mathanalysis_standardcalc101-109.ipynb

{
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   "execution_count": 1,
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    "%%html\n",
    "<style>\n",
    "@import url('https://fonts.googleapis.com/css?family=Roboto&effect=3d|brick-sign|');\n",
    "span {font-family:Roboto; color:black; text-shadow:4px 4px 4px #aaa;}  \n",
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  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "import math, sympy, numpy\n",
    "sympy.init_printing(use_unicode=True)\n",
    "from sympy import *\n",
    "import pylab as plt\n",
    "%matplotlib inline"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "[GeoGebra](https://www.geogebra.org/graphing?lang=en)\n",
    "\n",
    "[LaTeX/Mathematics](https://en.wikibooks.org/wiki/LaTeX/Mathematics)\n",
    "\n",
    "[SymPy Tutorial. Calculus](https://docs.sympy.org/latest/tutorial/calculus.html)\n",
    "\n",
    "[ SymPy Modules Reference. Calculus](https://docs.sympy.org/latest/modules/calculus/index.html)\n",
    "\n",
    "[Sympy : Symbolic Mathematics in Python](https://www.scipy-lectures.org/advanced/sympy.html)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Вариант 22\n",
    "# Пределы"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Доказать, что $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\frac { 5n }{8n + 3}  = \\frac{5}{8} }}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "По определению, предел последовательности $\\large{\\displaystyle{ a_n = \\frac{5n}{8n + 3} }}$ при $n \\to \\infty$ равен $\\large{\\frac{5}{8}}$, если\n",
    "\n",
    "для любого $\\varepsilon > 0$ существует такой номер $N(\\varepsilon)$, что для всех номеров $n > N$ выполняется неравенство \n",
    "\n",
    "$ \\large{\\displaystyle{\\bigg| \\frac{5n}{8n + 3} - \\frac{5}{8} \\bigg| < \\varepsilon }}$ (1), т.е.\n",
    "\n",
    "$ \\large{\\displaystyle{\\forall \\varepsilon > 0 \\ \\exists N = N(\\varepsilon) \\in \\mathbb{N}: \\forall n > N \\implies \\bigg| \\frac{5n}{8n + 3} - \\frac{5}{8}}} \\bigg| < \\varepsilon$.\n",
    "\n",
    "Зафиксируем произвольное  $\\varepsilon > 0$ и найдем номер  $N(\\varepsilon)$, для которого выполняется неравенство (1).\n",
    "\n",
    "$ \\large{\\displaystyle{\\bigg| \\frac{5n}{8n + 3} - \\frac{5}{8} \\bigg|  = \\bigg| \\frac{40n - 40n -15}{8 * (8n + 3)} \\bigg| =   \\bigg| \\frac{ -15}{64n + 24} \\bigg|  = \\bigg| \\frac{ 15}{64n + 24} \\bigg| = \\frac{ 15}{64n + 24} }}$, т.к. $n \\in \\mathbb{N}$.\n",
    "\n",
    "Неравенство (1) примет вид:\n",
    "\n",
    "$ \\large{\\displaystyle{\\frac{ 15}{64n + 24} < \\varepsilon \\iff 64n + 24 > \\frac{15}{\\varepsilon}  \\iff n >  \\frac{15}{64\\varepsilon} - \\frac{3}{8} \\iff n >  \\frac{15 - 24\\varepsilon}{64\\varepsilon} (2) }}$\n",
    "\n",
    "Для выполнения неравенства (1) достаточно подобрать такое $N = N(\\varepsilon) \\in \\mathbb{N}$, что (2) будет выполнено для всех $n > N$.\n",
    "\n",
    "Функция $\\large{\\displaystyle{\\frac{15 -24\\varepsilon}{64\\varepsilon} }}$ определена для любого $\\varepsilon > 0$, \n",
    "\n",
    "следовательно $N = N(\\varepsilon) \\in \\mathbb{N}$, при которых будет выполнено (2) для всех $n > N$, можно найти всегда.\n",
    "\n",
    "Например, при \n",
    "\n",
    "$\\large{\\displaystyle{\\varepsilon = \\frac {15}{88} :  n >  \\frac{15 * 88 - 24 * 15}{64 * 15} }}\\iff n > 1 \\implies N = 1$;\n",
    "\n",
    "$\\large{\\displaystyle{\\varepsilon = 0.125 :  n >  \\frac{15 - 24 * 0.125}{64 * 0.125} }}\\iff n > 1.5 \\implies N = 2$ и т.д.\n",
    "\n",
    "В качестве  $N = N(\\varepsilon) \\in \\mathbb{N}$ можно взять любое натуральное число, превышающее $\\large{\\displaystyle{\\frac{15 -24\\varepsilon}{64\\varepsilon} : n > N > \\frac{15 -24\\varepsilon}{64\\varepsilon} }}$ .\n",
