Portfolio 3 Ciclo

Porfílias 3° ciclo\n Aluna: Daniel Karina Jiller RA 1173053\n1) f(x) = \\frac{x^2 - 9}{x + 3}\n f(-3,1) = \\frac{(-3,1)^2 - 9}{-3,1 + 3} = \\frac{9,61 - 9}{-3,1 + 3} = \\frac{0,61}{-0,1} = -6,1\n f(-3,01) = \\frac{(-3,01)^2 - 9}{-3,01 + 3} = \\frac{9,0601 - 9}{-3,01 + 3} = \\frac{0,0601}{-0,01} = -6,01\n\n f(-3,001) = \\frac{(-3,001)^2 - 9}{-3,001 + 3} = \\frac{9,006001 - 9}{-3,001 + 3} = \\frac{0,006001}{-0,001} = -6,001\n\n lim f(x) = lim \\frac{x^2 - 9}{x + 3} \n x \\to -3 = \\frac{-3^2 - 9}{-3 + 3} = 0 \n\n lim f(x) = lim \\frac{x^2 - 9}{x + 3} = D(-3) \\\\ = \\frac{-3^2 - 9}{-3 + 3} = \\frac{4 - 14 + 10}{0}\n\n lim f(x) = lim \\frac{(x^2 - 9)(x + 3)}{(x - 3)(x - 3)}\n x \\to -3 = 0 \n\n lim h(x) = lim \\frac{x^2 - 9}{x - 3} \\\n x \\to -3 = 9 \n\n lim h(x) = lim \\frac{-59}{59} \\\n x \\to -2 \\\n = -1\n\n (5^2 - 2 \\cdot 4)\n\n lim 5x^2 - 2x < 1 \\\n x \\to 1\n\n b) lim \\sqrt[3]{x^2 + 8} = \\frac{\\sqrt[3]{2^3 + 2^2 + 8}}{x^2 + 8}\n \\to \\cdots = 3 \n\n 3/2 \\newline = lim = 2 \n x \\to -2 lim \\frac{x^2 + 7x + 10}{x^2 -x -2}\n x \\to 0 = \\frac{(x + 2)(x + 5)}{(x - 2)(x + 1)} = 0 \\newline -2 \n = \\frac{0}{0} \n\n d) lim \\frac{x^3 - 2x}{3x + 1}\n x \\to 1 = \\frac{2}{2 (1^2 + 2^2 + 2 + 1 + 3)}= 0 \n\n e) lim \\sqrt[3]{x^4 + 8} = \sqrt[3]{2^3 + 2^2 + 8} = 3 \n\n (1)lim \\sqrt{x + 1 - 1} = 0\n = \\mathbb{R} \n\n (2) a) lim \\frac{x + x}{x + 1} \\to x+1\n lim \\frac{x + 1}{n + 5} = \\frac{x + o + 1}{n + 1}\n = \\quad 4 \n\n b) lim \\sqrt{x + 1} = \\frac{ \\sqrt{2} + 2}{2}\n = lim \\frac{x + 5}{x - 2} = 0 - 70 = -1