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Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 1 Caro aluno, aprender exige esforΓ§o e perseveranΓ§a. Grandes conquistas exige grandes esforΓ§os. Bons estudos! A resolução dos exercΓcios propostos demonstra a possibilidade de resolução apenas aplicando a substituição do valor Γ qual o limite tende. 1. πππ π₯β2 (3π₯2 β 5π₯ + 2) = Para resolver o limite, vamos substituir o "x" pelo 1. πππ π₯β2 (3 β 22 β 5 β 2 + 2) = 4 Observe o grΓ‘fico e verifique a resolução β΄ πππ πβπ ( ππ π β ππ + π) = π __________________________________________________________________________________________ 2. πππ π₯ββ1 π₯2 + 2π₯ β 3 4π₯ β 3 = πππ π₯ββ1 (β1)2 + 2 β (β1) β 3 4 β (β1)β 3 = β4 β7 = 4 7 β΄ πππ πββπ π π + ππ β π ππ β π = π π __________________________________________________________________________________________ 3. πππ π₯β1 ( 2π₯2 β π₯ + 1 3π₯ β 2 ) 2 = πππ π₯β1 ( 2 β 12 β 1 + 1 3 β 1 β 2 ) 2 = 4 β΄ πππ πβπ ( ππ π β π + π ππ β π ) π = π __________________________________________________________________________________________ 4. πππ π₯ββ2 β π₯3 + 2π₯2 β 3π₯ + 2 π₯2 + 4π₯ + 3 = 3 πππ π₯ββ2 β (β2)3 + 2 β (β2)2 β 3 β (β2) + 2 (β2)2 + 4 β (β2) + 3 = β2 3 β΄ πππ πββπ β π π + ππ π β ππ + π π π + ππ + π = βπ π __________________________________________________________________________________________ 5. πππ π₯β1 4π₯2 β 7π₯ + 5 = πππ π₯β1 4 β 12 β 7 β 1 + 5 = 2 β΄ πππ πβπ ππ π β ππ + π = π __________________________________________________________________________________________ 6. πππ π₯ββ1 π₯3 β 2π₯2 β 4π₯ + 3 = πππ π₯ββ1 (β1)3 β 2 β (β1)2 β 4 β (β1) + 3 = 4 β΄ πππ πββπ π π β ππ π β ππ + π = π
2.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 2 __________________________________________________________________________________________ 7. πππ π₯β2 3π₯ + 2 π₯2 β 6π₯ + 5 = πππ π₯β2 3 β 2 + 2 22 β 6 β 2 + 5 = β 8 3 β΄ πππ πβπ ππ + π π π β ππ + π = β π π _________________________________________________________________________________________ 8. πππ π₯ββ1 π₯2 β 5π₯ + 4 2π₯ + 1 = πππ π₯ββ1 3 β (β1)2 β 5 β (β1)+ 4 2 β (β1)+ 1 = 12 β΄ πππ πββπ π π β ππ + π ππ + π = ππ __________________________________________________________________________________________ 9. πππ π₯ββ3 π₯2 + 2π₯ β 3 5 β 3π₯ = πππ π₯ββ3 (β3)2 + 2 β (β3) β 3 5 β 3 β (β3) = 0 14 = 0 β΄ πππ πββπ π π + ππ β π π β ππ = π __________________________________________________________________________________________ 10. πππ π₯β2 ( 3π₯2 β 2π₯ β 5 βπ₯2 + 3π₯ + 4 ) 3 = πππ π₯β2 ( 3 β 22 β 2 β 2 β 5 β22 + 3 β 2 + 4 ) 3 = ( 3 6 ) 3 = ( 1 2 ) 3 = 1 8 β΄ πππ πβπ ( ππ π β ππ β π βπ π + ππ + π ) π = π π __________________________________________________________________________________________ 11. πππ π₯β4 ( π₯3 β 3π₯2 β 2π₯ β 5 2π₯2 β 9π₯ + 2 ) 2 = πππ π₯β4 ( 43 β 3 β 42 β 2 β 4 β 5 2 β 42 β 9 β 4 + 2 ) 2 = ( 3 2 ) 2 = 9 4 β΄ πππ πβπ ( π π β ππ π β ππ β π ππ π β ππ + π ) π = π π __________________________________________________________________________________________ 12. πππ π₯ββ1 β 2π₯2 + 3π₯ β 4 5π₯ β 4 = πππ π₯ββ1 β 2 β (β1)2 + 3 β (β1) β 4 5 β (β1) β 4 = β β5 β9 = β5 3 β΄ πππ πββπ β ππ π + ππ β π ππ β π = β π π __________________________________________________________________________________________ 13. lim π₯ββ2 β 3π₯3 β 5π₯2 β π₯ + 2 4π₯ + 3 3 = lim π₯ββ2 β 3 β (β2)3 β 5 β (β2)2 β (β2) + 2 4(β2)+ 3 3 = β β40 β5 3 = β8 3 = 2
3.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 3 β΄ πππ πββπ β ππ π β ππ π β π + π ππ + π π = π __________________________________________________________________________________________ 14. πππ π₯β2 β2π₯2 + 3π₯ + 2 6 β 4π₯ = πππ π₯β2 β2 β 22 + 3 β 2 + 2 6 β 4 β 2 = 4 β2 = β2 β΄ πππ πβπ βππ π + ππ + π π β ππ = βπ __________________________________________________________________________________________ 15. lim π₯β2 π₯2 β 4 π₯2 β 2π₯ = lim π₯β2 22 β 4 22 β 2 β 2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador. π₯2 β 4 = ( π₯ + 2)( π₯ β 2) π π₯2 β 2π₯ = π₯(π₯ β 2) , assim temos: πππ π₯β2 π₯2 β 4 π₯2 β 2π₯ = πππ π₯β2 ( π₯ + 2)(π₯ β 2) π₯(π₯ β 2) = πππ π₯β2 ( π₯ + 2) π₯ = πππ π₯β2 (2 + 2) 2 = 2 β΄ πππ πβπ π π β π π π β ππ = π __________________________________________________________________________________________ 16. πππ π₯β1 π₯2 β 1 π₯ β 1 = πππ π₯β1 12 β 1 1 β 1 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador. π₯2 β 1 = ( π₯ + 1)( π₯ β 1) , ππ π ππ π‘ππππ : πππ π₯β1 π₯2 β 1 π₯ β 1 = πππ π₯β1 ( π₯ + 1)(π₯ β 1) (π₯ β 1) = πππ π₯β1 (π₯ + 1) = 2 β΄ πππ πβπ π π β π π β π = π __________________________________________________________________________________________ 17. πππ π₯ββ2 4 β π₯2 2 + π₯ = πππ π₯ββ2 4 β (β2)2 2 + (β2) = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador. 4 β π₯2 = (2 + π₯)(2 β π₯)
4.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 4 πππ π₯ββ2 4 β π₯2 2 + π₯ = πππ π₯ββ2 (2 + π₯)(2β π₯) (2 + π₯) = πππ π₯ββ2 (2 + 2) = 4 β΄ πππ πββπ π β π π π + π = π __________________________________________________________________________________________ 18. πππ π₯β 3 2 4π₯2 β 9 2π₯ β 3 = πππ π₯β 3 2 4( 3 2 ) 2 β 9 2 ( 3 2 ) β 3 = πππ π₯β 3 2 4 β 9 4 β 9 2 β 3 2 β 3 = πππ π₯β 3 2 9 β 9 3 β 3 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador. 4π₯2 β 9 = (2π₯ + 3)(2π₯ β 3) , ππ π ππ π‘ππππ : πππ π₯β 3 2 4π₯2 β 9 2π₯ β 3 = πππ π₯β 3 2 (2π₯ + 3)(2π₯ β 3) (2π₯ β 3) = πππ π₯β 3 2 (2 β 3 2 + 3) = 6 β΄ πππ πβ π π ππ π β π ππ β π = π __________________________________________________________________________________________ 19. lim π₯β3 π₯2 β 4π₯ + 3 π₯2 β π₯ β 6 = lim π₯β3 32 β 4 β 3 + 3 32 β 3 β 6 = 0 0 ( π‘ππππ π’ππ ππππππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador. π₯2 β 4π₯ + 3 = ( π₯ β 3)( π₯ β 1) π π₯2 β π₯ β 6 = (π₯ β 3)(π₯ + 2) , ππ π ππ π‘ππππ : πππ π₯β3 π₯2 β 4π₯ + 3 π₯2 β π₯ β 6 = πππ π₯β3 ( π₯ β 3)( π₯ β 1) (π₯ β 3)(π₯ + 2) = πππ π₯β3 ( π₯ β 1) (π₯ + 2) = = πππ π₯β3 (3 β 1) (3 + 2) = 2 5 β΄ πππ πβπ π π β ππ + π π π β π β π = π π __________________________________________________________________________________________ 20. πππ π₯β 1 2 2π₯2 + 5π₯ β 3 2π₯2 β 5π₯ + 2 = πππ π₯β 1 2 2 β ( 1 2) 2 + 5 β 1 2 β 3 2 β ( 1 2 ) 2 β 5 β 1 2 + 2 = πππ π₯β 1 2 2 β 1 4 + 5 β 1 2 β 3 2 β 1 4 β 5 β 1 2 + 2 = = πππ π₯β 1 2 1 2 + 5 2 β 3 1 2 β 5 2 + 2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador.
