5 casos para calcular limites
TRANSCRIPT
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5 CASOS PARA CALCULAR LIMITES
EN FUNCIONES Y LIMITES EN
FUNCIONES TRIGONOMETRICAS
CALCULO DIFERENCIAL
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CUANDO EVALUAMOS UNA FUNCION MEDIANTE LIMITES LOS RESULTADOS SON
SENCILLOS PERO DA LA CASUALIDAD EN QUE ALGUNAS FUNCIONES TIENEN RAIZ
CUADRADA, OTRAS TIENEN QUE SER FACTORIZADAS U OTRAS TIENEN QUE ESTAR
DERIVADAS (POR METODO DE LโHOSPITAL O REGLA DE LOS 4 PASOS).
VEAMOS ALGUNOS EJEMPLOS PARA ENTENDER ESTE TEMAโฆ
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lim๐ฅโ2
3๐ฅ
lim๐ฅโ2
3๐ฅ = 3 2 = 6
lim๐ฅโโ2
๐ฅ2 + 8๐ฅ โ 1
lim๐ฅโโ2
๐ฅ2 + 8๐ฅ โ 1 = โ2 2 + 8 โ2 โ 1 = 4 โ 16 โ 1 = โ13
lim๐โ2
8๐
lim๐โ2
8๐ = 8 2 = 16 = 4
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lim๐ฅโ
23
7๐ฅ โ 10
lim๐ฅโ
23
7๐ฅ โ 10 = 72
3โ 10 =
14
3โ 10 = โ
16
3
lim๐ฅโ0
3๐ฅ2 + 7๐ฅ โ 3
2๐ฅ โ 1
lim๐ฅโ0
3๐ฅ2 + 7๐ฅ โ 3
2๐ฅ โ 1=3 0 2 + 7 0 โ 3
2 0 โ 1=0 + 0 โ 3
0 โ 1=โ3
โ1= 3
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lim๐ฅโ0
4๐ฅ
lim๐ฅโ0
4๐ฅ = 4 0 = 0
lim๐ฅโ2
๐ฅ
3๐ฅ โ 2
lim๐ฅโ2
๐ฅ
3๐ฅ โ 2=
0
3 0 โ 2=
0
0 โ 2=
0
โ2= 0
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lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2
lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2= lim
๐ฅโ2
๐ฅ + 2 ๐ฅ โ 2
๐ฅ โ 2= lim
๐ฅโ2๐ฅ + 2 = 2 + 2 = 4
lim๐ฅโ3
๐ฅ2 + 4๐ฅ โ 21
๐ฅ โ 3
lim๐ฅโ3
๐ฅ2 + 4๐ฅ โ 21
๐ฅ โ 3=
3 2 + 4 3 โ 21
3 โ 3=9 + 12 โ 21
0=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ3
๐ฅ2 + 4๐ฅ โ 21
๐ฅ โ 3= lim
๐ฅโ3
๐ฅ โ 3 ๐ฅ + 7
๐ฅ โ 3= lim
๐ฅโ3๐ฅ + 7 = 3 + 7 = 10
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lim๐ฅโ1
๐ฅ3 โ 1
๐ฅ โ 1
lim๐ฅโ1
๐ฅ3 โ 1
๐ฅ โ 1=1 โ 1
1 โ 1=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ1