    "\n",
    "Таким образом, $ \\large{\\displaystyle{N = \n",
    "\\begin {cases}\n",
    "1 , \\varepsilon \\geq \\frac{15}{88} \\\\\n",
    "\\bigg[ \\frac{15 -24\\varepsilon}{64\\varepsilon} \\bigg]  + 1, \\varepsilon < \\frac{15}{88}\n",
    "\\end{cases} }}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
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\n",
      "text/plain": [
       "<Figure size 432x432 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "def N(e): return (15 - 24 * e) / (64 * e)\n",
    "f = plt.figure(figsize=(6,6))\n",
    "ax = f.add_subplot(111)\n",
    "ax.grid(c='steelblue', ls='--', lw=1)\n",
    "e = numpy.linspace(0.1,1)\n",
    "n = [N(el) for el in e]\n",
    "ax.plot(e, n, c='darkblue', lw=2)\n",
    "ax.axvline(15/88, 0.05, 0.95, c='darkred',ls='--')\n",
    "plt.xlabel('ε', fontsize=15); plt.ylabel('N(ε)', fontsize=15);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAAAsAAAArBAMAAABcLm8jAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAMhAi783diatmdpm7RFShYZRwAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAArUlEQVQYGWNgEGBgMmBgYPAPTVMAUmnfioAkQwOIQKc6V20ECR5isL8Aohl4EsAU92cgM4CB+yMDA0sCA8cHBgZ2IOsAAwPzBoYlCkAl1js2gRWSR/wHAaChRAD15XtBqjYwcDxgYOCcwMAUAHTLATDF+0eBewJQrv/XPpAS9v/BIOq5zf8LQJUbGN47AZ2swMD8j4FBHyjeCeQZMDAcZmDgDWLgXAAUkdgNNBoA8fgrMzm/66wAAAAASUVORK5CYII=\n",
      "text/latex": [
       "$$\\frac{5}{8}$$"
      ],
      "text/plain": [
       "5/8"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = symbols('n')\n",
    "limit( 5 * n / (8 * n + 3), n, oo)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Вычислить предел последовательности  $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\frac{n^2 \\sqrt[5]{13n + 1} + \\sqrt{81n^6 -13} }{(2n - 5 \\sqrt[3]{n}) \\sqrt{4n^4 + 7n + 8}} }}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "Числитель и знаменатель дроби являются бесконечно большими последовательностями, \n",
    "\n",
    "значит при вычислении предела имеется неопределенность вида $\\large{[\\frac {\\infty}{\\infty}]}$.\n",
    "\n",
    "Для ее раскрытия преобразуем формулу общего члена исходной последовательности.\n",
    "\n",
    "Заметим, что $\\large{n^2 \\sqrt[5]{13n + 1} + \\sqrt{81n^6 -13} \\sim n^2 \\sqrt[5]{13n} + \\sqrt{81n^6} \\sim  n^2 \\sqrt[5]{n} + \\sqrt{n^6} \\sim n^3} $ при $n \\to \\infty$.\n",
    "\n",
    "Числитель исходной последовательности подобен $n^3$, поэтому разделим и числитель, и знаменатель на $n^3$.\n",
    "\n",
    "$\\large{\\displaystyle{a_n = \\frac{(n^2 \\sqrt[5]{13n + 1} + \\sqrt{81n^6 -13} ) / n^3 }{(2n - 5 \\sqrt[3]{n}) \\sqrt{4n^4 + 7n + 8} / n^3}  = }}$\n",
    "\n",
    "$\\large{\\displaystyle{= \\frac{\\sqrt[5]{\\frac{13}{n^4} + \\frac{1}{n^5}} + \\sqrt{81 - \\frac{13}{n^6}}  }{(2 - \\frac{5}{\\sqrt[3]{n^2}}) \\sqrt{4 + \\frac{7}{n^3} + \\frac{8}{n^4}}} }}$\n",
    "\n",
    "Последовательности $\\large{\\displaystyle{ \\left\\{ \\frac{13}{n^4} \\right\\} , \\left\\{ \\frac{1}{n^5} \\right\\}, ...}}$ являются бесконечно малыми  при $n \\to \\infty$.\n",
    "\n",
    " Применяя арифметические свойства пределов последовательности, получим:\n",
    " \n",
    " $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\frac{\\sqrt[5]{\\frac{13}{n^4} + \\frac{1}{n^5}} + \\sqrt{81 - \\frac{13}{n^6}}  }{(2 - \\frac{5}{\\sqrt[3]{n^2}}) \\sqrt{4 + \\frac{7}{n^3} + \\frac{8}{n^4}}} }} = \\frac{\\sqrt{81}}{2 * \\sqrt{4}} = 2.25$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAACMAAAAPBAMAAACcmWsnAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAIpnNuzKriUQQ7912ZlSugimNAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAArUlEQVQYGU2NMQrCUAyGv4KtYCuIg4uLvAvUwc3BwdHFM7hJh/YozxO4uKiDozgVPIe7B+jiINTkRcFA+JIvkB/+K4YdRG62VhvYa+Y3GNF9qQpM66uHLZxUBSY6cfGUgx9NFVDqs0BTstXSWjXJ4x6y4rcZYcdHISudmDKudHFmvswlqz8xpVxCLs+eRHIwniUe4orMiwms4ABj5wr2U+OQ7AiLtm2QsMDIbTwfDacqZAVwsPIAAAAASUVORK5CYII=\n",
