1 Ejercicio
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1 Ejercicio: lim (
π₯β0
πΏπ (π₯) ) =0 πΆππ‘ (π₯)
π(π₯) = πΏπ (π₯) πΒ΄(π₯) =
1 π₯
π(π₯) = πΆππ‘(π₯) πΒ΄(π₯) = βπΆπ π 2 (π₯) 1 1 1 π₯ lim ( ) => ( ) = (β ) β π΄πππππ πππππππππππ ππ ππππππππππ 2 2 π₯β0 βπΆπ π (π₯) βπ₯πΆπ π (π₯) π₯πΆπ π 2 (π₯) lim (1)
β
π₯β0
lim (π₯πΆπ π 2 (π₯)) π₯β0
=β
1 =0 β
2 Ejercicio: π(π₯) = πΒ΄(π₯) =
(5π₯ 2 + 7π₯ + 4)2 π₯2 + 6
((5π₯ 2 + 7π₯ + 4)2 )(π₯ 2 + 6) β (π₯ 2 + 6)(5π₯ 2 + 7π₯ + 4)2 ) (π₯ 2 + 6)2 (5π₯ 2 + 7π₯ + 4)2 = 2(5π₯ 2 + 7π₯ + 4)2 (10π₯ + 7) (π₯ 2 + 6) = 2π₯
πΒ΄(π₯) =
2(5π₯ 2 + 7π₯ + 4)(10π₯ + 7)(π₯ 2 + 6) β 2π₯(5π₯ 2 + 7π₯ + 4) (π₯ 2 + 6)2
2(5π₯ 2 + 7π₯ + 4)(10π₯ + 7)(π₯ 2 + 6) β 2π₯(5π₯ 2 + 7π₯ + 4)2 : 50π₯ 5+70π₯ 4 + 600π₯ 3 + 1204π₯ 2 + 1039π₯ + 336
πΒ΄(π₯) =
50π₯ 5 + 70π₯ 4 + 600π₯ 3 + 1204π₯ 2 + 1039π₯ + 336 (π₯ 2 + 6)2
3 Ejercicio: π(π₯) = (3π₯ 2 + π₯)2 (2π₯ 2 ) π(π₯) = (3π₯ 2 + π₯)2 (2π₯ 2 ) => πΒ΄(π₯) = 2(54π₯ 5 + 30π₯ 4 + 4π₯ 3 ) (3π₯ 2 + π₯)2 = 2(3π₯ 2 + π₯)(6π₯ + 1) (π₯ 2 ) = 2π₯
2(3π₯ 2 + π₯)(6π₯ + 1)π₯ 2 + 2π₯(3π₯ 2 + π₯ 2 )): (54π₯ 5 + 30π₯ 4 + 4π₯ 3 ) πΒ΄(π₯) = 2(54π₯ 5 + 30π₯ 4 + 4π₯ 3 )
4 Ejercicio:
6π₯π¦ + 4π₯π¦ 3 = 4π¦π₯ 6π¦Β΄ + 6π₯π¦Β΄ + 4π¦ 3 + (4π₯)(3π¦ 2 )(π¦Β΄) = 4π¦ + 4π₯π¦Β΄ 6π¦ + 6π₯π¦Β΄ + 4π¦ 3 + 12π₯π¦ 2 π¦Β΄ = 4π¦ + 4π₯π¦Β΄
5 Ejercicio:
PROBLEMA 1
π₯β0
πΏπ (π₯) ) =0 πΆππ‘ (π₯)
π(π₯) = πΏπ (π₯) πΒ΄(π₯) =
1 π₯
π(π₯) = πΆππ‘(π₯) πΒ΄(π₯) = βπΆπ π 2 (π₯) 1 1 1 π₯ lim ( ) => ( ) = (β ) β π΄πππππ πππππππππππ ππ ππππππππππ 2 2 π₯β0 βπΆπ π (π₯) βπ₯πΆπ π (π₯) π₯πΆπ π 2 (π₯) lim (1)
β
π₯β0
lim (π₯πΆπ π 2 (π₯)) π₯β0
=β
1 =0 β
2 Ejercicio: π(π₯) = πΒ΄(π₯) =
(5π₯ 2 + 7π₯ + 4)2 π₯2 + 6
((5π₯ 2 + 7π₯ + 4)2 )(π₯ 2 + 6) β (π₯ 2 + 6)(5π₯ 2 + 7π₯ + 4)2 ) (π₯ 2 + 6)2 (5π₯ 2 + 7π₯ + 4)2 = 2(5π₯ 2 + 7π₯ + 4)2 (10π₯ + 7) (π₯ 2 + 6) = 2π₯
πΒ΄(π₯) =
2(5π₯ 2 + 7π₯ + 4)(10π₯ + 7)(π₯ 2 + 6) β 2π₯(5π₯ 2 + 7π₯ + 4) (π₯ 2 + 6)2
2(5π₯ 2 + 7π₯ + 4)(10π₯ + 7)(π₯ 2 + 6) β 2π₯(5π₯ 2 + 7π₯ + 4)2 : 50π₯ 5+70π₯ 4 + 600π₯ 3 + 1204π₯ 2 + 1039π₯ + 336
πΒ΄(π₯) =
50π₯ 5 + 70π₯ 4 + 600π₯ 3 + 1204π₯ 2 + 1039π₯ + 336 (π₯ 2 + 6)2
3 Ejercicio: π(π₯) = (3π₯ 2 + π₯)2 (2π₯ 2 ) π(π₯) = (3π₯ 2 + π₯)2 (2π₯ 2 ) => πΒ΄(π₯) = 2(54π₯ 5 + 30π₯ 4 + 4π₯ 3 ) (3π₯ 2 + π₯)2 = 2(3π₯ 2 + π₯)(6π₯ + 1) (π₯ 2 ) = 2π₯
2(3π₯ 2 + π₯)(6π₯ + 1)π₯ 2 + 2π₯(3π₯ 2 + π₯ 2 )): (54π₯ 5 + 30π₯ 4 + 4π₯ 3 ) πΒ΄(π₯) = 2(54π₯ 5 + 30π₯ 4 + 4π₯ 3 )
4 Ejercicio:
6π₯π¦ + 4π₯π¦ 3 = 4π¦π₯ 6π¦Β΄ + 6π₯π¦Β΄ + 4π¦ 3 + (4π₯)(3π¦ 2 )(π¦Β΄) = 4π¦ + 4π₯π¦Β΄ 6π¦ + 6π₯π¦Β΄ + 4π¦ 3 + 12π₯π¦ 2 π¦Β΄ = 4π¦ + 4π₯π¦Β΄
5 Ejercicio:
PROBLEMA 1
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