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Engenharia de Computação ·
Cálculo 1
· 2021/2
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a) \lim_{x \to 2} \frac{x^2 - 4x + 4}{x^2 - 3x + 4} = \frac{2^2 - 4.2 + 4}{2^2 - 3.2 + 4} = \frac{4 - 8 + 4}{4 - 6 + 4} = \frac{0}{2} = 0 note que: como \frac{x^2 - 4x + 4}{x^2 - 3x + 4} e continua basta substituir x por 2. b) \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} = \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} \cdot \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}} = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x} + \sqrt{3})}{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x} + \sqrt{3})}{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} = \lim_{x \to 3} \frac{x - 3}{(\sqrt{x} - \sqrt{3})} = 2\sqrt{3} logo \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} = 2\sqrt{3} c) \lim_{x \to \pi} \frac{1 - \sin(\frac{x}{2})}{\pi - x} = \lim_{x \to \pi} \frac{\cos(\frac{x}{2})(\frac{1}{2})}{-1}\) = \lim_{x \to \pi} \frac{1}{2} \cos(\frac{x}{2}) = 0 \hspace{1cm} \text{REGA DE L'HOSPITAL} logo \lim_{x \to L} \frac{\sin(\pi x)}{x - L} = -\pi\) note que: \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \text{ e o limite fundamental} f'(x) = \lim_{x \to 1} \frac{f(x) - f(L)}{x - 1}\) e f(x) = \begin{cases} x + 1, \text{ se } x \ge 1\\ 2x, \text{ se } x < 1 \end{cases}\) f_L'(1) = \lim_{x \to 1^+} \frac{x + 1 - 2}{x - 1} = \lim_{x \to 1^+} \frac{x - 1}{x - 1} = 1 f_L'(1) = \lim_{x \to 1^-} \frac{2x - 2}{x - 1} = \lim_{x \to 1^-} \frac{2(x - 1)}{x - 1} = 2 f_L'(L) = \lim_{x \to 1} \frac{f(x) - f(L)}{x - 1} = \text{não existe, pois os limites laterais são distintos.} Como f(1) = 1^5 + 1 + 1 = 3, f(-1) = (-1)^5 + 1 + 1 = -1 e sendo f(x) = x^5 + x + 1 uma função contínua no intervalo [-1, 1], então pelo Teorema do Valor Intermediário existe c \in [-1, 1] tal que f(c) = 0, ou seja, c é uma raiz de f(x) = x^5 + x + 1. OBS: Teorema do Valor Intermediário. Se f é contínua no intervalo fechado [a, b], então para todo K entre f(a) e f(b) existe c \in (a, b) tal que K = f(c). lim_{h -> 0} (f(x+h) - f(x))/h , onde f(x) = x^2 + \sen x lim_{h -> 0} [(x+h)^2 + \sen(x+h) - [x^2 + \sen x]]/h = lim_{h -> 0} x^2 + 2xh + h^2 + \sen(x+h) - x^2 - \sen x / h = lim_{h -> 0} 2x + h + lim_{h -> 0} \sen(x+h) - \sen x / h = lim_{h -> 0} 2x + lim_{h -> 0} \sen x \cos h + \cos x \sen h - \sen x / h = 2x + lim_{h -> 0} \cos x \sen h / h + lim_{h -> 0} \sen x (\cos h - 1) / h = 2x + \cos x lim_{h -> 0} \sen h / h + \sen x lim_{h -> 0} (\cos h - 1) / h = 2x + \cos x + \sen x lim_{h -> 0} \sen h / h = 2x + \cos x + \sen x · 0 = 2x + \cos x Logo lim_{h -> 0} (f(x+h) - f(x))/h = 2x + \cos x mostre que se a função f(x) for derivável no ponto P, então f(x) será contínua em P. Temos que: lim_{x -> P} (f(x) - f(P)) = lim_{x -> P} (x - P) ( (f(x) - f(P)) / (x - P) ) = lim_{x -> P} (x - P) lim_{x -> P} ( (f(x) - f(P)) / (x - P) ) = 0 · f'(P) . Deste modo lim_{x -> P} f(x) - f(P) = 0 e com isto lim_{x -> P} f(x) - lim_{x -> P} f(P) = 0 lim_{x -> P} f(x) = lim_{x -> P} f(P) = f(P) Logo lim_{x -> P} f(x) = f(P) => f é contínua em P. Y(x) = e^-x \cos(2x) d^2Y/dx^2 + 2dY/dx + 5Y = 0 dY/dx = -e^-x (\cos(2x) - \sen(2x) · 2e^-x) dY/dx = -e^-x (\cos(2x) + 2\sen(2x)) d^2Y/dx^2 = e^-x (\cos(2x) + 2\sen(2x)) + (-\sen(2x) · 2 + 2\cos(2x) · 2)(-e^-x) d^2Y/dx^2 = e^-x (\cos(2x) + 2\sen(2x)) + e^-x (2\sen(2x) - 4\cos(2x)) d^2Y/dx^2 = e^-x (-3\cos(2x) + 4\sen(2x)) Por fim d^2Y/dx^2 + 2dY/dx + 5Y = e^-x (-3\cos(2x) + 4\sen(2x)) - 2e^-x (\cos(2x) + 2\sen(2x)) + 5(e^-x\cos(2x)) = 0 = e^-x (-3\cos(2x) + 4\sen(2x) - 2\cos(2x) - 4\sen(2x) + 5\cos(2x)) = e^-x (0) = 0 , logo d^2Y/dx^2 + 2dY/dx + 5Y = 0 a) f(x) = x \cos x (1 + \ln x) f'(x) = (x \cos x)'(1 + \ln x) + (1 + \ln x)'(x \cos x) f'(x) = (\cos x - x \sin x)(1 + \ln x) + \left(\frac{1}{x}\right)(x \cos x) f'(x) = (\cos x - x \sin x)(1 + \ln x) + \cos x = \cos x - x \sin x + \cos x \ln x - x \sin x \ln x + \cos x = 2 \cos x - x \sin x + \cos x \ln x - x \sin x \ln x b) f(x) = -\left(\frac{x + \ln x}{x - \cos x}\right) f'(x) = -\frac{(x + \ln x)'(x - \cos x) - (x - \cos x)'(x + \ln x)}{(x - \cos x)^2} f'(x) = -\frac{(1 + \cos x)(x - \cos x) - (1 + \ln x)(x + \ln x)}{(x - \cos x)^2} f'(x) = -\frac{(x - \cos x + x \cos x - \cos^2 x - x - \ln x - x \ln x - \ln^2 x + \cos x \ln x)}{(x - \cos x)^2} f'(x) = -\frac{(-\cos x + x \cos x - \ln x - x \ln x - \ln^2 x + \cos x^2 x)}{(x - \cos x)^2} c) f(x) = \frac{x e^{2x}}{\ln(3x + 1)} f'(x) = \frac{(e^{2x} + e^{2x}(2)x)}{\ln(3x + 1)} = \frac{(e^{2x}(1 + 2x))}{\ln(3x + 1)} d) f(x) = e - x \sec(x^2) f'(x) = -\sec(x^2) - x \sec(x^2) \tan(x^2) \cdot 2x f'(x) = -\sec(x^2) - 2x^2 \sec(x^2) \tan(x^2) e) f(x) = (x^2 + \cotg(x))^3 f'(x) = 3(x^2 + \cotg(x))^2(2x + \csc^2(x))
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a) \lim_{x \to 2} \frac{x^2 - 4x + 4}{x^2 - 3x + 4} = \frac{2^2 - 4.2 + 4}{2^2 - 3.2 + 4} = \frac{4 - 8 + 4}{4 - 6 + 4} = \frac{0}{2} = 0 note que: como \frac{x^2 - 4x + 4}{x^2 - 3x + 4} e continua basta substituir x por 2. b) \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} = \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} \cdot \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}} = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x} + \sqrt{3})}{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x} + \sqrt{3})}{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} = \lim_{x \to 3} \frac{x - 3}{(\sqrt{x} - \sqrt{3})} = 2\sqrt{3} logo \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} = 2\sqrt{3} c) \lim_{x \to \pi} \frac{1 - \sin(\frac{x}{2})}{\pi - x} = \lim_{x \to \pi} \frac{\cos(\frac{x}{2})(\frac{1}{2})}{-1}\) = \lim_{x \to \pi} \frac{1}{2} \cos(\frac{x}{2}) = 0 \hspace{1cm} \text{REGA DE L'HOSPITAL} logo \lim_{x \to L} \frac{\sin(\pi x)}{x - L} = -\pi\) note que: \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \text{ e o limite fundamental} f'(x) = \lim_{x \to 1} \frac{f(x) - f(L)}{x - 1}\) e f(x) = \begin{cases} x + 1, \text{ se } x \ge 1\\ 2x, \text{ se } x < 1 \end{cases}\) f_L'(1) = \lim_{x \to 1^+} \frac{x + 1 - 2}{x - 1} = \lim_{x \to 1^+} \frac{x - 1}{x - 1} = 1 f_L'(1) = \lim_{x \to 1^-} \frac{2x - 2}{x - 1} = \lim_{x \to 1^-} \frac{2(x - 1)}{x - 1} = 2 f_L'(L) = \lim_{x \to 1} \frac{f(x) - f(L)}{x - 1} = \text{não existe, pois os limites laterais são distintos.