      "text/latex": [
       "$$2.25$$"
      ],
      "text/plain": [
       "2.25000000000000"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = symbols('n')\n",
    "limit((n**2*(13*n + 1)**0.2 + (81*n**6 - 13)**0.5)/((2*n - 5*n**(1/3))*(4*n**4 + 7*n + 8)**0.5), n, oo)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №3\n",
    "\n",
    "Вычислить предел последовательности  $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\sqrt{n^5} (\\sqrt{6 + n^5} - \\sqrt{4 + n ^5}) }}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "$\\large{\\displaystyle{ \\sqrt{n^5}\\sqrt{6 + n^5} ,   \\sqrt{n^5}\\sqrt{4 + n ^5}  \\sim  \\sqrt{n^5} \\sqrt{n ^5}}}  \\sim n^5 $ при $n \\to \\infty$. \n",
    "\n",
    "Значит, требуется раскрыть необределенность вида $[\\infty - \\infty]$.\n",
    "\n",
    "Применим умножение на сопряженное выражение.\n",
    "\n",
    " $\\large{\\displaystyle{ \\sqrt{n^5} (\\sqrt{6 + n^5} - \\sqrt{4 + n ^5}) = \\frac{\\sqrt{n^5} (\\sqrt{6 + n^5} - \\sqrt{4 + n ^5})(\\sqrt{6 + n^5} + \\sqrt{4 + n ^5})}{\\sqrt{6 + n^5}  + \\sqrt{4 + n ^5}} = }}$\n",
    " \n",
    " $\\large{\\displaystyle{ = \\frac{\\sqrt{n^5} (6 + n^5 - 4  - n ^5)}{\\sqrt{6 + n^5}  + \\sqrt{4 + n ^5}} = \\frac{2\\sqrt{n^5}}{\\sqrt{6 + n^5}  + \\sqrt{4 + n ^5}} = }}$\n",
    " \n",
    " $\\large{\\displaystyle{ = \\frac{ 2 \\sqrt{n^5} \\sqrt{n^{-5}} }{ \\sqrt{n^{-5}} (\\sqrt{6 + n^5}  + \\sqrt{4 + n ^5}) } = \\frac{ 2 }{ \\sqrt{n^{-5}} (\\sqrt{6 + n^5}  + \\sqrt{4 + n ^5}) } = }} $\n",
    " \n",
    " $\\large{\\displaystyle{ = \\frac{ 2 }{ \\sqrt{\\frac{6}{n^5} + 1}  + \\sqrt{\\frac{4}{n^5} + 1} }  }} $\n",
    " \n",
    " Последовательности $\\large{\\displaystyle{ \\left\\{ \\frac{6}{n^5} \\right\\} , \\left\\{ \\frac{4}{n^5} \\right\\} }}$ являются бесконечно малыми  при $n \\to \\infty$.\n",
    " \n",
    " Применяя арифметические свойства пределов последовательности, получим:\n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{n \\to \\infty} \\frac{2}{\\sqrt{\\frac{6}{n^5} + 1}  + \\sqrt{\\frac{4}{n^5} + 1}} =  \\frac{2}{1 + 1} = 1}}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAABkAAAAOBAMAAAAoFKpzAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAMt0Qq5nNdrvviSJUZkQZkPbxAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAdklEQVQIHWNgYFRmAINVpRsYGEzCPoM5bAeYtYEMdgjvEgPDCgSvhYHhvAFc7isDw/sDMB4ziHcBxuP+xcAgX4DCWwDjMQPl3sPlGID6zsNNYehgYLiPsOEmA0MY1Hb7BAamA8wTGRhYOj+5MzAFMDAvLxFgAACKdSe0qRxCpwAAAABJRU5ErkJggg==\n",
      "text/latex": [
       "$$1.0$$"
      ],
      "text/plain": [
       "1.00000000000000"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = symbols('n')\n",
    "limit(n ** 2.5 * ((6 + n ** 5) ** 0.5 - (4 + n ** 5) ** 0.5), n, oo)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №4\n",
    "Вычислить предел последовательности  $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\frac{4 + 8 + 12 + ... + 4n}{\\sqrt[3]{n^2 + 5} + \\sqrt{4n^4 + 6}}  }}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "Применим формулу суммы арифметической прогрессии и преобразуем числитель дроби:\n",
    "\n",
    "$\\large{\\displaystyle{\\frac{4 + 8 + 12 + ... + 4n}{\\sqrt[3]{n^2 + 5} + \\sqrt{4n^4 + 6}}  =  \\frac{(2 * 4 + 4 *(n - 1)) * n / 2 }{\\sqrt[3]{n^2 + 5} + \\sqrt{4n^4 + 6}} =  \\frac{2 n^2 + 2n }{\\sqrt[3]{n^2 + 5} + \\sqrt{4n^4 + 6}} }}$\n",
    "\n",
    "Числитель и знаменатель дроби являются бесконечно большими последовательностями.\n",
    "\n",
    "При вычислении предела раскроем неопределенность вида $\\large{[\\frac {\\infty}{\\infty}]}$.\n",