5.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 5 2π₯2 + 5π₯ β 3 = ( π₯ + 3)(π₯ β 1 2 ) π 2π₯2 β 5π₯ + 2 = ( π₯ β 2)(π₯ β 1 2 ) , πππ‘Γ£π π‘ππππ : πππ π₯β 1 2 2π₯2 + 5π₯ β 3 2π₯2 β 5π₯ + 2 = πππ π₯β 1 2 ( π₯ + 3) (π₯ β 1 2 ) ( π₯ β 2) (π₯ β 1 2 ) = πππ π₯β 1 2 ( π₯ + 3) ( π₯ β 2) = πππ π₯β 1 2 ( 1 2 + 3) ( 1 2 β 2) = = πππ π₯β 1 2 ( 1 2 + 3) ( 1 2 β 2) = πππ π₯β 1 2 7 2 β 3 2 = β 14 6 = β 7 3 β΄ πππ π β π π ππ π + ππ β π ππ π β ππ + π = β π π __________________________________________________________________________________________ 21. lim π₯ββ 3 2 6π₯2 + 11π₯ + 3 2π₯2 β 5π₯ β 12 = lim π₯ββ 3 2 6 β (β 3 2 ) 2 + 11 β (β 3 2 ) + 3 2 β (β 3 2 ) 2 β 5 β (β 3 2 ) β 12 = lim π₯ββ 3 2 6 β 9 4 β 33 2 + 3 2 β 9 4 + 15 2 β 12 = = lim π₯ββ 3 2 27 2 β 33 2 + 3 9 2 + 15 2 β 12 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador. 6π₯2 + 11π₯ + 3 = (3π₯ + 1)(2π₯ + 3) π 2π₯2 β 5π₯ β 12 = ( π₯ β 4)(2π₯ + 3), ππ π ππ π‘ππππ : πππ π₯ββ 3 2 6π₯2 + 11π₯ + 3 2π₯2 β 5π₯ β 12 = πππ π₯ββ 3 2 (3π₯ + 1)(2π₯ + 3) ( π₯ β 4)(2π₯ + 3) = πππ π₯ββ 3 2 (3π₯ + 1) ( π₯ β 4) = = πππ π₯ββ 3 2 3 β (β 3 2 ) + 1 β 3 2 β 4 = πππ π₯ββ 3 2 β 9 2 + 1 β 3 2 β 4 = β 7 2 β 11 2 = 14 22 = 7 11 β΄ πππ πββ π π ππ π + πππ + π ππ π β ππ β ππ = π ππ __________________________________________________________________________________________ 22. πππ π₯β1 π₯3 β 1 π₯2 β 1 = πππ π₯β1 13 β 1 12 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador. π₯3 β 1 = ( π₯ β 1)( π₯2 + π₯ + 1) π π₯2 β 1 = ( π₯ + 1)( π₯ β 1), ππ π ππ π‘ππππ : πππ π₯β1 π₯3 β 1 π₯2 β 1 = πππ π₯β1 ( π₯ β 1)( π₯2 + π₯ + 1) ( π₯ + 1)( π₯ β 1) = πππ π₯β1 ( π₯2 + π₯ + 1) ( π₯ + 1) = = πππ π₯β1 12 + 1 + 1 1 + 1 = 3 2 β΄ πππ πβπ π π β π π π β π = π π
6.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 6 __________________________________________________________________________________________ 23. lim π₯ββ2 8 + π₯3 4 β π₯2 = lim π₯ββ2 8 + (β2)3 4 β (β2)2 = lim π₯ββ2 8 β 8 4 β 4 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador. 8 + π₯3 = (2 + π₯)(4 β 2π₯ + π₯2) π 4 β π₯2 = (2 + π₯)(2 β π₯) , ππ π ππ π‘ππππ : lim π₯ββ2 8 + π₯3 4 β π₯2 = lim π₯ββ2 (2 + π₯)(4β 2π₯ + π₯2 ) (2 + π₯)(2 β π₯) = lim π₯ββ2 (4 β 2π₯ + π₯2 ) (2 β π₯) = = πππ π₯ββ2 4 β 2 β (β2) + (β2)2 2 β (β2) = 3 β΄ πππ πββπ π + π π π β π π = π __________________________________________________________________________________________ 24. πππ π₯β2 π₯4 β 16 8 β π₯3 = πππ π₯β2 24 β 16 8 β 23 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador. π₯4 β 16 = ( π₯2 + 4)( π₯ + 2)( π₯ β 2) π 8 β π₯3 = (2 β π₯)(4 + 2π₯ + π₯2) , ππ π ππ π‘ππππ : πππ π₯β2 π₯4 β 16 8 β π₯3 = πππ π₯β2 ( π₯2 + 4)( π₯ + 2)( π₯ β 2) β( π₯ β 2)(4 + 2π₯ + π₯2) = πππ π₯β2 β ( π₯2 + 4)( π₯ + 2) (4 + 2π₯ + π₯2) = = πππ π₯β2 β (22 + 4)(2 + 2) (4 + 2 β 2 + 2) = β 32 12 = β 8 3 β΄ πππ πβπ π π β ππ π β π π = β π π __________________________________________________________________________________________ 25. πππ π₯β1 π₯2 β 3π₯ + 2 π₯ β 1 = πππ π₯β1 12 β 3 β 1 + 2 1 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador. π₯2 β 3π₯ + 2 = ( π₯ β 2)(π₯ β 1), assim temos: πππ π₯β1 π₯2 β 3π₯ + 2 π₯ β 1 = πππ π₯β1 ( π₯ β 2)(π₯ β 1) (π₯ β 1) = πππ π₯β1 1 β 2 = β1 β΄ πππ πβπ π π β ππ + π π β π = βπ __________________________________________________________________________________________
7.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 7 26. πππ π₯β1 2π₯3 + π₯2 β 4π₯ + 1 π₯3 β 3π₯2 + 5π₯ β 3 = πππ π₯β1 2 β (1)3 + 12 β 4 β 1 + 1 13 β 3 β (1)2 + 5 β 1 β 3 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. Veja a demonstração no site em Aulas Slides - Fatoração e Produtos NotΓ‘veis. 2π₯3 + π₯2 β 4π₯ + 1 = ( π₯ β 1)(2π₯2 + 3π₯ β 1) π₯3 β 3π₯2 + 5π₯ β 3 = ( π₯ β 1)( π₯2 β 2π₯ + 3), ππ π ππ π‘ππππ : πππ π₯β1 2π₯3 + π₯2 β 4π₯ + 1 π₯3 β 3π₯2 + 5π₯ β 3 = πππ π₯β1 ( π₯ β 1)(2π₯2 + 3π₯ β 1) ( π₯ β 1)( π₯2 β 2π₯ + 3) = πππ π₯β1 (2π₯2 + 3π₯ β 1) ( π₯2 β 2π₯ + 3) = = πππ π₯β1 2 β (1)2 + 3 β (1) β 1 (1)2 β 2 β (1)+ 3 = 4 2 = 2 β΄ πππ πβπ ππ π + π π β ππ + π π π β ππ π + ππ β π = π __________________________________________________________________________________________ 27. πππ π₯ββ1 π₯3 + 3π₯2 β π₯ β 3 π₯3 β π₯2 + 2 = πππ π₯ββ1 (β1)3 + 3(β1)2 β (β1)β 3 (β1)β (β1)2 + 2 = = πππ π₯ββ1 (β1)3 + 3(β1)2 β (β1) β 3 (β1) β (β1)2 + 2 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯3 + 3π₯2 β π₯ β 3 = ( π₯ + 1)(π₯2 + 2π₯ β 3) π₯3 β π₯2 + 2 = ( π₯ + 1)(π₯2 β 2π₯ + 2), assim temos: πππ π₯ββ1 π₯3 + 3π₯2 β π₯ β 3 π₯3 β π₯2 + 2 = πππ π₯ββ1 ( π₯ + 1)(π₯2 + 2π₯ β 3) ( π₯ + 1)(π₯2 β 2π₯ + 2) = πππ π₯ββ1 π₯2 + 2π₯ β 3 π₯2 β 2π₯ + 2 = = πππ π₯ββ1 (β1)2 + 2 β (β1)β 3 (β1)2 β 2 β (β1)+ 2 = β 4 5 β΄ πππ πββπ π π + ππ π β π β π π π β π π + π = β π π __________________________________________________________________________________________