๐ฅ โ 1 ๐ฅ2 + ๐ฅ + 1
๐ฅ โ 1= lim
๐ฅโ1๐ฅ2 + ๐ฅ + 1 = 1 2 + 1 + 1 = 1 + 1 + 1 = 3
lim๐ฅโ0
๐ฅ2 + 9๐ฅ
๐ฅ
lim๐ฅโ0
๐ฅ2 + 9๐ฅ
๐ฅ=02 + 9 0
0=0 + 0
0=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ0
๐ฅ2 + 9๐ฅ
๐ฅ= lim
๐ฅโ0
๐ฅ ๐ฅ + 9
๐ฅ= lim
๐ฅโ0๐ฅ + 9 = 0 + 9 = 9
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FORMULA DE LโHOPITAL
lim๐ฅโ๐
๐ ๐ฅ
๐ ๐ฅ= lim
๐ฅโ๐
๐โฒ ๐ฅ
๐โฒ ๐ฅ= lim
๐ฅโ๐
๐๐๐ฅ
๐ ๐ฅ
๐๐๐ฅ
๐ ๐ฅ
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lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2
lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2= lim
๐ฅโ2
๐๐๐ฅ
๐ฅ2 โ 4
๐๐๐ฅ
๐ฅ โ 2= lim
๐ฅโ2
2๐ฅ
1= 2 2 = 4
lim๐ฅโ2
๐ฅ2 โ 4
๐ฅ โ 2
lim๐ฅโ3
๐ฅ2 + 4๐ฅ โ 21
๐ฅ โ 3=
3 2 + 4 3 โ 21
3 โ 3=9 + 12 โ 21
0=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ3
๐ฅ2 + 4๐ฅ โ 21
๐ฅ โ 3= lim
๐ฅโ3
๐๐๐ฅ
๐ฅ2 + 4๐ฅ โ 21
๐๐๐ฅ
๐ฅ โ 3= lim
๐ฅโ3
2๐ฅ + 4
1= 2 3 + 4 = 6 + 4 = 10
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lim๐ฅโ1
๐ฅ3 โ 1
๐ฅ โ 1
lim๐ฅโ1
๐ฅ3 โ 1
๐ฅ โ 1=1 โ 1
1 โ 1=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ1
๐ฅ3 โ 1
๐ฅ โ 1= lim
๐ฅโ1
๐๐๐ฅ
๐ฅ3 โ 1
๐๐๐ฅ
๐ฅ โ 1= lim
๐ฅโ1
3๐ฅ2
1= 3 1 2 = 3 1 = 3
lim๐ฅโ0
๐ฅ2 + 9๐ฅ
๐ฅ
lim๐ฅโ0
๐ฅ2 + 9๐ฅ
๐ฅ=02 + 9 0
0=0 + 0
0=0
0= ๐ผ๐๐๐๐ก๐๐๐๐๐๐๐๐
lim๐ฅโ0
๐ฅ2 + 9๐ฅ
๐ฅ= lim
๐ฅโ0
๐๐๐ฅ
๐ฅ2 + 9๐ฅ
๐๐๐ฅ
๐ฅ= lim
๐ฅโ0
2๐ฅ + 9
1= 2 0 + 9 = 0 + 9 = 9
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lim๐ฅโ2
4 โ ๐ฅ2
3 โ ๐ฅ2 + 5
= lim๐ฅโ2
4 โ ๐ฅ2
3 โ ๐ฅ2 + 5
3 + ๐ฅ2 + 5
3 + ๐ฅ2 + 5= lim
๐ฅโ2
4 โ ๐ฅ2 3 + ๐ฅ2 + 5
9 โ ๐ฅ2 + 5
= lim๐ฅโ2
4 โ ๐ฅ2 3 + ๐ฅ2 + 5
9 โ ๐ฅ2 โ 5= lim
๐ฅโ2
4 โ ๐ฅ2 3 + ๐ฅ2 + 5
4 โ ๐ฅ2
= lim๐ฅโ2
3 + ๐ฅ2 + 5 = 3 + 2 2 + 5 = 3 + 4 + 5 = 3 + 9 = 3 + 3
= 6
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lim๐ฅโ2
2๐ฅ โ 3
2๐ฅ โ 3
= lim๐ฅโ2
2๐ฅ โ 3
2๐ฅ โ 3
2๐ฅ + 3
2๐ฅ + 3= lim
๐ฅโ2
2๐ฅ โ 3
