Transcripccion calculo (C)-F.docx - L\u00edmites y continuidad 1 y f(x)=| x| 1 x 1 0 0 2 2 1-1 2-2 2 Funci\u00f3n Continua 1-2-1 1 2 x 2 x-z \u2264 0 Si f(x)=x |x

Transcripccion calculo (C)-F.docx - Lu00edmites y...

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Límites y continuidad 1) FunciónContinua 2) Si x-z 0 x<z f(x) = x –x +z f(x)=z Si x –z > 0 f(x) = 2x –z 3) y 2 1 -2 -1 1 2 x f(x)=| x| x 1 1 0 0 2 2 1 -1 2 -2 y 6 5 4 3 2 1 -2 -1 1 2 3 4 5 x f(x)=x +|x –z| x 2 0 2 1 2 2 4 3 2 -1 2 -2 2 -3 8 5 f(x)= x 2 -| x| x 0 1 0 0 2 2 2 -2 6 3 6 -3 0 -1 Y 6 5 4 3 2 1 -2 -1 1 2 3 4 5 x
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4) f(x) = 1 x lim x→ 0 1 x = 1 0 = La función no es continua en x=0 5) x 3 x 1 f(x) = 2 x + 3 x>1 lim x→ 1 x 3 = lim x → 1 2 x + 3 1 3 =− 2 ( 1 ) + 3 1 = 1 Si es continua 6) f ( x ) = 1 x 2 ¿Es f continua en x=0? lim x→ 0 1 x 2 = 1 0 2 = La función no está definida en x=0
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3.2.1 lim x→ 5 25 x 3 375 x 2 + 1839 x 2945 = 0 | x 5 | < δ ¿ 25 x 3 375 x 2 + 1839 x 2945 ¿ ε 25 x 3 375 x 2 + 1839 x 2945 =( x 5 )( 25 x 2 250 x + 589 ) | ( x 5 ) ( 25 x 2 250 x + 589 ) | < ε Por otro lado | x 5 | < δ δ = 1 | x 5 | < 1 1 < x 5 < 1 4 < x < 6 16 < x 2 < 36 400 < 25 x 2 < 900 1000 < 250 x < 1500 1000 >− 250 x >− 1500 1500 250 x 1000 1100 < 25 x 2 250 x 100 589 1100 < 25 x 2 250 x + 589 < 589 100 511 < 25 x 2 250 x + 589 < 489 ¿ 25 x 2 250 x + 5891 ¿ 511 (1) Como ( x 5 ) ( 25 x 2 250 x + 589 ) < ε (2) en (2) ( x 5 ) ( 511 ) < ε δ ( 511 ) = ε 25 25 -375 1839 -2945 5 125 -1250 2945 25 -250 589 0
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δ = ε 511
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Continuidad de funciones 1) f(x) = x 2 9 x 3 x 3 A x=3 lim x→ 3 x 2 9 x 3 = lim x → 3 A lim x→ 3 ( x + 3 )( x 3 ) x 3 = A 3 + 3 = A 6 = A 2) 2 x + 5 x ¿ 0 f(x) = A + 2 0 ≤x < 5 2 4 x B x≥ 5 2 lim x→ 0 2 x + 5 = lim x → 0 A + 2 2 ( 0 ) + 5 = A + 2 3 = A lim x → 5 2 A + 2 = lim x→ 5 2 4 x B A + 2 = 4 ( 5 2 ) B 3 + 2 = 10 B B = 5 3) T 2 x 2 + x 1 x ¿ 1 f(x) = x 5 x≥ 5 2 lim T → 1 T 2 x 2 + x 1 = lim x→ 1 x 5 T 2 ( 1 ) 2 + 1 1 = 1 5 T 2 =− 4 T = 4 Imaginario NO HAY SOLUCION!!