8.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 8 28. πππ π₯β3 π₯3 β 6π₯ β 9 π₯3 β 8π₯ β 3 = πππ π₯β3 33 β 6 β 3 β 9 33 β 8 β 3 β 3 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯3 β 6π₯ β 9 = ( π₯ β 3)(π₯2 + 3π₯ + 3) π₯3 β 8π₯ β 3 = ( π₯ β 3)(π₯2 + 3π₯ + 1) , assim temos: πππ π₯β3 π₯3 β 6π₯ β 9 π₯3 β 8π₯ β 3 = πππ π₯β3 ( π₯ β 3)(π₯2 + 3π₯ + 3) ( π₯ β 3)(π₯2 + 3π₯ + 1) = πππ π₯β3 π₯2 + 3π₯ + 3 π₯2 + 3π₯ + 1 = = πππ π₯β3 32 + 3 β 3 + 3 32 + 3 β 3 + 1 = 21 19 β΄ πππ πβπ π π β ππ β π π π β ππ β π = ππ ππ __________________________________________________________________________________________ 29. πππ π₯β1 π₯3 β 3π₯2 + 6π₯ β 4 π₯3 β 4π₯2 + 8π₯ β 5 = πππ π₯β1 13 β 3 β 12 + 6 β 1 β 4 13 β 4 β 12 + 8 β 1 β 5 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯3 β 3π₯2 + 6π₯ β 4 = ( π₯ β 1)(π₯2 β 2π₯ + 4) π₯3 β 4π₯2 + 8π₯ β 5 = ( π₯ β 1)(π₯2 β 3π₯ + 5) πππ π₯β1 π₯3 β 3π₯2 + 6π₯ β 4 π₯3 β 4π₯2 + 8π₯ β 5 = πππ π₯β1 ( π₯ β 1)(π₯2 β 2π₯ + 4) ( π₯ β 1)(π₯2 β 3π₯ + 5) = πππ π₯β1 (π₯2 β 2π₯ + 4) (π₯2 β 3π₯ + 5) = = πππ π₯β1 12 β 2 β 1 + 4 12 β 3 β 1 + 5 = 3 3 = 1 β΄ πππ πβπ π π β ππ π + ππ β π π π β ππ π + ππ β π = π __________________________________________________________________________________________ 30. πππ π₯β2 π₯4 β 10π₯ + 4 π₯3 β 2π₯2 = πππ π₯β2 24 β 10 β 2 + 4 23 β 2 β 22 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯4 β 10π₯ + 4 = ( π₯ β 2)(π₯3 + 2π₯2 + 4π₯ β 2) π₯3 β 2π₯2 = ( π₯ β 2) π₯2 , assim temos:
9.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 9 πππ π₯β2 π₯4 β 10π₯ + 4 π₯3 β 2π₯2 = πππ π₯β2 ( π₯ β 2)(π₯3 + 2π₯2 + 4π₯ β 2) ( π₯ β 2) π₯2 = πππ π₯β2 π₯3 + 2π₯2 + 4π₯ β 2 π₯2 = = πππ π₯β2 23 + 2 β 22 + 4 β 2 β 2 22 = 22 4 = 11 2 β΄ πππ πβπ π π β πππ + π π π β ππ π = ππ π __________________________________________________________________________________________ 31. πππ π₯β1 3π₯3 β 4π₯2 β π₯ + 2 2π₯3 β 3π₯2 + 1 = πππ π₯β1 3 β (1)3 β 4 β (1)2 β 1 + 2 2 β (1)2 β 3 β (1)2 + 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. 3π₯3 β 4π₯2 β π₯ + 2 = ( π₯ β 1)(3π₯2 β π₯ β 2) 2π₯3 β 3π₯2 + 1 = ( π₯ β 1)(2π₯2 β π₯ β 1), ππ π ππ π‘ππππ : πππ π₯β1 3π₯3 β 4π₯2 β π₯ + 2 2π₯3 β 3π₯2 + 1 = πππ π₯β1 ( π₯ β 1)(3π₯2 β π₯ β 2) ( π₯ β 1)(2π₯2 β π₯ β 1) = πππ π₯β1 3π₯2 β π₯ β 2 2π₯2 β π₯ β 1 = = πππ π₯β1 3 β (1)2 β 1 β 2 2 β (1)2 β 1 β 1 = 0 0 (ππππ‘πππ’ππππ πππ π πππππ‘ππππππçãπ) Vamos fazer novamente a fatoração utilizando Bhaskara. 3π₯2 β π₯ β 2 = ( π₯ β 1)(3π₯ + 2) π 2π₯2 β π₯ β 1 = ( π₯ β 1)(2π₯ + 1), ππ π ππ π‘ππππ : πππ π₯β1 3π₯2 β π₯ β 2 2π₯2 β π₯ β 1 = πππ π₯β1 ( π₯ β 1)(3π₯ + 2) ( π₯ β 1)(2π₯ + 1) = πππ π₯β1 3π₯ + 2 2π₯ + 1 = = πππ π₯β1 3 β 1 + 2 2 β 1 + 1 = 5 3 β΄ πππ πβπ ππ π β ππ π β π + π ππ π β ππ π + π = π π __________________________________________________________________________________________ 32. πππ π₯β1 π₯3 β 3π₯ + 2 π₯4 β 4π₯ + 3 = πππ π₯β1 (1)3 β 3 β (1) + 2 (1)4 β 4 β (1) + 3 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯3 β 3π₯ + 2 = ( π₯ β 1)( π₯2 + π₯ β 2) π π₯4 β 4π₯ + 3 = ( π₯ β 1)( π₯3 + π₯2 + π₯ β 3), ππ π ππ π‘ππππ : πππ π₯β1 π₯3 β 3π₯ + 2 π₯4 β 4π₯ + 3 = πππ π₯β1 ( π₯ β 1)( π₯2 + π₯ β 2) ( π₯ β 1)( π₯3 + π₯2 + π₯ β 3) = πππ π₯β1 π₯2 + π₯ β 2 π₯3 + π₯2 + π₯ β 3 = = πππ π₯β1 (1)2 + 1 β 2 (1)3 + (1)2 + 1 β 3 = 0 0 (ππππ‘πππ’ππππ πππ π πππππ‘ππππππçãπ)
10.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 10 Vamos fazer novamente a fatoração utilizando Bhaskara. π₯2 + π₯ β 2 = ( π₯ β 1)( π₯ + 2) π π₯3 + π₯2 + π₯ β 3 = ( π₯ β 1)( π₯2 + 2π₯ + 3), ππ π ππ π‘ππππ : πππ π₯β1 π₯2 + π₯ β 2 π₯3 + π₯2 + π₯ β 3 = πππ π₯β1 ( π₯ β 1)( π₯ + 2) ( π₯ β 1)( π₯2 + 2π₯ + 3) = πππ π₯β1 π₯ + 2 π₯2 + 2π₯ + 3 = = πππ π₯β1 1 + 2 12 + 2 β 1 + 3 = 3 6 = 1 2 β΄ πππ πβπ π π β ππ + π π π β ππ + π = π π __________________________________________________________________________________________ 33. πππ π₯ββ2 π₯4 + 4π₯3 + π₯2 β 12π₯ β 12 2π₯3 + 7π₯2 + 4π₯ β 4 = πππ π₯ββ2 (β2)4 + 4 β (β2)3 + (β2)2 β 12 β (β2)β 12 2 β (β2)3 + 7 β (β2)2 + 4 β (β2) β 4 = = πππ π₯ββ2 16 β 32 + 4 + 24 β 12 β16 + 28 β 8 β 4 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯4 + 4π₯3 + π₯2 β 12π₯ β 12 = ( π₯ + 2)(π₯3 + 2π₯2 β 3π₯ β 6) 2π₯3 + 7π₯2 + 4π₯ β 4 = ( π₯ + 2)(2π₯2 + 3π₯ β 2), ππ π ππ π‘ππππ : πππ π₯ββ2 π₯4 + 4π₯3 + π₯2 β 12π₯ β 12 2π₯3 + 7π₯2 + 4π₯ β 4 = πππ π₯ββ2 ( π₯ + 2)(π₯3 + 2π₯2 β 3π₯ β 6) ( π₯ + 2)(2π₯2 + 3π₯ β 2) = πππ π₯ββ2 (π₯3 + 2π₯2 β 3π₯ β 6) (2π₯2 + 3π₯ β 2) = = πππ π₯ββ2 (β2)3 + 2 β (β2)2 β 3 β (β2) β 6 2 β (β2)2 + 3 β (β2)β 2 = 0 0 (ππππ‘πππ’ππππ πππ π πππππ‘ππππππçãπ) Vamos fazer novamente a fatoração utilizando o Dispositivo de Briot-Ruffini e Bhaskara. π₯3 + 2π₯2 β 3π₯ β 6 = ( π₯ + 2)(π₯2 β 3) 2π₯2 + 3π₯ β 2 = ( π₯ + 2)(2π₯ β 1) , assim temos: πππ π₯ββ2 (π₯3 + 2π₯2 β 3π₯ β 6) (2π₯2 + 3π₯ β 2) = πππ π₯ββ2 ( π₯ + 2)(π₯2 β 3) ( π₯ + 2)(2π₯ β 1) = πππ π₯ββ2 (π₯2 β 3) (2π₯ β 1) = = πππ π₯ββ2 π₯2 β 3 2π₯ β 1 = πππ π₯ββ2 (β2)2 β 3 2 β (β2) β 1 = β 1 5 β΄ πππ πββπ π π + ππ π + π π β πππ β ππ ππ π + ππ π + ππ β π = β π π __________________________________________________________________________________________ 34. πππ π₯ββ1 π₯4 β π₯3 β π₯2 + 5π₯ + 4 π₯3 + 4π₯2 + 5π₯ + 2 = πππ π₯ββ1 (β1)4 β (β1)3 β (β1)2 + 5 β (β1) + 4 (β1)3 + 4 β (β1)2 + 5 β (β1) + 2 =
