Download - 5 casos para calcular limites
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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 LβHOSPITAL O REGLA DE LOS 4 PASOS).
VEAMOS ALGUNOS EJEMPLOS PARA ENTENDER ESTE TEMAβ¦
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limπ₯β2
3π₯
limπ₯β2
3π₯ = 3 2 = 6
limπ₯ββ2
π₯2 + 8π₯ β 1
limπ₯ββ2
π₯2 + 8π₯ β 1 = β2 2 + 8 β2 β 1 = 4 β 16 β 1 = β13
limπβ2
8π
limπβ2
8π = 8 2 = 16 = 4
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limπ₯β
23
7π₯ β 10
limπ₯β
23
7π₯ β 10 = 72
3β 10 =
14
3β 10 = β
16
3
limπ₯β0
3π₯2 + 7π₯ β 3
2π₯ β 1
limπ₯β0
3π₯2 + 7π₯ β 3
2π₯ β 1=3 0 2 + 7 0 β 3
2 0 β 1=0 + 0 β 3
0 β 1=β3
β1= 3
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limπ₯β0
4π₯
limπ₯β0
4π₯ = 4 0 = 0
limπ₯β2
π₯
3π₯ β 2
limπ₯β2
π₯
3π₯ β 2=
0
3 0 β 2=
0
0 β 2=
0
β2= 0
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limπ₯β2
π₯2 β 4
π₯ β 2
limπ₯β2
π₯2 β 4
π₯ β 2=0
0= πΌππππ‘ππππππππ
limπ₯β2
π₯2 β 4
π₯ β 2= lim
π₯β2
π₯ + 2 π₯ β 2
π₯ β 2= lim
π₯β2π₯ + 2 = 2 + 2 = 4
limπ₯β3
π₯2 + 4π₯ β 21
π₯ β 3
limπ₯β3
π₯2 + 4π₯ β 21
π₯ β 3=
3 2 + 4 3 β 21
3 β 3=9 + 12 β 21
0=0
0= πΌππππ‘ππππππππ
limπ₯β3
π₯2 + 4π₯ β 21
π₯ β 3= lim
π₯β3
π₯ β 3 π₯ + 7
π₯ β 3= lim
π₯β3π₯ + 7 = 3 + 7 = 10
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limπ₯β1
π₯3 β 1
π₯ β 1
limπ₯β1
π₯3 β 1
π₯ β 1=1 β 1
1 β 1=0
0= πΌππππ‘ππππππππ
limπ₯β1
π₯ β 1 π₯2 + π₯ + 1
π₯ β 1= lim
π₯β1π₯2 + π₯ + 1 = 1 2 + 1 + 1 = 1 + 1 + 1 = 3
limπ₯β0
π₯2 + 9π₯
π₯
limπ₯β0
π₯2 + 9π₯
π₯=02 + 9 0
0=0 + 0
0=0
0= πΌππππ‘ππππππππ
limπ₯β0
π₯2 + 9π₯
π₯= lim
π₯β0
π₯ π₯ + 9
π₯= lim
π₯β0π₯ + 9 = 0 + 9 = 9
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FORMULA DE LβHOPITAL
limπ₯βπ
π π₯
π π₯= lim
π₯βπ
πβ² π₯
πβ² π₯= lim
π₯βπ
πππ₯
π π₯
πππ₯
π π₯
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limπ₯β2
π₯2 β 4
π₯ β 2
limπ₯β2
π₯2 β 4
π₯ β 2=0
0= πΌππππ‘ππππππππ
limπ₯β2
π₯2 β 4
π₯ β 2= lim
π₯β2
πππ₯
π₯2 β 4
πππ₯
π₯ β 2= lim
π₯β2
2π₯
1= 2 2 = 4
limπ₯β2
π₯2 β 4
π₯ β 2
limπ₯β3
π₯2 + 4π₯ β 21
π₯ β 3=
3 2 + 4 3 β 21
3 β 3=9 + 12 β 21
0=0
0= πΌππππ‘ππππππππ
limπ₯β3
π₯2 + 4π₯ β 21
π₯ β 3= lim
π₯β3
πππ₯
π₯2 + 4π₯ β 21
πππ₯
π₯ β 3= lim
π₯β3
2π₯ + 4
1= 2 3 + 4 = 6 + 4 = 10
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limπ₯β1
π₯3 β 1
π₯ β 1
limπ₯β1
π₯3 β 1
π₯ β 1=1 β 1
1 β 1=0
0= πΌππππ‘ππππππππ
limπ₯β1
π₯3 β 1
π₯ β 1= lim
π₯β1
πππ₯
π₯3 β 1
πππ₯
π₯ β 1= lim
π₯β1
3π₯2
1= 3 1 2 = 3 1 = 3
limπ₯β0