    "\n",
    "Числитель подобен $n^2$ при $n \\to \\infty$, значит необходимо поделить и числитель, и знаменатель на $n^2$.\n",
    "\n",
    "$\\large{\\displaystyle{\\frac{(2 n^2 + 2n) / n^2}{\\bigg(\\sqrt[3]{n^2 + 5} + \\sqrt{4n^4 + 6}\\bigg) / n^2} = \\frac{\\frac{2 n^2} {n^2} + \\frac{2n}{n^2}}{\\frac{\\sqrt[3]{n^2 + 5}}{n^2} + \\frac{\\sqrt{4n^4 + 6}}{n^2}} = }}$.\n",
    "\n",
    "$\\large{\\displaystyle{= \\frac{2 + \\frac{2}{n}}{\\sqrt[3]{\\frac{n^2 + 5}{n^6}} + \\sqrt{\\frac{4n^4 + 6}{n^4}}} = \\frac{2 + \\frac{2}{n}}{\\sqrt[3]{\\frac{1}{n^4} + \\frac{5}{n^6}} + \\sqrt{4 + \\frac{6}{n^4}}} }}$\n",
    "\n",
    "Последовательности $\\large{\\displaystyle{ \\left\\{ \\frac{2}{n} \\right\\} , \\left\\{ \\frac{1}{n^4} \\right\\}, ...}}$ являются бесконечно малыми  при $n \\to \\infty$.\n",
    "\n",
    " Применяя арифметические свойства пределов последовательности, получим:\n",
    " \n",
    " $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\frac{2 + \\frac{2}{n}}{\\sqrt[3]{\\frac{1}{n^4} + \\frac{5}{n^6}} + \\sqrt{4 + \\frac{6}{n^4}}} = \\frac {2 + 0} {\\sqrt[3]{0 + 0} + \\sqrt{4 + 0}} = 1}}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAABkAAAAOBAMAAAAoFKpzAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAMt0Qq5nNdrvviSJUZkQZkPbxAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAdklEQVQIHWNgYFRmAINVpRsYGEzCPoM5bAeYtYEMdgjvEgPDCgSvhYHhvAFc7isDw/sDMB4ziHcBxuP+xcAgX4DCWwDjMQPl3sPlGID6zsNNYehgYLiPsOEmA0MY1Hb7BAamA8wTGRhYOj+5MzAFMDAvLxFgAACKdSe0qRxCpwAAAABJRU5ErkJggg==\n",
      "text/latex": [
       "$$1.0$$"
      ],
      "text/plain": [
       "1.00000000000000"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = symbols('n')\n",
    "limit((2 * n ** 2 + 2 * n) / ((n ** 2 + 5 ) ** (1/3) + (4 * n ** 4 + 6) ** 0.5), n, oo)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №5\n",
    "Вычислить предел последовательности  $\\large{\\displaystyle{\\lim_{n \\to \\infty} \\Bigg( \\frac{n^2 + 4n - 3}{n^2 - n + 6} \\Bigg)^{3n}  }}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "1) Докажем, что $\\large{\\displaystyle{\\lim_{n \\to \\infty}  \\frac{n^2 + 4n - 3}{n^2 - n + 6} = 1}}$.\n",
    "\n",
    "Числитель и знаменатель дроби являются бесконечно большими последовательностями.\n",
    "\n",
    "При вычислении предела необходимо раскрыть неопределенность вида $\\large{[\\frac {\\infty}{\\infty}]}$.\n",
    "\n",
    "Числитель подобен $n^2$ при $n \\to \\infty$, значит нужно поделить и числитель, и знаменатель на $n^2$.\n",
    "\n",
    "$\\large{\\displaystyle{ \\frac{n^2 + 4n - 3}{n^2 - n + 6}  =  \\frac{(n^2 + 4n - 3)  / n^2}{(n^2 - n + 6) / n^2}  =  \\frac{1 + \\frac{4}{n} - \\frac{3}{n^2}}{1 - \\frac{1}{n} + \\frac{6}{ n^2}} }}$\n",
    "\n",
    "Последовательности $\\large{\\displaystyle{ \\left\\{ \\frac{4}{n} \\right\\} , \\left\\{ \\frac{-3}{n^2} \\right\\}, ...}}$ являются бесконечно малыми  при $n \\to \\infty$.\n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{n \\to \\infty}  \\frac{n^2 + 4n - 3}{n^2 - n + 6} = \\frac {1 + 0 - 0}{1 - 0 + 0} = 1}}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2) $\\large{\\displaystyle{3n \\to \\infty}}$  при $\\large{\\displaystyle{n \\to \\infty}}$. \n",
    "\n",
    "Значит, исходная последовательность под знаком предела является неопределенным выражением вида  $\\large{\\displaystyle{1^{\\infty} }}$.\n",
    "\n",
    "Воспользуемся формулой второго замечательного предела: \n",
    "\n",
    "### $\\large{\\displaystyle{\\lim_{n \\to \\infty}  \\Bigg( 1 + \\alpha_n \\Bigg) ^ {\\frac{1}{\\alpha_n}} = e }}$,\n",
    "\n",
    "где $\\large{\\displaystyle{ \\{ \\alpha_n \\} }}$ - бесконечно малая последовательность.\n",
    "\n",