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4) 4 x + 5 k + x 0 < x < 5 f(x)= k 2 + x x≥ 5 lim x→ 5 4 x + 5 k + x = lim x → 5 k 2 + x 20 + 5 k + x = k 2 + 5 25 = ( k + 5 ) ( k 2 + 5 ) 25 = k 3 + 5 k + 5 k 2 + 25 k 3 + 5 k + 5 k 2 = 0 k ( k 2 + 5 k + 5 ) = 0 k = 0 k = 5 ± 25 20 2 k = 5 ± 5 2 3.2.3 Calculo de límites 1) lim x→ 0 x 10 x + 1 x 2 1 = 0 10 0 + 1 0 2 1 = 1 1 =− 1 2) lim x→ 0 3 x 3 x 3 x + 3 x = 3 0 3 0 3 0 + 3 0 = 1 1 1 + 1 = 0 3) lim x→ 2 x 4 4 x 2 5 x + 6 = lim x → 2 ( x 2 ) ( x + 2 ) ( x 3 ) ( x 2 ) = 2 + 2 2 3 =− 4 4) lim x→ 1 ( x 1 x 2 + 8 3 ) ( x 2 + 8 + 3 x 2 + 8 + 3 ) = lim x → 1 ( x 1 ) ( x 2 + 8 + 3 ) x 2 + 8 9 = lim x → 1 ( x 1 ) ( x 2 + 8 + 3 ) x 2 1 = lim x→ 1 ( x 1 ) ( x 2 + 8 + 3 ) ( x 1 ) ( x + 1 ) = 1 5) lim x→ 0 x = lim x → 0 x + 1 0 = 0 + 1 0 1 NO EXISTE 6) lim x→ 7 ( 5 3 x + 4 x 2 49 ) ( 5 + 3 x + 4 5 + 3 x + 4 ) = lim x → 7 ( 5 ) 2 ( 3 x + 4 ) 2 ( x 2 49 ) ( 5 + 3 x + 4 ) = lim x→ 7 25 3 x 4 ( x 2 49 ) ( 5 + 3 x + 4 ) = lim x→ 7 21 3 x ( x 2 49 ) ( 5 + 3 x + 4 ) = ( x
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Remplazando x=7 lim x→ 7 3 ( 7 + 7 ) ( 5 + 21 + 4 ) = 3 ( 14 ) ( 10 ) = 3 140 7) lim x → 27 ( x 27 3 x 3 ) ( 3 x 2 + 3 3 x + 9 3 x 2 + 3 3 x + 9 ) = lim x→ 27 ( x 27 ) ( 3 x 2 + 3 3 x + 9 ) 3 x 3 3 3 = lim x→ 27 ( x 27 ) ( 3 x 2 + 3 3 x + 9 ) x 27 = 3 27 2 + 3 3 27 + 9 = 8) lim x→ a ( x a x a ) ( x a x a ) = lim x →a ( x ) 2 ( a ) 2 ( x a ) ( x + a ) = lim x→ a x a ( x a ) ( x + a ) = 1 a + a = 1 2 a 9) lim x→ a ( x 1 / n a 1 / n x a ) x a = ( x 1 n a 1 n )( x n 1 n + x n 2 n a 1 n + x n 3 n a 2 n + x n 4 n a 3 n + ... + a n 1 n ) lim x→ a ( x 1 / n a 1 / n ( x 1 n a 1 n )( x n 1 n + x n 2 n a 1 n + x n 3 n a 2 n + x n 4 n a 3 n + + a n 1 n ) ) = 1 a n 1 n + a n 2 n a 1 n + a n 3 n a 2 n + + a n 1 n = a n 1 n + n veces 10) lim x→ 3 ( x 2 2 x + 6 x 2 + 2 x 6 x 2 4 x + 3 ) ( x 2 2 x + 6 + x 2 + 2 x 6 x 2 2 x + 6 + x 2 + 2 x 6 ) = lim x→ 3 ( ( x 2 2 x + 6 ) 2 ( x 2 + 2 x 6 ) 2 ( x 2 4 x + 3 ) ( x 2 2 x + 6 + x 2 + 2 x 6 ) 11) lim x→ 1 ( x + 1 x 2 + 1 x 1 ) ( x + 1 + x 2 + 1 x + 1 + x 2 + 1 ) = lim x → 1 ( ( x + 1 ) 2 ( x 2 + 1 ) 2 ( x 1 ) ( 2 + 2 ) ) = lim x → 1 ( x ( x 1 ) ( x 1 ) ( 2 2 ) ) = 1 2 2 = 2 4 12) lim x → 2 ( x 2 2 x 2 + 2 2 ) ( x 2 + 2 + 2 x 2 + 2 + 2 ) = lim x → 2 ( x 2 2 ) ( x 2 + 2 +
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