π₯2 + 9π₯
π₯
limπ₯β0
π₯2 + 9π₯
π₯=02 + 9 0
0=0 + 0
0=0
0= πΌππππ‘ππππππππ
limπ₯β0
π₯2 + 9π₯
π₯= lim
π₯β0
πππ₯
π₯2 + 9π₯
πππ₯
π₯= lim
π₯β0
2π₯ + 9
1= 2 0 + 9 = 0 + 9 = 9
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limπ₯β2
4 β π₯2
3 β π₯2 + 5
= limπ₯β2
4 β π₯2
3 β π₯2 + 5
3 + π₯2 + 5
3 + π₯2 + 5= lim
π₯β2
4 β π₯2 3 + π₯2 + 5
9 β π₯2 + 5
= limπ₯β2
4 β π₯2 3 + π₯2 + 5
9 β π₯2 β 5= lim
π₯β2
4 β π₯2 3 + π₯2 + 5
4 β π₯2
= limπ₯β2
3 + π₯2 + 5 = 3 + 2 2 + 5 = 3 + 4 + 5 = 3 + 9 = 3 + 3
= 6
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limπ₯β2
2π₯ β 3
2π₯ β 3
= limπ₯β2
2π₯ β 3
2π₯ β 3
2π₯ + 3
2π₯ + 3= lim
π₯β2
2π₯ β 3
2π₯ β 3 2π₯ + 3= lim
π₯β2
1
2π₯ + 3
=1
2(2) + 3=
1
4 + 3
=1
2 + 3
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REGLA DE LOS 4 PASOS
USANDO LA FORMULA SIGUIENTE:
limββ0
π π₯ + β β π π₯
β
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HALLAR EL LIMITE DE LA FUNCION
π¦ = 3π₯2
π π₯ + β = 3 π₯ + β 2 = 3 π₯2 + 2π₯β + β2 = 3π₯2 + 6π₯β + 3β2
π π₯ = 3π₯2
limββ0
π π₯ + β β π π₯
β= lim
ββ0
3π₯2 + 6π₯β + 3β2 β 3π₯2
β= lim
ββ0
6π₯β + 3β2
β
= limββ0
β(6π₯ + 3β)
β= lim
ββ06π₯ + 3β = 6π₯ + 3 0
= 6π₯
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HALLAR EL LIMITE DE LA FUNCION
π π₯ = π₯ + 5
π π₯ + β = π₯ + β + 5
π π₯ = π₯ + 5
limββ0
π π₯ + β β π π₯
β= lim
ββ0
π₯ + β + 5 β π₯ + 5
β
= limββ0
π₯ + β + 5 β π₯ + 5
β
π₯ + β + 5 + π₯ + 5
π₯ + β + 5 + π₯ + 5
= limββ0
π₯ + β + 5 β (π₯ + 5)
β π₯ + β + 5 + π₯ + 5= lim
ββ0
β
β π₯ + β + 5 + π₯ + 5
= limββ0
1
π₯ + β + 5 + π₯ + 5=
1
π₯ + 0 + 5 + π₯ + 5=
1
π₯ + 5 + π₯ + 5
=1
2 π₯ + 5
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limπΌβ0
π ππ πΌ
tanπΌ
limπΌβ0
π ππ πΌ
tanπΌ=π ππ 0
tan 0=0
0= πΌππ·πΈππΈπ ππΌππ΄π·π
limπΌβ0
π ππ πΌ
tanπΌ= lim
πΌβ0
πππ₯
π ππ πΌ
πππ₯
tan πΌ= lim
πΌβ0
cos πΌ
(sec πΌ)2=
cos 0
(sec 0)2=
1
1 2=1
1
= 1
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limπΌβ0
ππ π πΌ
1 β πππ 2 πΌ
limπΌβ0
ππ π πΌ
1 β πππ 2 πΌ=
0
1 β 1=0
0= πΌππ·πΈππΈπ ππΌππ΄π·π
limπΌβ0
ππ π πΌ
1 β πππ 2 πΌ= lim
πΌβ0
ππ π πΌ
π ππ2 πΌ= lim
πΌβ0(csc πΌ) csc πΌ 2 = lim
πΌβ0csc πΌ 3 = csc 0 3
= 0
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BIBLIOGRAFIAS
W. SWOKOWSKI, Earl, CΓ‘lculo con GeometrΓa AnalΓtica, 2da. EdiciΓ³n,
Panamericana, Colombia, 1989, 1097 pΓ‘gs.
AGUILAR SΓ‘nchez, Gerardo y CASTRO PΓ©rez, Jaime, βProblemas de
CΓ‘lculo integralβ 1ra EdiciΓ³n, Tec de Monterrey, MΓ©xico, DF., Julio
2003, 127 pΓ‘gs.