    "$\\large{\\displaystyle{\\Bigg( \\frac{n^2 + 4n - 3}{n^2 - n + 6} \\Bigg)^{3n}  = \\Bigg( 1 + \\frac{5n - 9}{n^2 - n + 6} \\Bigg)^{3n} =  \\Bigg[ \\bigg( 1 + \\frac{5n - 9}{n^2 - n + 6} \\bigg)^ \\frac{n^2 - n + 6}{5n - 9} \\Bigg] ^ {\\frac{3n * (5n - 9)}{n^2 - n + 6}} }}$.\n",
    "\n",
    " Последовательность $\\large{\\displaystyle{\\left\\{ \\alpha_n = \\frac{5n - 9}{n^2 - n + 6} \\right\\} }}$ является бесконечно малой при $\\large{\\displaystyle{n \\to \\infty}}$.\n",
    " \n",
    " $\\large{\\displaystyle{\\lim_{n \\to \\infty}  \\bigg( 1 + \\frac{5n - 9}{n^2 - n + 6} \\bigg) ^ \\frac{n^2 - n + 6}{5n - 9} = e }}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "3) Вычислим $\\large{\\displaystyle{\\lim_{n \\to \\infty}  \\frac{3n * (5n - 9)}{n^2 - n + 6} }}$.\n",
    "\n",
    "$\\large{\\displaystyle{\\frac{3n * (5n - 9)}{n^2 - n + 6} = \\frac{15n^2 - 27n}{n^2 - n + 6} = \\frac{(15n^2 - 27n) / n^2}{(n^2 - n + 6) / n^2} = \\frac{15 - \\frac{27}{n} }{1 - \\frac{1}{n} + \\frac{6}{ n^2} } }}$\n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{n \\to \\infty}  \\frac{15 - \\frac{27}{n} }{1 - \\frac{1}{n} + \\frac{6}{ n^2} } = \\frac{15 - 0}{1 - 0 + 0} = 15 \\implies }}$\n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{n \\to \\infty} \\Bigg[ \\bigg( 1 + \\frac{5n - 9}{n^2 - n + 6} \\bigg)^ \\frac{n^2 - n + 6}{5n - 9} \\Bigg] ^ {\\frac{3n * (5n - 9)}{n^2 - n + 6}} = e^{15} \\approx 3 \\ 269 \\ 017}}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAABoAAAATBAMAAAB8awA1AAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAEIl2mUTdMiJmu6tUze/kkN0jAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAi0lEQVQYGWNgAAPJCwwcqpEBEA67cQID6/skCIeBgQfIK4BxwLzSgzAuSG6H9AEoF8hjYGD/gODxMDB+RfD4Czh+IHhMDMwbIDwO3S8GjMpKQA6jsokAVAWQCitgV4DzOH4yMAfAeVyfjVThHAZ+qEEQEX4FhAwDAxuQdwAuwHmBQaQBzmMwMjaAcwDwnBeLQKQJSwAAAABJRU5ErkJggg==\n",
      "text/latex": [
       "$$e^{15}$$"
      ],
      "text/plain": [
       " 15\n",
       "ℯ  "
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = symbols('n')\n",
    "limit(((n ** 2 + 4 * n - 3) / (n ** 2 - n + 6) ) ** (3 * n), n, oo)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$$3269017.372472108$$"
      ],
      "text/plain": [
       "3269017.372472108"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "math.e ** 15"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №6\n",
    "\n",
    "Используя определение предела функции по Коши, доказать равенство ( найти $\\delta (\\varepsilon) $ ): \n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{x \\to 3} ( -2x^2 - 4 ) =  -22 }}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "Запишем определение по Коши для данного случая.\n",
    "\n",
    "Число $-22$ является пределом функции $f(x) = -2x^2 - 4$ при $x \\to 3$, если\n",
    "\n",
    "для любого $\\varepsilon > 0$ существует такое число $\\delta(\\varepsilon) > 0$, \n",
    "\n",
    "что для всех  $x$, удовлетворяющих условию $0 < |x - 3| < \\delta \\ (1)$ \n",
    "\n",
    "выполняется неравенство $|(-2x^2 - 4) - (-22) | < \\varepsilon \\iff |-2x^2 + 18 | < \\varepsilon \\ (2)$, т.е.\n",
    "\n",
    "$ \\large{\\displaystyle{\\forall \\varepsilon > 0 \\ \\exists \\delta = \\delta(\\varepsilon) >0: \\{ \\forall x: 0 < \\mid x - 3 \\mid < \\delta  \\implies \\mid -2x^2 + 18 \\mid < \\varepsilon \\} }}$.\n",
    "\n",
    "Зафиксируем произвольное число $\\varepsilon > 0$ и найдем такое $\\delta(\\varepsilon) > 0$, \n",
    "\n",
    "что для всех  $x$, удовлетворяющих условию  (1), выполнялось неравенство (2)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "***I способ.***\n",
    "\n",
    "$\\large{\\displaystyle{ 0 < |x - 3| < \\delta \\iff -\\delta < x - 3 < \\delta, x \\neq 3 \\ (1*) }}$\n",
    "\n",