11.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 11 πππ π₯ββ1 1 + 1 β 1 β 5 + 4 β1 + 4 β 5 + 2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. Vamos efetuar as fatoraçáes e fazer a verificação, onde pode ser notado a necessidade de efetuar novamente a fatoração. Numerador: π₯4 β π₯3 β π₯2 + 5π₯ + 4 = ( π₯ + 1)(π₯3 β 2π₯2 + π₯ + 4), se substituirmos x= - 1, notaremos que o polinΓ΄mio terΓ‘ como valor zero, assim hΓ‘ a necessidade de nova fatoração. π₯3 β 2π₯2 + π₯ + 4 = ( π₯ + 1)(π₯2 β 3π₯ + 4) , se substituirmos x= - 1, notaremos que o polinΓ΄mio terΓ‘ como solução um nΓΊmero diferente de zero. Sendo assim nosso polinΓ΄mio fatorado resulta: π₯4 β π₯3 β π₯2 + 5π₯ + 4 = ( π + π)(π + π)(π π β ππ + π) Denominador: (repetir o processo) π₯3 + 4π₯2 + 5π₯ + 2 = ( π + π)( π+ π)( π + π) Como chegamos Γ simplificação final, vamos resolver o limite. πππ π₯ββ1 π₯4 β π₯3 β π₯2 + 5π₯ + 4 π₯3 + 4π₯2 + 5π₯ + 2 = πππ π₯ββ1 ( π₯ + 1)(π₯ + 1)(π₯2 β 3π₯ + 4) ( π₯ + 1)( π₯ + 1)( π₯ + 2) = = πππ π₯ββ1 (π₯2 β 3π₯ + 4) ( π₯ + 2) = πππ π₯ββ1 (β1)2 β 3(β1) + 4 β1 + 2 = 8 1 β΄ πππ πββπ π π β π π β π π + ππ + π π π + ππ π + ππ + π = π Caro aluno, quero deixar claro que existe outras formas de resolução, cabe a vocΓͺ efetuar pesquisas para melhorar seu entendimento. 35. πππ π₯ββ2 π₯4 + 2π₯3 β 5π₯2 β 12π₯ β 4 2π₯4 + 7π₯3 + 2π₯2 β 12π₯ β 8 = πππ π₯ββ2 (β2)4 + 2 β (β2)3 β 5(β2)2 β 12 β (β2)β 4 2 β (β2)4 + 7 β (β2)3 + 2 β (β2)2 β 12 β (β2) β 8 = = πππ π₯ββ2 16 β 16 β 20 + 24 β 4 32 β 56 + 8 + 24 β 8 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ)
12.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 12 Para resolver este limite temos que fatorar o numerador e o denominador, utilizaremos o Dispositivo de Briot- Ruffini. π₯4 + 2π₯3 β 5π₯2 β 12π₯ β 4 = ( π₯ + 2)( π₯ + 2)(π₯2 β 2π₯ β 1) 2π₯4 + 7π₯3 + 2π₯2 β 12π₯ β 8 = ( π₯ + 2)( π₯ + 2)(2π₯2 β π₯ β 2) πππ π₯ββ2 π₯4 + 2π₯3 β 5π₯2 β 12π₯ β 4 2π₯4 + 7π₯3 + 2π₯2 β 12π₯ β 8 = πππ π₯ββ2 ( π₯ + 2)( π₯ + 2)(π₯2 β 2π₯ β 1) ( π₯ + 2)( π₯ + 2)(2π₯2 β π₯ β 2) = πππ π₯ββ2 (π₯2 β 2π₯ β 1) (2π₯2 β π₯ β 2) = = πππ π₯ββ2 (β2)2 β 2(β2) β 1 2(β2)2 β (β2) β 2 = 7 8 β΄ πππ πββπ π π + ππ π β ππ π β πππ β π ππ π + ππ π + ππ π β πππ β π = π π ______________________________________________________________________________________ Digite a equação aqui.____ 36. πππ π₯β3 β1 + π₯ β 2 π₯ β 3 = πππ π₯β3 β1 + 3 β 2 3 β 3 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β3 β1 + π₯ β 2 π₯ β 3 = πππ π₯β3 (β1+ π₯ β 2)(β1+ π₯ + 2) ( π₯ β 3)(β1 + π₯ + 2) = πππ π₯β3 (β1 + π₯)2 β 22 ( π₯ β 3)(β1+ π₯ + 2) = = πππ π₯β3 1 + π₯ β 4 ( π₯ β 3)(β1 + π₯ + 2) = πππ π₯β3 (π₯ β 3) ( π₯ β 3)(β1 + π₯ + 2) = πππ π₯β3 1 (β1+ π₯ + 2) = = πππ π₯β3 1 (β1+ 3 + 2) = 1 4 β΄ πππ πβπ β π + π β π π β π = π π __________________________________________________________________________________________ 37. πππ π₯β1 β π₯ β 1 π₯ β 1 = πππ π₯β1 β1 β 1 1 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 β π₯ β 1 π₯ β 1 = πππ π₯β1 (β π₯ β 1)(β π₯ + 1) (π₯ β 1)(β π₯ + 1) = πππ π₯β1 (β π₯) 2 β 12 (π₯ β 1)(β π₯ + 1) = πππ π₯β1 (π₯ β 1) (π₯ β 1)(β π₯ + 1) = πππ π₯β1 1 (β π₯ + 1) = 1 2 β΄ πππ πβπ β π β π π β π = π π 38. πππ π₯β0 1 β β1 β π₯ π₯ = πππ π₯β0 1 β β1 β 0 0 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) = πππ π₯β0 1 β β1 β π₯ π₯ = πππ π₯β0 (1 β β1 β π₯)(1 + β1 β π₯) π₯(1 + β1 β π₯) = πππ π₯β0 12 β (β1+ π₯) 2 π₯(1 + β1 + π₯) =
13.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 13 = πππ π₯β0 1 β 1 + π₯ π₯(1 + β1 + π₯) = = πππ π₯β0 π₯ π₯(1 + β1 + π₯) = πππ π₯β0 1 (1 + β1 + π₯) = = πππ π₯β0 1 (1 + β1 + 0) = 1 2 β΄ πππ πβπ π β β π β π π = π π __________________________________________________________________________________________ 39. πππ π₯β1 β π₯ + 3 β 2 π₯ β 1 = πππ π₯β1 β1 + 3 β 2 1 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 β π₯ + 3 β 2 π₯ β 1 = πππ π₯β1 (β π₯ + 3 β 2)(β π₯ + 3 + 2) (π₯ β 1)(β π₯ + 3 + 2) = πππ π₯β1 (β π₯ + 3) 2 β 22 (π₯ β 1)(β π₯ + 3 + 2) = = πππ π₯β1 π₯ + 3 β 4 (π₯ β 1)(β π₯ + 3 + 2) = πππ π₯β1 (π₯ β 1) (π₯ β 1)(β π₯ + 3 + 2) = πππ π₯β1 1 (β π₯ + 3 + 2) = = πππ π₯β1 1 (β1 + 3 + 2) = 1 4 β΄ πππ πβπ β π + π β π π β π = π π __________________________________________________________________________________________ 40. πππ π₯β0 β1 β 2π₯ β π₯2 β 1 π₯ = πππ π₯β0 β1 β 2 β 0 β 02 β 1 0 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β0 β1 β 2π₯ β π₯2 β 1 π₯ = πππ π₯β0 (β1 β 2π₯ β π₯2 β 1)(β1β 2π₯ β π₯2 + 1) π₯(β1β 2π₯ β π₯2 + 1) = πππ π₯β0 (β1 β 2π₯ β π₯2) 2 β 12 π₯(β1β 2π₯ β π₯2 + 1) = = πππ π₯β0 1 β 2π₯ β π₯2 β 1 π₯(β1 β 2π₯ β π₯2 + 1) = πππ π₯β0 β2π₯ β π₯2 π₯(β1β 2π₯ β π₯2 + 1) = πππ π₯β0 π₯(β2 β π₯) π₯(β1β 2π₯ β π₯2 + 1) = = πππ π₯β0 (β2 β π₯) (β1 β 2π₯ β π₯2 + 1) = πππ π₯β0 (β2 β 0) (β1β 2 β 0 β 0 + 1) = β 2 2 = β1 β΄ πππ πβπ βπ β ππ β π π β π π = β π __________________________________________________________________________________________ 41. πππ π₯β0 β1 + π₯ β β1 β π₯ π₯ = πππ π₯β0 β1 + 0 β β1 β 0 0 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β0 β1 + π₯ β β1 β π₯ π₯ = πππ π₯β0 (β1+ π₯ β β1 β π₯)(β1 + π₯ + β1 β π₯) π₯(β1+ π₯ + β1 β π₯) = πππ π₯β0 (β1 + π₯) 2 β (β1 β π₯) 2 π₯(β1+ π₯ + β1 β π₯) =