} Como f(1) = 1^5 + 1 + 1 = 3, f(-1) = (-1)^5 + 1 + 1 = -1 e sendo f(x) = x^5 + x + 1 uma função contínua no intervalo [-1, 1], então pelo Teorema do Valor Intermediário existe c \in [-1, 1] tal que f(c) = 0, ou seja, c é uma raiz de f(x) = x^5 + x + 1. OBS: Teorema do Valor Intermediário. Se f é contínua no intervalo fechado [a, b], então para todo K entre f(a) e f(b) existe c \in (a, b) tal que K = f(c). lim_{h -> 0} (f(x+h) - f(x))/h , onde f(x) = x^2 + \sen x lim_{h -> 0} [(x+h)^2 + \sen(x+h) - [x^2 + \sen x]]/h = lim_{h -> 0} x^2 + 2xh + h^2 + \sen(x+h) - x^2 - \sen x / h = lim_{h -> 0} 2x + h + lim_{h -> 0} \sen(x+h) - \sen x / h = lim_{h -> 0} 2x + lim_{h -> 0} \sen x \cos h + \cos x \sen h - \sen x / h = 2x + lim_{h -> 0} \cos x \sen h / h + lim_{h -> 0} \sen x (\cos h - 1) / h = 2x + \cos x lim_{h -> 0} \sen h / h + \sen x lim_{h -> 0} (\cos h - 1) / h = 2x + \cos x + \sen x lim_{h -> 0} \sen h / h = 2x + \cos x + \sen x · 0 = 2x + \cos x Logo lim_{h -> 0} (f(x+h) - f(x))/h = 2x + \cos x mostre que se a função f(x) for derivável no ponto P, então f(x) será contínua em P. Temos que: lim_{x -> P} (f(x) - f(P)) = lim_{x -> P} (x - P) ( (f(x) - f(P)) / (x - P) ) = lim_{x -> P} (x - P) lim_{x -> P} ( (f(x) - f(P)) / (x - P) ) = 0 · f'(P) . Deste modo lim_{x -> P} f(x) - f(P) = 0 e com isto lim_{x -> P} f(x) - lim_{x -> P} f(P) = 0 lim_{x -> P} f(x) = lim_{x -> P} f(P) = f(P) Logo lim_{x -> P} f(x) = f(P) => f é contínua em P. Y(x) = e^-x \cos(2x) d^2Y/dx^2 + 2dY/dx + 5Y = 0 dY/dx = -e^-x (\cos(2x) - \sen(2x) · 2e^-x) dY/dx = -e^-x (\cos(2x) + 2\sen(2x)) d^2Y/dx^2 = e^-x (\cos(2x) + 2\sen(2x)) + (-\sen(2x) · 2 + 2\cos(2x) · 2)(-e^-x) d^2Y/dx^2 = e^-x (\cos(2x) + 2\sen(2x)) + e^-x (2\sen(2x) - 4\cos(2x)) d^2Y/dx^2 = e^-x (-3\cos(2x) + 4\sen(2x)) Por fim d^2Y/dx^2 + 2dY/dx + 5Y = e^-x (-3\cos(2x) + 4\sen(2x)) - 2e^-x (\cos(2x) + 2\sen(2x)) + 5(e^-x\cos(2x)) = 0 = e^-x (-3\cos(2x) + 4\sen(2x) - 2\cos(2x) - 4\sen(2x) + 5\cos(2x)) = e^-x (0) = 0 , logo d^2Y/dx^2 + 2dY/dx + 5Y = 0 a) f(x) = x \cos x (1 + \ln x) f'(x) = (x \cos x)'(1 + \ln x) + (1 + \ln x)'(x \cos x) f'(x) = (\cos x - x \sin x)(1 + \ln x) + \left(\frac{1}{x}\right)(x \cos x) f'(x) = (\cos x - x \sin x)(1 + \ln x) + \cos x = \cos x - x \sin x + \cos x \ln x - x \sin x \ln x + \cos x = 2 \cos x - x \sin x + \cos x \ln x - x \sin x \ln x b) f(x) = -\left(\frac{x + \ln x}{x - \cos x}\right) f'(x) = -\frac{(x + \ln x)'(x - \cos x) - (x - \cos x)'(x + \ln x)}{(x - \cos x)^2} f'(x) = -\frac{(1 + \cos x)(x - \cos x) - (1 + \ln x)(x + \ln x)}{(x - \cos x)^2} f'(x) = -\frac{(x - \cos x + x \cos x - \cos^2 x - x - \ln x - x \ln x - \ln^2 x + \cos x \ln x)}{(x - \cos x)^2} f'(x) = -\frac{(-\cos x + x \cos x - \ln x - x \ln x - \ln^2 x + \cos x^2 x)}{(x - \cos x)^2} c) f(x) = \frac{x e^{2x}}{\ln(3x + 1)} f'(x) = \frac{(e^{2x} + e^{2x}(2)x)}{\ln(3x + 1)} = \frac{(e^{2x}(1 + 2x))}{\ln(3x + 1)} d) f(x) = e - x \sec(x^2) f'(x) = -\sec(x^2) - x \sec(x^2) \tan(x^2) \cdot 2x f'(x) = -\sec(x^2) - 2x^2 \sec(x^2) \tan(x^2) e) f(x) = (x^2 + \cotg(x))^3 f'(x) = 3(x^2 + \cotg(x))^2(2x + \csc^2(x))