    "$\\large{\\displaystyle{|-2x^2 + 18 | < \\varepsilon \\iff |x^2 - 9 | < \\frac{\\varepsilon}{2} \\iff 9 - \\frac{\\varepsilon}{2} < x ^2 < 9 + \\frac{\\varepsilon}{2} \\iff}}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\iff \n",
    "\\begin {cases}\n",
    "\\sqrt{9 - \\frac{\\varepsilon}{2}} < x < \\sqrt{9 + \\frac{\\varepsilon}{2}}, \\forall \\varepsilon \\leq 18 \\\\\n",
    "-\\sqrt{9 + \\frac{\\varepsilon}{2}} < x < \\sqrt{9 + \\frac{\\varepsilon}{2}}, \\forall \\varepsilon > 18\n",
    "\\end{cases} \\iff }}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\iff \n",
    "\\begin {cases}\n",
    "\\sqrt{9 - \\frac{\\varepsilon}{2}} - 3 < x - 3 < \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3, \\forall \\varepsilon \\leq 18 \\\\\n",
    "-\\sqrt{9 + \\frac{\\varepsilon}{2}} - 3 < x - 3 < \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3, \\forall \\varepsilon > 18\n",
    "\\end{cases} \\iff }}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\iff \n",
    "\\begin {cases}\n",
    "-(3 - \\sqrt{9 - \\frac{\\varepsilon}{2}}) < x - 3 < \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3, \\forall \\varepsilon \\leq 18 \\\\\n",
    "-(\\sqrt{9 + \\frac{\\varepsilon}{2}} + 3) < x - 3 < \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3, \\forall \\varepsilon > 18\n",
    "\\end{cases} \\ (2*) }}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "При $0 < \\varepsilon \\leq 18$ выберем $\\delta =\\min \\{ 3 - \\sqrt{9 - \\frac{\\varepsilon}{2}}, \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3 \\}$.\n",
    "\n",
    "Сравним эти числа.\n",
    "\n",
    " $\\large{\\displaystyle{ \\frac{ 3 - \\sqrt{9 - \\frac{\\varepsilon}{2}} } {\\sqrt{9 + \\frac{\\varepsilon}{2}} - 3} =  }} $\n",
    " \n",
    " $ \\large{\\displaystyle{= \\frac{ 3 - \\sqrt{9 - \\frac{\\varepsilon}{2}} } {\\sqrt{9 + \\frac{\\varepsilon}{2}} - 3} * \\frac{3 + \\sqrt{9 - \\frac{\\varepsilon}{2}}} {3 + \\sqrt{9 - \\frac{\\varepsilon}{2}}} * \\frac{\\sqrt{9 + \\frac{\\varepsilon}{2}} + 3} {\\sqrt{9 + \\frac{\\varepsilon}{2}} + 3}  = }}$\n",
    " \n",
    " $\\large{\\displaystyle{ = \\frac{\\sqrt{9 + \\frac{\\varepsilon}{2}} + 3} {3 + \\sqrt{9 - \\frac{\\varepsilon}{2}}} * \\frac{\\varepsilon}{2} * \\frac{2}{\\varepsilon} =  }} $\n",
    " \n",
    " $\\large{\\displaystyle{ = \\frac{\\sqrt{9 + \\frac{\\varepsilon}{2}} + 3} {\\sqrt{9 - \\frac{\\varepsilon}{2}} + 3} > 1 \\implies }} $\n",
    " \n",
    " $\\delta =\\sqrt{9 + \\frac{\\varepsilon}{2}} - 3,  \\forall \\varepsilon: 0 < \\varepsilon \\leq 18$.\n",
    " \n",
    " При $\\varepsilon > 18$ выберем $\\delta =\\min \\{ \\sqrt{9 + \\frac{\\varepsilon}{2}} + 3, \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3 \\} = \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3$.\n",
    " \n",
    " Выражение полностью идентично предыдущему случаю $\\implies \\delta = \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3,  \\forall \\varepsilon > 0$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "***II способ.***\n",
    "\n",
    "$\\large{\\displaystyle{ 0 < |x - 3| < \\delta \\ (1*) }}$.\n",
    "\n",
    "$\\large{\\displaystyle{| -2x^2 + 18 | = 2 * | x^2 - 9 | = 2 * |x - 3| * |x + 3| = 2 * |x - 3| * |x - 3 + 6| \\ (2*) }}$.\n",
    "\n",
    "$\\large{\\displaystyle{ (1*), (2*) \\implies  | -2x^2 + 18 | < 2 * \\delta * (\\delta + 6) = 2\\delta^2 + 12\\delta}}$.\n",
    "\n",
    "Неравенство $(2)$ выполняется при $2\\delta^2 + 12\\delta \\leq \\varepsilon$.\n",
    "\n",
    "Решим это неравенство при фиксированном $\\varepsilon > 0$ относительно $\\delta > 0$:\n",
    "\n",
    "$2\\delta^2 + 12\\delta \\leq \\varepsilon \\iff \\delta ^2 + 6\\delta - \\frac{\\varepsilon}{2} \\leq 0$\n",
    "\n",
    "$D = 6^2 - 4 * (- \\frac{\\varepsilon}{2}) = 36 + 2\\varepsilon$\n",