14.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 14 = πππ π₯β0 1 + π₯ β (1 β π₯) π₯(β1 + π₯ + β1β π₯) = πππ π₯β0 1 + π₯ β 1 + π₯ π₯(β1+ π₯ + β1 β π₯) = πππ π₯β0 2π₯ π₯(β1 + π₯ + β1 β π₯) = = πππ π₯β0 2 β π₯ π₯(β1 + π₯ + β1β π₯) = πππ π₯β0 2 (β1+ 0 + β1 β 0) = 2 2 = 1 β΄ πππ πβπ β π + π β β π β π π = π __________________________________________________________________________________________ 42. πππ π₯β1 β2π₯ β β π₯ + 1 π₯ β 1 = πππ π₯β1 β2 β 1 β β1 + 1 1 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 β2π₯ β β π₯ + 1 π₯ β 1 = πππ π₯β1 (β2π₯ β β π₯ + 1)(β2π₯ + β π₯ + 1) (π₯ β 1)(β2π₯ + β π₯ + 1) = πππ π₯β1 (β2π₯) 2 β (β π₯ + 1) 2 (π₯ β 1)(β2π₯ + β π₯ + 1) = = πππ π₯β1 2π₯ β (π₯ + 1) (π₯ β 1)(β2π₯ + β π₯ + 1) = πππ π₯β1 2π₯ β π₯ β 1 (π₯ β 1)(β2π₯ + β π₯ + 1) = πππ π₯β1 (π₯ β 1) (π₯ β 1)(β2π₯ + β π₯ + 1) = = πππ π₯β1 1 (β2 β 1 + β1 + 1) = 1 2β2 = β2 4 β΄ πππ πβπ β ππ β β π + π π β π = β π π __________________________________________________________________________________________ 43. πππ π₯β1 3 β β10 β π₯ π₯2 β 1 = πππ π₯β1 3 β β10 β 1 12 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 3 β β10 β π₯ π₯2 β 1 = πππ π₯β1 (3 β β10 β π₯)(3 + β10β π₯) ( π₯ + 1)(π₯ β 1)(3+ β10 β π₯) = πππ π₯β1 (3)2 β (β10 β π₯) 2 ( π₯ + 1)(π₯ β 1)(3+ β10 β π₯) = = πππ π₯β1 9 β 10 + π₯ ( π₯ + 1)(π₯ β 1)(3 + β10 β π₯) = πππ π₯β1 (β1 + π₯) ( π₯ + 1)(π₯ β 1)(3+ β10β π₯) = πππ π₯β1 1 ( π₯ + 1)(3+ β10β π₯) = = πππ π₯β1 1 (1 + 1)(3 + β10β 1) = 1 12 β΄ πππ πβπ π β β ππ β π π π β π = π ππ __________________________________________________________________________________________ 44. πππ π₯β3 2 β β π₯ + 1 π₯2 β 9 = πππ π₯β3 2 β β3 + 1 32 β 9 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β3 2 β β π₯ + 1 π₯2 β 9 = πππ π₯β3 (2 β β π₯ + 1)(2+ β π₯ + 1) ( π₯ + 3)(π₯ β 3)(2+ β π₯ + 1) = πππ π₯β3 (2)2 β (β π₯ + 1) 2 ( π₯ + 3)(π₯ β 3)(2 + β π₯ + 1) =
15.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 15 = πππ π₯β3 4 β π₯ β 1 ( π₯ + 3)(π₯ β 3)(2 + β π₯ + 1) = πππ π₯β3 β(π₯ β 3) ( π₯ + 3)(π₯ β 3)(2+ β π₯ + 1) = πππ π₯β3 β1 ( π₯ + 3)(2+ β π₯ + 1) = = πππ π₯β3 β1 ( π₯ + 3)(2 + β π₯ + 1) = β 1 24 β΄ πππ πβπ π β β π + π π π β π = β π ππ __________________________________________________________________________________________ 45. πππ π₯β1 β π₯ + 3 β 2 π₯2 β 3π₯ + 2 = πππ π₯β1 β1 + 3 β 2 12 β 31 + 2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 β π₯ + 3 β 2 π₯2 β 3π₯ + 2 = πππ π₯β1 (β π₯ + 3 β 2)(β π₯ + 3 + 2) ( π₯ β 1)(π₯ β 2)(β π₯ + 3 + 2) = πππ π₯β1 (β π₯ + 3) 2 β (2)2 ( π₯ β 1)(π₯ β 2)(β π₯ + 3 + 2) = = πππ π₯β1 π₯ + 3 β 4 ( π₯ β 1)(π₯ β 2)(β π₯ + 3 + 2) = πππ π₯β1 (π₯ β 1) ( π₯ β 1)(π₯ β 2)(β π₯ + 3 + 2) = πππ π₯β1 1 (π₯ β 2)(β π₯ + 3 + 2) = = πππ π₯β1 1 (1 β 2)(β1+ 3 + 2) = β 1 4 β΄ πππ πβπ β π + π β π π π β ππ + π = β π π __________________________________________________________________________________________ 46. πππ π₯β2 π₯2 β 4 β π₯ + 2 β β3π₯ β 2 = πππ π₯β2 22 β 4 β2 + 2 β β3 β 2 β 2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β2 π₯2 β 4 β π₯ + 2 β β3π₯ β 2 = πππ π₯β2 ( π₯ + 2)(π₯ β 2)(β π₯ + 2 + β3π₯ β 2) (β π₯ + 2 β β3π₯ β 2)(β π₯ + 2 + β3π₯ β 2) = = πππ π₯β2 ( π₯ + 2)(π₯ β 2)(β π₯ + 2 + β3π₯ β 2) (β π₯ + 2) 2 β (β3π₯ β 2) 2 = πππ π₯β2 ( π₯ + 2)(π₯ β 2)(β π₯ + 2 + β3π₯ β 2) π₯ + 2 β 3π₯ + 2 = = πππ π₯β2 ( π₯ + 2)(π₯ β 2)(β π₯ + 2 + β3π₯ β 2) β2π₯ + 4 = πππ π₯β2 ( π₯ + 2)(π₯ β 2)(β π₯ + 2 + β3π₯ β 2) β2(π₯ β 2) = = πππ π₯β2 ( π₯ + 2)(β π₯ + 2 + β3π₯ β 2) β2 = πππ π₯β2 β (2 + 2)(β2+ 2 + β3 β 2 β 2) 2 = = πππ π₯β2 β (2 + 2)(β2+ 2 + β3β 2 β 2) 2 = β 16 2 = β8 β΄ πππ πβπ π π β π β π + π β β ππ β π = βπ __________________________________________________________________________________________ 47. πππ π₯β1 βπ₯2 β 3π₯ + 3 β βπ₯2 + 3π₯ β 3 π₯2 β 3π₯ + 2 = πππ π₯β1 β12 β 3 β 1 + 3 β β12 + 3 β 1 β 3 1 β 3 β 1 + 2 = 0 0
16.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 16 (temos uma indeterminação) πππ π₯β1 βπ₯2 β 3π₯ + 3 β βπ₯2 + 3π₯ β 3 π₯2 β 3π₯ + 2 = πππ π₯β1 (βπ₯2 β 3π₯ + 3 β βπ₯2 + 3π₯ β 3)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) ( π₯ β 1)(π₯ β 2) = = πππ π₯β1 (βπ₯2 β 3π₯ + 3 β βπ₯2 + 3π₯ β 3)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) ( π₯ β 1)(π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 (βπ₯2 β 3π₯ + 3) 2 β (β π₯2 + 3π₯ β 3) 2 ( π₯ β 1)(π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 π₯2 β 3π₯ + 3 β π₯2 β 3π₯ + 3 ( π₯ β 1)(π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 β3π₯ + 3 β 3π₯ + 3 ( π₯ β 1)(π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 β6π₯ + 6 ( π₯ β 1)(π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 β6(π₯ β 1) ( π₯ β 1)(π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 β 6 (π₯ β 2)(βπ₯2 β 3π₯ + 3 + βπ₯2 + 3π₯ β 3) = = πππ π₯β1 β 6 (1 β 2)(β12 β 3 β 1 + 3 + β12 + 3 β 1 β 3) = 6 2 = 3 β΄ πππ πβπ βπ π β ππ + π β βπ π + ππ β π π π β ππ + π = π 48. πππ π₯β2 β3π₯ β 2 β 2 β4π₯ + 1 β 3 = πππ π₯β2 β3 β 2 β 2 β 2 β4 β 2 + 1 β 3 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β2 β3π₯ β 2 β 2 β4π₯ + 1 β 3 = πππ π₯β2 (β3π₯ β 2 β 2)(β4π₯ + 1 + 3) (β4π₯ + 1 β 3)(β4π₯ + 1 + 3) = πππ π₯β2 (β3π₯ β 2 β 2)(β4π₯ + 1 + 3) (β4π₯ + 1) 2 β (3)2 =
