Lista - Fixação Cálculo 1 - 2023-2

Cuiabá, 13 de março de 2024. Docente: Elaynne Xavier Souza Araujo Disciplina: Matemática — Agronomia 2023/2 12ª Lista de Exercícios de fixação Calcule o limite: 1) lim (2x/(x−1)) x→1 2) lim (x²/(x−2)) x→2 3) lim (√4+3x²)/x³ x→0 4) lim (x²−5x+1)/(x²−5x−2x) x→3 5) lim (2x²+3x−2)/(x²−3x−4) x→3 6) lim (√25−x²)/(x−5) x→5 7) lim (x²−1)/(x²+11) x→∞ 8) lim (x+1/x−2) x→2 9) lim (x²−4)/(x+2) x→−2 10) lim (x⁴−16x)/x³ x→4 11) lim x→2 x²+x+1 12) lim (5x²+7x+3)/x² x→∞ 13) lim (7x²−3x)/(x³−100) x→0 14) lim (2x²+5x+10)/x² x→−5 15) lim (x⁶/(x⁶+100)) x→∞ 16) lim (6x²/(3x+5)) x→∞ 17) lim (5x²−3x)/x² x→∞ 18) lim ((√x²−5x−1)/(3x)) x→∞ 19) lim (7x³−15x²/(13x)) x→+∞ d) \lim_{e \to 5} \frac{\sqrt{25-t^2}}{4-1} = \lim_{t \to 5} \frac{\sqrt{5+t}\sqrt{5-t}}{\sqrt{5+t}-\sqrt{5-t}} = \frac{\sqrt{10}}{\infty} = -\infty 7) \lim_{x \to 1^-} \frac{x-1}{|x^2-1|} = \lim_{x \to 1^-} \frac{x-1}{-(x^2-1)} = -1 = -1 8) \lim_{x \to 2^-} \frac{x^2+1}{x-2} = \frac{5}{0} = -\infty 9) \lim_{z \to 2^+} \frac{z^2+1}{z-2} = \frac{5}{0} = +\infty 10) \lim_{x \to -\infty} \frac{1+6x}{-2+x} = 6 11) \lim_{x \to -\infty} \frac{2x^2+x+1}{-4x^2+5x+10} = -\frac{1}{2} 12) \lim_{x \to \infty} \frac{5x^2-7x+3}{8x^2+5x+7} = \frac{5}{8} 13) \lim_{x \to -\infty} \frac{7x^3+3x+1}{x^3-2x+3} = 7 14) \lim_{x \to \infty} \frac{x^{100}+x}{x^{101}-x^{100}} = 0 15) \lim_{x \to \infty} \frac{8x}{\sqrt[4]{3x^4+5}} = \frac{8}{\sqrt[4]{3}} 12 Lista 1) \lim_{x \to 1^+} \frac{2x}{x-1} = \frac{2(1)}{0^+} = +\infty 2) \lim_{x \to 2^-} \frac{x^2}{x-2} = -\infty 3) \lim_{x \to 0^+} \frac{\sqrt{4+3x^2}}{50^x} = +\infty n) \lim_{z \to 1} \frac{z^2+4z+3}{z^2-1} = \lim_{z \to 1} \frac{(z+3)(z+1)}{(z-1)(z+1)} = \frac{-1+3}{-1-2} = \frac{-2}{-1} = 1 s) \lim_{t \to 7} \frac{t^2-49}{t+7} = \lim_{t \to 7} \frac{(t+7)(t-7)}{t+7} = -7-7 = -14 t) \lim_{h \to 0} \frac{(3+h)^2-9}{h} = \lim_{h \to 0} \frac{9+6h+h^2-9}{h} = \lim_{h \to 0} \frac{6h+h^2}{h} = \lim_{h \to 0} \frac{6h}{h} = \lim_{h \to 0} 6h = 6 u) \lim_{h \to 0} \frac{2-\sqrt{4-h}}{h} = \lim_{h \to 0} \frac{(2-\sqrt{4-h})(2+\sqrt{4-h})}{h(2+\sqrt{4-h})} = \lim_{h \to 0} \frac{4-(4-h)}{h(2+\sqrt{4-h})} = \lim_{h \to 0} \frac{h}{h(2+\sqrt{4-h})} = \lim_{h \to 0} \frac{1}{2+\sqrt{4}} = \frac{1}{4} v) \lim_{x \to 1} \frac{\sqrt{8x+1}}{x+3} = \frac{\sqrt{8(1)+1}}{1+3} = \frac{\sqrt{9}}{4} = \frac{3}{2} 3) a) \lim_{x\to 2} [f(x) + g(x)] = \lim_{x\to 2} f(x) + \lim_{x\to 2} g(x) = 4 + 3 = 7 b) \lim_{x\to 2} [f(x) - g(x)] = \lim_{x\to 2} f(x) - \lim_{x\to 2} g(x) = 4 - 3 = 1 c) \lim_{x\to 2} [f(x) \cdot g(x)] = (\lim_{x\to 2} f(x)) \cdot (\lim_{x\to 2} g(x)) = 4 \cdot 3 = 12 d) \lim_{x\to 2} \left[ \frac{f(x)}{g(x)} \right] = \frac{\lim_{x\to 2} f(x)}{\lim_{x\to 2} g(x)} = \frac{4}{3} e) \lim_{x\to 2} \sqrt{f(x) \cdot g(x)} = \sqrt{(\lim_{x\to 2} f(x))(\lim_{x\to 2} g(x))} = \sqrt{4 \cdot 3} = 2\sqrt{3} f) \lim_{x\to 2} \left[ \frac{f(x)}{g(x)} \right] = \frac{\lim_{x\to 2} f(x)}{\lim_{x\to 2} g(x)} = \frac{4}{3} 4) a) \lim_{x\to 1} (5 - 3x - x^2) = 5 - 3(1) - (1)^2 = 5 - 3 - 1 = 1 b) \lim_{x\to 3} 5x^2 - 7x - 3 = 5(3)^2 - 7(3) - 3 = 45 - 21 - 3 = 21 c) \lim_{x\to 2} \frac{x^2 + x + 1}{x^2 + 2x} = \frac{2^2 + 2 + 1}{2^2 + 2(2)} = \frac{7}{8} d) \lim_{y\to -1} (3y^3 - 2y^2 + 5y - 1) = 3(-1)^3 - 2(-1)^2 + 5(-1) - 1 = -11 e) \lim_{x\to 5/2} \frac{4x^2 - 25}{2x - 5} = \lim_{x\to 5/2} \frac{(2x + 5)(2x - 5)}{2x - 5} = 2(\frac{5}{2}) + 5 = 10 f) \lim_{x\to 3} \frac{x^2 - 2x - 3}{x - 3} = \lim_{x\to 3} \frac{(x - 3)(x + 1)}{x - 3} = 3 + 1 = 4 11^a Lista 1) a) \lim_{x\to 4} 3x = 3(4) = 12 b) \lim_{x\to 1} (3x - 6) = 3(1) - 6 = 3 - 6 = -3 c) \lim_{x\to 2} (2 - 3x) = 2 - 3(-2) = 2 + 6 = 8 d) \lim_{x\to 5} \frac{2}{x} = \frac{2}{5} e) \lim_{x\to 1/2} |1 - 2x| = |1 - 2(1/2)| = |1 - 1| = 0 f) \lim_{x\to 3} \frac{x^2 - 9}{x - 3} = \lim_{x\to 3} \frac{(x - 3)(x + 3)}{x - 3} = 3 + 3 = 6 2) a) \lim_{x\to 2} \frac{x^2 - 5x + 6}{x - 2} = \lim_{x\to 2} \frac{(x - 2)(x - 3)}{x - 2} = 2 - 3 = -1 b) \lim_{t\to 0} \frac{t^2 + 2t + 1}{t + 5} = \frac{1}{5} c) \lim_{x\to -2} \frac{x^2 - 4}{x + 2} = \lim_{x\to -2} \frac{(x + 2)(x - 2)}{x + 2} = -2 - 2 = -4 d) \lim_{x\to 1} \frac{x^2 - 2x + 1}{x - 1} = \lim_{x\to 1} \frac{(x - 1)^2}{x - 1} = \lim_{x\to 1} x - 1 = 1 - 1 = 0 e) \lim_{x\to 1} \frac{x^3 - 1}{x^2 - 1} = \lim_{x\to 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{1 + 1 + 1}{1 + 1} = \frac{3}{2} f) \lim_{x\to 3} \frac{x^2 - 2x - 3}{x - 3} = \lim_{x\to 3} \frac{(x - 3)(x + 1)}{x - 3} = 3 + 1 = 4