    "\n",
    "$\\implies \\delta = \\sqrt{9 + \\frac{\\varepsilon}{2}} - 3,  \\forall \\varepsilon > 0$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAACMAAAAOBAMAAABXxbiCAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAEM3dMiKZu6uJRO92ZlRfaCDxAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAfUlEQVQYGT3IMQrCYAyG4RdsB0GheAlHB93tEQoujp3+e0gPUO/j4HlcHUWoyfebBPKQvPCftux7RBSurD+VTDd4IDI9O+ZBZJpg7kUmO87+iKjN1y4RhdVop7M7+pzsK7YVP3y2Y+Cf5kU7IKI0dzadiMKllKmS6bAsb8QPt80gYOTroZYAAAAASUVORK5CYII=\n",
      "text/latex": [
       "$$-22$$"
      ],
      "text/plain": [
       "-22"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x = symbols('x')\n",
    "limit(-2 * x ** 2 - 4, x, 3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №7\n",
    "\n",
    "Вычислить предел функции \n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{x \\to -1} \\frac{x^3 + 5 x^2 + 8 x + 4}{x^3 + 7 x^2 + 16 x + 10} }}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "Числитель и знаменатель дроби стремятся к 0 при $x \\to -1$, \n",
    "\n",
    "значит дробь является неопределенным выражением вида $\\big[\\frac{0}{0}\\big]$ при $x \\to -1$.\n",
    "\n",
    "Для раскрытия неопределенности разложим числитель и знаменатель на множители и\n",
    "\n",
    "выполним тождественные преобразования при условии $x \\neq -1$:\n",
    "\n",
    "$\\large{\\displaystyle{ \\frac{x^3 + 5 x^2 + 8 x + 4}{x^3 + 7 x^2 + 16 x + 10} = \\frac{x^2 * (x + 1) + 4 x^2 + 8 x + 4}{x^2 * (x + 1) + 6 x^2 + 16 x + 10} = }}$\n",
    "\n",
    "$\\large{\\displaystyle{= \\frac{x^2 * (x + 1) + 4 x * (x + 1) + 4 * (x + 1)}{x^2 * (x + 1) + 6 x * (x + 1) + 10 * (x + 1)} = \\frac{(x + 1) * (x^2 + 4 x + 4) }{(x + 1) * (x^2 + 6 x + 10)} = }}$\n",
    "\n",
    "$\\large{\\displaystyle{ = \\frac{ x^2 + 4 x + 4}{x^2 + 6 x + 10} }}$.\n",
    "\n",
    "Используя арифметические свойства пределов функций, получим:\n",
    "\n",
    "$\\large{\\displaystyle{ \\lim_{x \\to -1} \\frac{ x^2 + 4 x + 4}{x^2 + 6 x + 10} = \\frac{(-1)^2 + 4 * (-1) + 4}{(-1)^2 + 6 * (-1) + 10} = 0.2}}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAAAsAAAAqBAMAAACXcryGAAAALVBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAADAOrOgAAAADnRSTlMAMt0Qq5nNdrvvIolmRIT9yeQAAAAJcEhZcwAADsQAAA7EAZUrDhsAAABuSURBVBgZY2BgEDJgYGBgds0DUQwMdVSkPOapFIANJYt4BwIP8GsVYGBeAFQxL00J5GylR5tByh1ABDrlUx4IEmxmWHcBRDNwKIAp1ldAZgID60MGBk4FBpYHDAzcQFYDAwNXAEMlyMylIcEMDAA1Fh2NT+ORhgAAAABJRU5ErkJggg==\n",
      "text/latex": [
       "$$\\frac{1}{5}$$"
      ],
      "text/plain": [
       "1/5"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x = symbols('x')\n",
    "limit((x ** 3 + 5 * x ** 2 + 8 * x + 4) / (x ** 3 + 7 * x ** 2 + 16 * x + 10), x, -1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №8\n",
    "\n",
    "Вычислить предел функции \n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{x \\to 0} \\frac{\\sqrt{2 + x} - \\sqrt{2 - x}}{\\sqrt[3]{2 + x} - \\sqrt[3]{2 - x}} }}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "Числитель и знаменатель дроби стремятся к 0 при $x \\to 0$, \n",
    "\n",
    "значит дробь является неопределенным выражением вида $\\big[\\frac{0}{0}\\big]$ при $x \\to 0$.\n",
    "\n",
    "Для раскрытия неопределенности \n",
    "\n",
    "- домножим числитель и знаменатель дроби на выражения, соответствующие формулам разности квадратов и кубов,\n",
    "\n",
    "- и выполним тождественные преобразования при условии $x \\neq 0$:\n",
    "\n",