17.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 17 = πππ π₯β2 (β3π₯ β 2 β 2)(β4π₯ + 1 + 3) (β4π₯ + 1) 2 β (3)2 = πππ π₯β2 (β3π₯ β 2 β 2)(β4π₯ + 1 + 3) (β4π₯ + 1) 2 β (3)2 = = πππ π₯β2 (β3π₯ β 2 β 2)(β4π₯ + 1 + 3) 4π₯ + 1 β 9 = πππ π₯β2 (β3π₯ β 2 β 2)(β4π₯ + 1 + 3) (4π₯ β 8) = = πππ π₯β2 (β3π₯ β 2 β 2)(β3π₯ β 2 + 2)(β4π₯ + 1 + 3) (4π₯ β 8)(β3π₯ β 2 + 2) = πππ π₯β2 [(β3π₯ β 2) 2 β (2)2 ](β4π₯ + 1 + 3) (4π₯ β 8)(β3π₯ β 2 + 2) = = πππ π₯β2 [(β3π₯ β 2) 2 β (2)2 ](β4π₯ + 1 + 3) (4π₯ β 8)(β3π₯ β 2 + 2) = πππ π₯β2 [3π₯ β 2 β 4](β4π₯ + 1 + 3) (4π₯ β 8)(β3π₯ β 2 + 2) = = πππ π₯β2 (3π₯ β 6)(β4π₯ + 1 + 3) (4π₯ β 8)(β3π₯ β 2 + 2) = πππ π₯β2 3(π₯ β 2)(β4π₯ + 1 + 3) 4(π₯ β 2)(β3π₯ β 2 + 2) = πππ π₯β2 3(β4π₯ + 1 + 3) 4(β3π₯ β 2 + 2) = = πππ π₯β2 3 β (β4 β 2 + 1 + 3) 4 β (β3 β 2 β 2 + 2) = 18 16 = 9 8 β΄ πππ πβπ β ππ β π β π β ππ + π β π = π π Obs.: aqui o conjugado foi aplicado separadamente, no prΓ³ximo limite aplicaremos de uma sΓ³ vez. Veja! __________________________________________________________________________________________ 49. πππ π₯β4 β2π₯ + 1 β 3 β π₯ β 2 β β2 = πππ π₯β4 β24 + 1 β 3 β4 β 2 β β2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β4 β2π₯ + 1 β 3 β π₯ β 2 β β2 = πππ π₯β4 (β2π₯ + 1 β 3)(β2π₯ + 1 + 3)(β π₯ β 2 + β2) (β π₯ β 2 β β2)(β π₯ β 2 + β2)(β2π₯ + 1 + 3) = = πππ π₯β4 [(β2π₯ + 1) 2 β (3)2 ](β π₯ β 2 + β2) [(β π₯ β 2) 2 β (β2) 2 ](β2π₯ + 1 + 3) = πππ π₯β4 [2π₯ + 1 β 9](β π₯ β 2 + β2) [ π₯ β 2 β 2](β2π₯ + 1 + 3) = = πππ π₯β4 (2π₯ β 8)(β π₯ β 2 + β2) (π₯ β 4)(β2π₯ + 1 + 3) = πππ π₯β4 2(π₯ β 4)(β π₯ β 2 + β2) (π₯ β 4)(β2π₯ + 1 + 3) = πππ π₯β4 2 β (β π₯ β 2 + β2) (β2π₯ + 1 + 3) = = πππ π₯β4 2 β (β π₯ β 2 + β2) (β2π₯ + 1 + 3) = = πππ π₯β4 2 β (β4β 2 + β2) (β2 β 4 + 1 + 3) = 4β2 6 = 2β2 3 β΄ πππ πβπ β ππ + π β π β π β π β β π = πβ π π __________________________________________________________________________________________ 50. πππ π₯β6 4 β β10 + π₯ 2 β β10 β π₯ = πππ π₯β6 4 β β10 + 6 2 β β10 β 6 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ)
18.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 18 πππ π₯β6 4 β β10 + π₯ 2 β β10 β π₯ = πππ π₯β6 (4 β β10 + π₯)(4 + β10+ π₯)(2 + β10 β π₯) (2 β β10 β π₯)(2 + β10β π₯)(4 + β10 + π₯) = = πππ π₯β6 [(4)2 β (β10 + π₯) 2 ](2 + β10β π₯) [(2)2 β (β10 β π₯) 2 ](4 + β10+ π₯) = πππ π₯β6 [16 β 10 β π₯](2+ β10 β π₯) [4 β 10 + π₯](4 + β10 + π₯) = = πππ π₯β6 (6 β π₯)(2 + β10 β π₯) β(6 β π₯)(4 + β10 + π₯) = πππ π₯β6 (6 β π₯)(2 + β10 β π₯) β(6 β π₯)(4 + β10 + π₯) = πππ β π₯β6 (2 + β10 β π₯) (4 + β10 + π₯) = = πππ β π₯β6 (2 + β10 β 6) (4 + β10 + 6) = β 4 8 = β 1 2 β΄ πππ πβπ π β β ππ + π π β β ππ β π = β π π __________________________________________________________________________________________ 51. πππ π₯β0 β3π₯ + 4 β β π₯ + 4 β π₯ + 1 β 1 = πππ π₯β0 β3 β 0 + 4 β β0 + 4 β0 + 1 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β0 (β π₯ + 1 + 1)(β3π₯ + 4 β β π₯ + 4)(β3π₯ + 4 + β π₯ + 4) (β π₯ + 1 β 1)(β π₯ + 1 + 1)(β3π₯ + 4 + β π₯ + 4) = = πππ π₯β0 (β π₯ + 1 + 1) [(β3π₯ + 4) 2 β (β π₯ + 4) 2 ] [(β π₯ + 1) 2 β (1)2](β3π₯ + 4 β β π₯ + 4) = πππ π₯β0 (β π₯ + 1 + 1)[3π₯ + 4 β π₯ β 4] [ π₯ + 1 β 1](β3π₯ + 4 + β π₯ + 4) = = πππ π₯β0 (β π₯ + 1 + 1)[3π₯ β π₯] π₯(β3π₯ + 4 + β π₯ + 4) = πππ π₯β0 (β π₯ + 1 + 1) β 2π₯ π₯ β (β3π₯ + 4 + β π₯ + 4) = πππ π₯β0 (β π₯ + 1 + 1) β 2 (β3π₯ + 4 + β π₯ + 4) = = πππ π₯β0 (β0 + 1 + 1) β 2 (β3β 0 + 4 + β0 + 4) = 4 4 = 1 β΄ πππ πβπ β ππ + π β β π + π β π + π β π = π __________________________________________________________________________________________ 52. πππ π₯β2 βπ₯2 + π₯ β 2 β βπ₯2 β π₯ + 2 β π₯ + 2 β 2 = πππ π₯β2 β22 + 2 β 2 β β22 β 2 + 2 β2 + 2 β 2 = 0 0 ( π‘ππππ π’ππ πππππ‘πππ. ) πππ π₯β2 βπ₯2 + π₯ β 2 β βπ₯2 β π₯ + 2 β π₯ + 2 β 2 = = πππ π₯β2 (β π₯ + 2 + 2)(βπ₯2 + π₯ β 2 β βπ₯2 β π₯ + 2)(βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) (β π₯ + 2 β 2)(β π₯ + 2 + 2)(βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) =
19.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 19 = πππ π₯β2 (β π₯ + 2 + 2)[(βπ₯2 + π₯ β 2) 2 β (βπ₯2 β π₯ + 2) 2 ] [(β π₯ + 2) 2 β (2)2](βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) = = πππ π₯β2 (β π₯ + 2 + 2)[ π₯2 + π₯ β 2 β π₯2 + π₯ β 2] [ π₯ + 2 β 4](βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) = πππ π₯β2 (β π₯ + 2 + 2)[2π₯ β 4] [ π₯ β 2](βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) = = πππ π₯β2 (β π₯ + 2 + 2)2[ π₯ β 2] [ π₯ β 2](βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) = πππ π₯β2 (β π₯ + 2 + 2) β 2 (βπ₯2 + π₯ β 2 + βπ₯2 β π₯ + 2) = = πππ π₯β2 (β2 + 2 + 2) β 2 (β22 + 2 β 2 + β22 β 2 + 2) = 8 4 = 2 β΄ πππ πβπ βπ π + π β π β βπ π β π + π β π + π β π = π __________________________________________________________________________________________ 53. πππ π₯β2 β2π₯2 β 3π₯ + 2 β 2 β3π₯2 β 5π₯ β 1 β 1 = πππ π₯β2 β2 β 22 β 3 β 2 + 2 β 2 β3 β 22 β 5 β 2 β 1 β 1 = 0 0 ( π‘ππππ π’ππ ππππππππππçãπ) πππ π₯β2 β2π₯2 β 3π₯ + 2 β 2 β3π₯2 β 5π₯ β 1 β 1 = πππ π₯β2 (β2π₯2 β 3π₯ + 2 β 2)(β2π₯2 β 3π₯ + 2 + 2)(β3π₯2 β 5π₯ β 1 + 1) (β3π₯2 β 5π₯ β 1 β 1)(β3π₯2 β 5π₯ β 1 + 1)(β2π₯2 β 3π₯ + 2 + 2) = = πππ π₯β2 [(β2π₯2 β 3π₯ + 2) 2 β (2)2 ](β3π₯2 β 5π₯ β 1 + 1) [(β3π₯2 β 5π₯ β 1) 2 β (1)2](β2π₯2 β 3π₯ + 2 + 2) = πππ π₯β2 [2π₯2 β 3π₯ + 2 β 4](β3π₯2 β 5π₯ β 1 + 1) [3π₯2 β 5π₯ β 1 β 1](β2π₯2 β 3π₯ + 2 + 2) = = πππ π₯β2 [2π₯2 β 3π₯ β 2](β3π₯2 β 5π₯ β 1 + 1) [3π₯2 β 5π₯ β 2](β2π₯2 β 3π₯ + 2 + 2) = πππ π₯β2 ( π₯ β 2)(2π₯ + 1)(β3π₯2 β 5π₯ β 1 + 1) ( π₯ β 2)(3π₯ + 1)(β2π₯2 β 3π₯ + 2 + 2) = = πππ π₯β2 [2 β (2) + 1](β3 β (2)2 β 5 β (2)β 1 + 1) [3 β (2) + 1](β2 β (2)2 β 3 β (2)+ 2 + 2) = πππ π₯β2 5 β (β12 β 10 β 1 + 1) 7 β (β8 β 6 + 2 + 2) = = πππ π₯β2 5 β (1 + 1) 7 β (2 + 2) = 5 β 2 7 β 4 = 10 28 = 5 14 β΄ πππ πβπ βππ π β ππ + π β π βππ π β ππ β π β π = π ππ __________________________________________________________________________________________ 54. πππ π₯ββ1 β3π₯2 + 4π₯ + 2 β 1 βπ₯2 + 3π₯ + 6 β 2 = πππ π₯ββ1 β3 β (β1)2 + 4 β (β1) + 2 β 1 β(β1)2 + 3 β (β1) + 6 β 2 = 0 0 (π‘ππππ π’ππ πππππ‘πππ. ) πππ π₯ββ1 β3π₯2 + 4π₯ + 2 β 1 βπ₯2 + 3π₯ + 6 β 2 = πππ π₯ββ1 (β3π₯2 + 4π₯ + 2 β 1)(β3π₯2 + 4π₯ + 2 + 1)((βπ₯2 + 3π₯ + 6 + 2)) (βπ₯2 + 3π₯ + 6 β 2)(βπ₯2 + 3π₯ + 6 + 2)(β3π₯2 + 4π₯ + 2 + 1) =