2๐ฅ โ 3 2๐ฅ + 3= lim
๐ฅโ2
1
2๐ฅ + 3
=1
2(2) + 3=
1
4 + 3
=1
2 + 3
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REGLA DE LOS 4 PASOS
USANDO LA FORMULA SIGUIENTE:
limโโ0
๐ ๐ฅ + โ โ ๐ ๐ฅ
โ
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HALLAR EL LIMITE DE LA FUNCION
๐ฆ = 3๐ฅ2
๐ ๐ฅ + โ = 3 ๐ฅ + โ 2 = 3 ๐ฅ2 + 2๐ฅโ + โ2 = 3๐ฅ2 + 6๐ฅโ + 3โ2
๐ ๐ฅ = 3๐ฅ2
limโโ0
๐ ๐ฅ + โ โ ๐ ๐ฅ
โ= lim
โโ0
3๐ฅ2 + 6๐ฅโ + 3โ2 โ 3๐ฅ2
โ= lim
โโ0
6๐ฅโ + 3โ2
โ
= limโโ0
โ(6๐ฅ + 3โ)
โ= lim
โโ06๐ฅ + 3โ = 6๐ฅ + 3 0
= 6๐ฅ
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HALLAR EL LIMITE DE LA FUNCION
๐ ๐ฅ = ๐ฅ + 5
๐ ๐ฅ + โ = ๐ฅ + โ + 5
๐ ๐ฅ = ๐ฅ + 5
limโโ0
๐ ๐ฅ + โ โ ๐ ๐ฅ
โ= lim
โโ0
๐ฅ + โ + 5 โ ๐ฅ + 5
โ
= limโโ0
๐ฅ + โ + 5 โ ๐ฅ + 5
โ
๐ฅ + โ + 5 + ๐ฅ + 5
๐ฅ + โ + 5 + ๐ฅ + 5
= limโโ0
๐ฅ + โ + 5 โ (๐ฅ + 5)
โ ๐ฅ + โ + 5 + ๐ฅ + 5= lim
โโ0
โ
โ ๐ฅ + โ + 5 + ๐ฅ + 5
= limโโ0
1
๐ฅ + โ + 5 + ๐ฅ + 5=
1
๐ฅ + 0 + 5 + ๐ฅ + 5=
1
๐ฅ + 5 + ๐ฅ + 5
=1
2 ๐ฅ + 5
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lim๐ผโ0
๐ ๐๐ ๐ผ
tan๐ผ
lim๐ผโ0
๐ ๐๐ ๐ผ
tan๐ผ=๐ ๐๐ 0
tan 0=0
0= ๐ผ๐๐ท๐ธ๐๐ธ๐ ๐๐ผ๐๐ด๐ท๐
lim๐ผโ0
๐ ๐๐ ๐ผ
tan๐ผ= lim
๐ผโ0
๐๐๐ฅ
๐ ๐๐ ๐ผ
๐๐๐ฅ
tan ๐ผ= lim
๐ผโ0
cos ๐ผ
(sec ๐ผ)2=
cos 0
(sec 0)2=
1
1 2=1
1
= 1
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lim๐ผโ0
๐๐ ๐ ๐ผ
1 โ ๐๐๐ 2 ๐ผ
lim๐ผโ0
๐๐ ๐ ๐ผ
1 โ ๐๐๐ 2 ๐ผ=
0
1 โ 1=0
0= ๐ผ๐๐ท๐ธ๐๐ธ๐ ๐๐ผ๐๐ด๐ท๐
lim๐ผโ0
๐๐ ๐ ๐ผ
1 โ ๐๐๐ 2 ๐ผ= lim
๐ผโ0
๐๐ ๐ ๐ผ
๐ ๐๐2 ๐ผ= lim
๐ผโ0(csc ๐ผ) csc ๐ผ 2 = lim
๐ผโ0csc ๐ผ 3 = csc 0 3
= 0
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BIBLIOGRAFIAS
W. SWOKOWSKI, Earl, Cรกlculo con Geometrรญa Analรญtica, 2da. Ediciรณn,
Panamericana, Colombia, 1989, 1097 pรกgs.
AGUILAR Sรกnchez, Gerardo y CASTRO Pรฉrez, Jaime, โProblemas de
Cรกlculo integralโ 1ra Ediciรณn, Tec de Monterrey, Mรฉxico, DF., Julio
2003, 127 pรกgs.