    "$\\large{\\displaystyle{ \\frac{\\sqrt{2 + x} - \\sqrt{2 - x}}{\\sqrt[3]{2 + x} - \\sqrt[3]{2 - x}}  = \\frac{(\\sqrt{2 + x} - \\sqrt{2 - x}) * (\\sqrt{2 + x} + \\sqrt{2 - x})} {(\\sqrt[3]{2 + x} - \\sqrt[3]{2 - x})  * (\\sqrt{2 + x} + \\sqrt{2 - x})} = }}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\frac{2x}{(\\sqrt[3]{2 + x} - \\sqrt[3]{2 - x})  * (\\sqrt{2 + x} + \\sqrt{2 - x})}  = }}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\frac{2x * (\\sqrt[3]{(2 + x)^2} + \\sqrt[3]{4 - x^2} + \\sqrt[3]{(2 - x)^2})}{(\\sqrt[3]{2 + x} - \\sqrt[3]{2 - x}) * (\\sqrt[3]{(2 + x)^2} + \\sqrt[3]{4 - x^2} + \\sqrt[3]{(2 - x)^2}) * (\\sqrt{2 + x} + \\sqrt{2 - x})} = }}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\frac{2x * (\\sqrt[3]{(2 + x)^2} + \\sqrt[3]{4 - x^2} + \\sqrt[3]{(2 - x)^2})}{2x * (\\sqrt{2 + x} + \\sqrt{2 - x})} = \\frac{\\sqrt[3]{(2 + x)^2} + \\sqrt[3]{4 - x^2} + \\sqrt[3]{(2 - x)^2}} {\\sqrt{2 + x} + \\sqrt{2 - x}} }}$\n",
    "\n",
    "$\\large{\\displaystyle{ \\lim_{x \\to 0} \\frac{\\sqrt[3]{(2 + x)^2} + \\sqrt[3]{4 - x^2} + \\sqrt[3]{(2 - x)^2}} {\\sqrt{2 + x} + \\sqrt{2 - x}} =  \\frac{\\sqrt[3]{(2 + 0)^2} + \\sqrt[3]{4 - 0} + \\sqrt[3]{(2 - 0)^2}} {\\sqrt{2 + 0} + \\sqrt{2 - 0}} = }}$\n",
    "\n",
    "$\\large{\\displaystyle{= \\frac{3 * 2^{\\frac{2}{3}}} {2^{\\frac{3}{2}}} = 3 * 2^{\\frac{2}{3} - \\frac{3}{2}} = \\frac {3}{\\sqrt[6]{32}} }}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$$1.68369307246406$$"
      ],
      "text/plain": [
       "1.68369307246406"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x = symbols('x')\n",
    "limit(((2 + x) ** 0.5 - (2 - x) ** 0.5) / ((2 + x) ** (1/3) - (2 - x) ** (1/3)), x, 0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$$1.6836930724640595$$"
      ],
      "text/plain": [
       "1.6836930724640595"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "3 * 32 ** (-1/6)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Типовой расчет №9\n",
    "\n",
    "Вычислить предел функции \n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{x \\to 0} \\frac{sin^2{x} - tg^2{x}} {x^4} }}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Решение.\n",
    "\n",
    "Числитель и знаменатель дроби стремятся к 0 при $x \\to 0$, \n",
    "\n",
    "значит дробь является неопределенным выражением вида $\\big[\\frac{0}{0}\\big]$ при $x \\to 0$.\n",
    "\n",
    "Для раскрытия неопределенности преобразуем функцию, используя тригонометрические формулы:\n",
    "    \n",
    "$\\large{\\displaystyle{ \\frac{sin^2{x} - tg^2{x}} {x^4} = \\frac{sin^2{x} - \\frac{sin^2{x}}{cos^2{x}}} {x^2 * x^2} = \\frac{sin^2{x}} {x^2} * \\frac{1 - \\frac{1}{cos^2{x}}} {x^2}  = }}$ \n",
    "\n",
    "$\\large{\\displaystyle{= - \\frac{sin^2{x}} {x^2} * \\frac{1 - cos^2{x}}{cos^2{x} * x^2} = - \\frac{sin^2{x}} {x^2} * \\frac{sin^2{x}}{cos^2{x} * x^2} = -  \\frac{sin^2{x}} {x^2} * \\frac{tg^2{x}}{x^2} }}$.\n",
    "\n",
    "Используем арифметические свойства пределов функций, \n",
    "\n",
    "а также формулы первого замечательного предела и его следствий:\n",
    "\n",
    "$\\large{\\displaystyle{\\lim_{x \\to 0}( -  \\frac{sin^2{x}} {x^2} * \\frac{tg^2{x}}{x^2}) = - \\lim_{x \\to 0}\\frac{sin^2{x}} {x^2} * \\lim_{x \\to 0} \\frac{tg^2{x}}{x^2} = }}$\n",
    "\n",
    "$\\large{\\displaystyle{= - \\Big(\\lim_{x \\to 0} \\frac{sin{x}} {x}\\Big)^2 * \\Big(\\lim_{x \\to 0} \\frac{tg{x}} {x}\\Big)^2 = - 1^2 * 1^2 = -1}}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAABgAAAAOBAMAAADH1sFNAAAAIVBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAADdcGRXAAAACnRSTlMAEM3dMquZdrvvHSJQDQAAAAlwSFlzAAAOxAAADsQBlSsOGwAAACZJREFUCB1jYIABFkMYi4FBNGkhgsPATg2OkDEImDBQbBpr5+JwAHRPC1OtyQ5cAAAAAElFTkSuQmCC\n",
      "text/latex": [
       "$$-1$$"
      ],
      "text/plain": [
       "-1"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x = symbols('x')\n",
    "limit(((sin(x)) ** 2 - (tan(x)) ** 2) / (x ** 4), x, 0)"
   ]
  }
 ],
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