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1. a) Si x ≥ 2, numerador y denominador son ≥ 0 , si x < −2

1. a) Si x ≥ 2, numerador y denominador son ≥ 0 , si x < −2

1. a) Si x ≥ 2, numerador y denominador son ≥ 0 , si x < −2

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68. Continua en R − {−1,0,1} . En x = −1 discontinua ( f(x) → −∞ ).En x = 0,1 continua: lím(x − 1) −x2 +o(x 2 )log|1−xx→0 x= 0 ; lím2 | 2xx→1 x/(x−1)= lím (x − 1) 2 2x(1−x)=x→1 1−x 2 límx→1Es derivable en x = 0 : f ′ (0) = lím(h − 1) log(1−h2 ) −h=h→0 h 2 −lím2 +o(h 2 )= 1 .h→0 h 21+x= 0 .f continua en [0,1] , derivable en (0,1) y f(0) = f(1) ⇒ existe c , 0 < c < 1 , con f ′ (c) = 0 (Rolle).69. f ′ (0) = − 1 2 . lím = ∞ , lím = 0 , f > 0 ∀x y f continua ∀x ⇒ imagen = (0,∞) .x→−∞ x→∞f ′ (x) = xe−x −1+e −xx 2= x+1−exx 2 e x< 0 ∀x ⇒ f decreciente en todo R, pues g(x) = x+1−e x < 0 ∀x(g(0) = 0 , g ′ (x) = 1 − e x po<strong>si</strong>tivo <strong>si</strong> x < 0 y negativo <strong>si</strong> x > 0 ⇒ g negativa <strong>si</strong> x < 0 y <strong>si</strong> x > 0 ).f(x) = 1 − x 2+ ···+(−1)n(n+1)! xn + ··· ⇒ f (n) (0) = n!(−1)n(n+1)!= (−1)n(n+1) . En particular, f (2003) (0) = − 12004 .10

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