\\Bigr)^{x}.\n\\]\nTake \\(x\\to\\infty\\).  The first factor tends to 1, while the second tends to Euler’s number \\(e\\).  Therefore\n\\[\n\\boxed{\\displaystyle\\lim_{x\\to\\infty}\\frac{k(x)}{x}=e}.\n\\]\n\n────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.873093",
        "was_fixed": false,
        "difficulty_analysis": "1. Multi-parameter limit.  Unlike the original one-variable expression, the problem now asks for a limit as \\((r,s)\\to(0,0)\\) in \\(\\mathbb R^{2}\\), so the candidate must prove the limit exists and is independent of the path—something that cannot be dealt with by a single application of l’Hôpital’s Rule.\n\n2. Functional reformulation.  The solution introduces the auxiliary function \\(F_x\\) and recognises the given expression as an exponential of a two-point finite difference quotient.  Identifying and manipulating this structure demands greater conceptual insight than the straightforward logarithm-and-differentiate tactic used in the original problem.\n\n3. Use of integral representation for path-independence.  Showing that \\(\\displaystyle\\frac{G(u)-G(v)}{u-v}\\to G'(a)\\) invokes an integral (mean-value) representation of a difference quotient; the justification requires a first-year graduate‐level analysis argument that is absent from the initial problem.\n\n4. Layered limits.  One must compute (i) the inner two-variable limit, (ii) convert it into a single-variable expression, and (iii) evaluate a second, outer limit \\(x\\to\\infty\\).  Coordinating these layers correctly is more intricate than the iterated single-variable limit in the original setting.\n\n5. Potential pitfalls.  Because the denominator of the exponent is \\(r-s\\), careless handling of signs, domains (\\(r\\neq s\\)), or path selection easily leads to indeterminate forms or divergence.  Establishing that everything stays strictly positive and well-defined imposes extra technical checks not present in the earlier versions.\n\nThese added analytical subtleties, the need for path-independence proofs, and the interplay between several limits together make the enhanced variant substantially more challenging than both the original problem and the existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "For every positive real number x define the two–parameter expression  \n\\[\nK_x(r,s)=\\frac{(x+1)^{\\,r+1}-x^{\\,r+1}}{(x+1)^{\\,s+1}-x^{\\,s+1}},\\qquad r\\neq s ,\n\\]\nand set  \n\\[\nk(x)=\\lim_{(r,s)\\to(0,0)}\\Bigl(K_x(r,s)\\Bigr)^{\\frac1{\\,r-s\\,}},\n\\]\nwhere the double‐limit is taken with \\((r,s)\\) approaching \\((0,0)\\) arbitrarily in the open first quadrant.  \nDetermine the value of  \n\\[\n\\boxed{\\displaystyle\\lim_{x\\to\\infty}\\frac{k(x)}{x}}.\n\\]\n\n────────────────────────────────────────",
      "solution": "Step 1.  Re–interpret the two-variable limit.  \nPut  \n\\[\nF_x(t)=\\log\\!\\bigl((x+1)^{\\,t+1}-x^{\\,t+1}\\bigr)\\quad(t>-1).\n\\]\nThen  \n\\[\n\\log K_x(r,s)=F_x(r)-F_x(s),\\qquad \n\\Bigl(K_x(r,s)\\Bigr)^{\\frac1{r-s}}\n       =\\exp\\!\\Bigl(\\tfrac{F_x(r)-F_x(s)}{\\,r-s}\\Bigr).\n\\]\nHence  \n\\[\nk(x)=\\exp\\!\\Bigl(\\lim_{(r,s)\\to(0,0)}\n                 \\frac{F_x(r)-F_x(s)}{\\,r-s}\\Bigr)=\\exp\\!\\bigl(F_x'(0)\\bigr),\n\\]\nprovided the directional derivative \\(F_x'(0)\\) exists and the limit defining \\(k(x)\\) is path–independent.  All that remains is to show:\n\n(1) \\(F_x\\) is differentiable at \\(t=0\\);\n\n(2) the two-variable limit really equals \\(F_x'(0)\\);\n\n(3) evaluate \\(F_x'(0)\\) explicitly and take \\(x\\to\\infty\\).\n\nStep 2.  Differentiability of \\(F_x\\) at the origin.  \nBecause both summands in \\((x+1)^{\\,t+1}-x^{\\,t+1}\\) are strictly positive for \\(t>-1\\), \\(F_x\\) is well–defined and \\(C^{\\infty}\\) on \\((-1,\\infty)\\).  In particular\n\\[\nF_x'(t)=\\frac{(x+1)^{\\,t+1}\\log(x+1)-x^{\\,t+1}\\log x}\n              {(x+1)^{\\,t+1}-x^{\\,t+1}},\n\\]\nso \\(F_x\\) is differentiable at \\(t=0\\).\n\nStep 3.  Path–independence of the two-variable limit.  \nFor any \\(C^{1}\\)-function \\(G\\) one has\n\\[\n\\lim_{(u,v)\\to(a,a)}\\frac{G(u)-G(v)}{u-v}=G'(a),\n\\]\nbecause the numerator factors as \\((u-v)\\int_{0}^{1}G'(v+\\theta(u-v))\\,d\\theta\\).\nApplying this with \\(G=F_x\\) and \\(a=0\\) proves that\n\\[\n\\lim_{(r,s)\\to(0,0)}\\frac{F_x(r)-F_x(s)}{r-s}=F_x'(0),\n\\]\nindependently of the path, so the limit defining \\(k(x)\\) exists and equals\n\\[\nk(x)=\\exp\\!\\bigl(F_x'(0)\\bigr).\n\\]\n\nStep 4.  Closed form for \\(k(x)\\).  \nUsing the formula for \\(F_x'(t)\\) and putting \\(t=0\\),\n\\[\nF_x'(0)=\n\\frac{(x+1)\\log(x+1)-x\\log x}{(x+1)-x}\n=(x+1)\\log(x+1)-x\\log x.\n\\]\nHence\n\\[\nk(x)=\\exp\\!\\bigl((x+1)\\log(x+1)-x\\log x\\bigr)\n     =\\frac{(x+1)^{x+1}}{x^{\\,x}}\n\\]\n—the same closed expression that arose in the original problem, but now obtained from a non-trivial two-parameter limit.\n\nStep 5.  Asymptotics for large x.  \nWrite\n\\[\n\\frac{k(x)}{x}\n     =\\frac{x+1}{x}\\Bigl(1+\\frac1x\\Bigr)^{x}\n     =\\Bigl(1+\\frac1x\\Bigr)\\Bigl(1+\\frac1x\\Bigr)^{x}.\n\\]\nTake \\(x\\to\\infty\\).  The first factor tends to 1, while the second tends to Euler’s number \\(e\\).  Therefore\n\\[\n\\boxed{\\displaystyle\\lim_{x\\to\\infty}\\frac{k(x)}{x}=e}.\n\\]\n\n────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.661385",
        "was_fixed": false,
        "difficulty_analysis": "1. Multi-parameter limit.  Unlike the original one-variable expression, the problem now asks for a limit as \\((r,s)\\to(0,0)\\) in \\(\\mathbb R^{2}\\), so the candidate must prove the limit exists and is independent of the path—something that cannot be dealt with by a single application of l’Hôpital’s Rule.\n\n2. Functional reformulation.  The solution introduces the auxiliary function \\(F_x\\) and recognises the given expression as an exponential of a two-point finite difference quotient.  Identifying and manipulating this structure demands greater conceptual insight than the straightforward logarithm-and-differentiate tactic used in the original problem.\n\n3. Use of integral representation for path-independence.  Showing that \\(\\displaystyle\\frac{G(u)-G(v)}{u-v}\\to G'(a)\\) invokes an integral (mean-value) representation of a difference quotient; the justification requires a first-year graduate‐level analysis argument that is absent from the initial problem.\n\n4. Layered limits.  One must compute (i) the inner two-variable limit, (ii) convert it into a single-variable expression, and (iii) evaluate a second, outer limit \\(x\\to\\infty\\).  Coordinating these layers correctly is more intricate than the iterated single-variable limit in the original setting.\n\n5. Potential pitfalls.  Because the denominator of the exponent is \\(r-s\\), careless handling of signs, domains (\\(r\\neq s\\)), or path selection easily leads to indeterminate forms or divergence.  Establishing that everything stays strictly positive and well-defined imposes extra technical checks not present in the earlier versions.\n\nThese added analytical subtleties, the need for path-independence proofs, and the interplay between several limits together make the enhanced variant substantially more challenging than both the original problem and the existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}