20.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 20 = πππ π₯ββ1 [(β3π₯2 + 4π₯ + 2) 2 β (1)2 ](βπ₯2 + 3π₯ + 6 + 2) [(βπ₯2 + 3π₯ + 6) 2 β (2)2](β3π₯2 + 4π₯ + 2 + 1) = = πππ π₯ββ1 [3π₯2 + 4π₯ + 2 β 1](βπ₯2 + 3π₯ + 6 + 2) [ π₯2 + 3π₯ + 6 β 4](β3π₯2 + 4π₯ + 2 + 1) = πππ π₯ββ1 [3π₯2 + 4π₯ + 1](βπ₯2 + 3π₯ + 6 + 2) [ π₯2 + 3π₯ + 2](β3π₯2 + 4π₯ + 2 + 1) = = πππ π₯ββ1 ( π₯ + 1)(3π₯ + 1)(βπ₯2 + 3π₯ + 6 + 2) ( π₯ + 1)( π₯ + 2)(β3π₯2 + 4π₯ + 2 + 1) = πππ π₯ββ1 (3π₯ + 1)(βπ₯2 + 3π₯ + 6 + 2) ( π₯ + 2)(β3π₯2 + 4π₯ + 2 + 1) = = πππ π₯ββ1 [3(β1) + 1](β(β1)2 + 3(β1) + 6 + 2) (β1 + 2)(β3(β1)2 + 4(β1)+ 2 + 1) = πππ π₯ββ1 [β3 + 1](β1 β 3 + 6 + 2) 1 β (β3 β 4 + 2 + 1) = = πππ π₯ββ1 [β3 + 1](2+ 2) 1 β (1 + 1) = β2 β 4 2 = β4 β΄ πππ πββπ βππ π + ππ + π β π βπ π + ππ + π β π = βπ __________________________________________________________________________________________ 55. πππ π₯β2 π₯ β 2 β3π₯ β 5 3 β 1 = πππ π₯β2 2 β 2 β3β 2 β 5 3 β 1 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) Atenção!! VocΓͺ jΓ‘ sabe. x3 - y3 = (x - y)(x2 + xy + y2) πππ π₯β2 π₯ β 2 β3π₯ β 5 3 β 1 = πππ π₯β2 (π₯ β 2) (β3π₯ β 5 3 β 1) = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) β 1 + 12 ] (β3π₯ β 5 3 β 1) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) β 1 + 12] = = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) β 1 + 12 ] (β3π₯ β 5 3 β 1) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] = = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + ( β3π₯ β 5 3 ) + 1] ( β3π₯ β 5 3 ) β (β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) β (β3π₯ β 5 3 ) + (β3π₯ β 5 3 ) β (β3π₯ β 5 3 ) 2 β (β3π₯ β 5 3 ) β 1 = = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] ( β3π₯ β 5 3 ) 3 + (β3π₯ β 5 3 ) 2 + ( β3π₯ β 5 3 ) β (β3π₯ β 5 3 ) 2 β ( β3π₯ β 5 3 ) β 1 = = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] ( β3π₯ β 5 3 ) 3 + (β3π₯ β 5 3 ) β (β3π₯ β 5 3 ) β 1 =
21.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 21 = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] (3π₯ β 5) + (β3π₯ β 5 3 ) β (β3π₯ β 5 3 ) β 1 = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] (3π₯ β 5) β 1 = = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] 3π₯ β 5 β 1 = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] 3π₯ β 6 = = πππ π₯β2 (π₯ β 2) β [(β3π₯ β 5 3 ) 2 + (β3π₯ β 5 3 ) + 1] 3(π₯ β 2) = = πππ π₯β2 [(β3 β 2 β 5 3 ) 2 + (β3β 2 β 5 3 ) + 1] 3 = 3 3 = 1 β΄ πππ πβπ π β π β ππ β π π β π = π _________________________________________________________________________________________ 56. πππ π₯β0 β π₯ + 1 3 β 1 π₯ = πππ π₯β0 β0+ 1 3 β 1 0 = 0 0 ( π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β0 β π₯ + 1 3 β 1 π₯ = πππ π₯β0 (β π₯ + 1 3 β 1) β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) β 1 + 12 ] π₯ β [( β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) β 1 + 12] = = πππ π₯β0 ( β π₯ + 1 3 β 1) β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) + 1] π₯ β [(β π₯ + 13 ) 2 + (β π₯ + 13 ) β 1 + 12] = = πππ π₯β0 ( β π₯ + 1 3 ) β (β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) β (β π₯ + 1 3 ) + ( β π₯ + 1 3 ) β 1 β (β π₯ + 1 3 ) 2 β (β π₯ + 1 3 ) β 1 π₯ β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) + 1] = = πππ π₯β0 ( β π₯ + 1 3 ) 3 + (β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) β (β π₯ + 1 3 ) 2 β (β π₯ + 1 3 ) β 1 π₯ β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) + 1] = = πππ π₯β0 ( β π₯ + 1 3 ) 3 + (β π₯ + 1 3 ) β (β π₯ + 1 3 ) β 1 π₯ β [(β π₯ + 1 3 ) 2 + ( β π₯ + 1 3 ) + 1] = πππ π₯β0 ( β π₯ + 1 3 ) 3 β 1 π₯ β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) + 1] = = πππ π₯β0 π₯ + 1 β 1 π₯ β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) + 1] = πππ π₯β0 π₯ π₯ β [(β π₯ + 1 3 ) 2 + (β π₯ + 1 3 ) + 1] = = πππ π₯β0 1 [(β π₯ + 1 3 ) 2 + ( β π₯ + 1 3 ) + 1] = πππ π₯β0 1 [(β0 + 1 3 ) 2 + ( β0+ 1 3 ) + 1] = 1 3 β΄ πππ πβπ β π + π π β π π = π π
22.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 22 __________________________________________________________________________________________ 57. πππ π₯ββ1 π₯ + 1 β2π₯ + 3 3 β 1 = πππ π₯ββ1 β1 + 1 β2(β1)+ 3 3 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯ββ1 π₯ + 1 β2π₯ + 3 3 β 1 = πππ π₯ββ1 ( π₯ + 1) (β2π₯ + 3 3 β 1) = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 )(1) + (1)2 ] (β2π₯ + 3 3 β 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 )(1) + (1)2] = = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] ( β2π₯ + 3 3 β 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] = = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] ( β2π₯ + 33 )(β2π₯ + 33 ) 2 + (β2π₯ + 33 )( β2π₯ + 33 ) + (β2π₯ + 33 ) β 1(β2π₯ + 33 ) 2 β ( β2π₯ + 33 ) β 1 = = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] ( β2π₯ + 3 3 ) 3 + (β2π₯ + 3 3 ) 2 + ( β2π₯ + 3 3 ) β (β2π₯ + 3 3 ) 2 β (β2π₯ + 3 3 ) β 1 = = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] (2π₯ + 3) + ( β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) β ( β2π₯ + 3 3 ) 2 β (β2π₯ + 3 3 ) β 1 = = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] (2π₯ + 3) β 1 = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] 2π₯ + 3 β 1 = = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 33 ) 2 + (β2π₯ + 33 ) + 1] 2π₯ + 2 = πππ π₯ββ1 ( π₯ + 1) β [(β2π₯ + 33 ) 2 + (β2π₯ + 33 ) + 1] 2( π₯ + 1) = = πππ π₯ββ1 [(β2π₯ + 3 3 ) 2 + (β2π₯ + 3 3 ) + 1] 2 = πππ π₯ββ1 [(β2 β (β1) + 3 3 ) 2 + (β2β (β1)+ 3 3 ) + 1] 2 = = πππ π₯ββ1 [(ββ2+ 3 3 ) 2 + (ββ2 + 3 3 ) + 1] 2 = πππ π₯ββ1 [(1)2 + 1 + 1] 2 = 3 2 β΄ πππ πββπ π + π β ππ+ π π β π = π π __________________________________________________________________________________________ 58. πππ π₯β0 β8 β 2π₯ + π₯23 β 2 π₯ β π₯2 = πππ π₯β0 β8β 2(0)+ (0)23 β 2 (0)β (0)2 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ)
23.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 23 πππ π₯β0 β8 β 2π₯ + π₯23 β 2 π₯ β π₯2 = πππ π₯β0 (β8 β 2π₯ + π₯23 β 2)[(β8 β 2π₯ + π₯23 ) 2 + ( β8β 2π₯ + π₯23 )(2)+ (2)2 ] π₯(1 β π₯)[(β8 β 2π₯ + π₯23 ) 2 + (β8 β 2π₯ + π₯23 )(2) + (2)2] = = πππ π₯β0 ( β8β 2π₯ + π₯23 β 2) [(β8β 2π₯ + π₯23 ) 2 + 2(β8 β 2π₯ + π₯23 ) + 4] π₯(1 β π₯) [(β8β 2π₯ + π₯23 ) 2 + 2(β8β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 ( β8β 2π₯ + π₯23 ) β ( β8β 2π₯ + π₯23 ) 2 + ( β8β 2π₯ + π₯23 ) β 2( β8β 2π₯ + π₯23 ) + 4( β8β 2π₯ + π₯23 ) β 2 β ( β8β 2π₯ + π₯23 ) 2 β 4 β ( β8β 2π₯ + π₯23 ) β 8 π₯(1β π₯)[( β8β 2π₯ + π₯23 ) 2 + 2( β8β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 ( β8β 2π₯ + π₯23 ) 3 + 2( β8β 2π₯ + π₯23 ) 2 + 4( β8 β 2π₯ + π₯23 ) β 2 β ( β8β 2π₯ + π₯23 ) 2 β 4 β ( β8 β 2π₯ + π₯23 ) β 8 π₯(1 β π₯) [( β8β 2π₯ + π₯23 ) 2 + 2( β8β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 ( β8β 2π₯ + π₯23 ) 3 + 2( β8β 2π₯ + π₯23 ) 2 + 4( β8 β 2π₯ + π₯23 ) β 2 β ( β8β 2π₯ + π₯23 ) 2 β 4 β ( β8 β 2π₯ + π₯23 ) β 8 π₯(1 β π₯) [( β8β 2π₯ + π₯23 ) 2 + 2( β8β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 (β8 β 2π₯ + π₯23 ) 3 β 8 π₯(1 β π₯)[(β8 β 2π₯ + π₯23 ) 2 + 2(β8 β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 8 β 2π₯ + π₯2 β 8 π₯(1 β π₯)[(β8 β 2π₯ + π₯23 ) 2 + 2(β8 β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 β2π₯ + π₯2 π₯(1 β π₯)[(β8 β 2π₯ + π₯23 ) 2 + 2(β8 β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 π₯(β2 + π₯) π₯(1 β π₯) [(β8β 2π₯ + π₯23 ) 2 + 2(β8β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 (β2 + π₯) (1 β π₯) [(β8β 2π₯ + π₯23 ) 2 + 2(β8β 2π₯ + π₯23 ) + 4] = = πππ π₯β0 (β2 + 0) (1 β 0) [(β8β 2 β 0 + 023 ) 2 + 2(β8 β 2 β 0 + 023 ) + 4] = = πππ π₯β0 β2 [(β8 3 ) 2 + 2(β8 3 ) + 4] = β2 4 + 4 + 4 = β 2 12 = β 1 6 β΄ πππ πβπ βπ β ππ + π ππ β π π β π π = β π π __________________________________________________________________________________________ 59. πππ π₯β0 1 β β1β π₯ 3 1 + β3π₯ β 1 3 = πππ π₯β0 1 β β1β 0 3 1 + β3 β 0 β 1 3 = 0 0 (π‘ππππ π’ππ ππππππππππçãπ)
24.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 24 πππ π₯β0 1 β β1 β π₯ 3 1 + β3π₯ β 1 3 = πππ π₯β0 (1 β β1β π₯ 3 ) β [(1)2 + (1)( β1 β π₯ 3 ) + ( β1 β π₯ 3 ) 2 ] β [(1)2 β (1)( β3π₯ β 1 3 ) + ( β3π₯ β 1 3 ) 2 ] (1 + β3π₯ β 1 3 ) β [(1)2 β (1)( β3π₯ β 1 3 ) + ( β3π₯ β 1 3 ) 2 ] β [(1)2 + (1)( β1 β π₯ 3 ) + ( β1 β π₯ 3 ) 2 ] = = πππ π₯β0 (1 β β1 β π₯ 3 ) β [1 + ( β1β π₯ 3 ) + (β1 β π₯ 3 ) 2 ] β [1 β (β3π₯ β 1 3 ) + (β3π₯ β 1 3 ) 2 ] (1 + β3π₯ β 1 3 ) β [1 β (β3π₯ β 1 3 ) + (β3π₯ β 1 3 ) 2 ] β [1 + ( β1β π₯ 3 ) + (β1 β π₯ 3 ) 2 ] = = πππ π₯β0 [1 + ( β1β π₯ 3 ) + ( β1β π₯ 3 ) 2 β ( β1β π₯ 3 ) β ( β1 β π₯ 3 ) 2 β ( β1 β π₯ 3 ) 3 ] β [1 β ( β3π₯ β 1 3 ) + ( β3π₯ β 1 3 ) 2 ] [1 β ( β3π₯ β 1 3 ) + ( β3π₯ β 1 3 ) 2 + ( β3π₯ β 1 3 ) β ( β3π₯ β 1 3 ) 2 + ( β3π₯ β 1 3 ) 3 ] β [1 + ( β1 β π₯ 3 ) + ( β1 β π₯ 3 ) 2 ] = = πππ π₯β0 [1 β (β1β π₯ 3 ) 3 ] β [1 β (β3π₯ β 1 3 ) + (β3π₯ β 1 3 ) 2 ] [1 + (β3π₯ β 1 3 ) 3 ] β [1 + (β1 β π₯ 3 ) + (β1β π₯ 3 ) 2 ] = = πππ π₯β0 [1 β 1 + π₯] β [1 β (β3π₯ β 1 3 ) + (β3π₯ β 1 3 ) 2 ] [1 + 3π₯ β 1] β [1 + (β1 β π₯ 3 ) + (β1β π₯ 3 ) 2 ] = = πππ π₯β0 π₯ β [1 β (β3π₯ β 1 3 ) + (β3π₯ β 1 3 ) 2 ] 3π₯ β [1 + (β1 β π₯ 3 ) + ( β1β π₯ 3 ) 2 ] = πππ π₯β0 [1 β (β3π₯ β 1 3 ) + ( β3π₯ β 1 3 ) 2 ] 3 β [1 + (β1 β π₯ 3 ) + ( β1β π₯ 3 ) 2 ] = = πππ π₯β0 [1 β (β3β 0 β 1 3 ) + (β3 β 0 β 1 3 ) 2 ] 3 β [1 + (β1β 0 3 ) + (β1 β 0 3 ) 2 ] = πππ π₯β0 [1 β (ββ1 3 ) + ( ββ1 3 ) 2 ] 3 β [1 + (β1 3 ) + (β1 3 ) 2 ] = πππ π₯β0 [1 β (ββ1 3 ) + (ββ1 3 ) 2 ] 3 β [1 + (β13 ) + (β13 ) 2 ] = 1 + 1 + 1 3 β (1 + 1 + 1) = 3 9 = 1 3 β΄ πππ πβπ π β β π β π π π + β ππ β π π = π π __________________________________________________________________________________________ 60. πππ π₯ββ2 β2β 3π₯3 β 2 1 + β2π₯ + 3 3 = πππ π₯ββ2 β2 β 3 β (β2)3 β 2 1 + β2 β (β2) + 3 3 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯ββ2 β2 β 3π₯ 3 β 2 1 + β2π₯ + 3 3 = = πππ π₯ββ2 (β2 β 3π₯3 β 2) β [(β2 β 3π₯3 ) 2 + (β2 β 3π₯3 )(2) + (2)2] β [(1)2 β (1)(β2π₯ + 33 ) + (β2π₯ + 33 ) 2 ] (1 + β2π₯ + 33 ) β [(1)2 β (1)(β2π₯ + 33 ) + (β2π₯ + 33 ) 2 ] β [(β2 β 3π₯3 ) 2 + (β2 β 3π₯3 )(2) + (2)2] = = πππ π₯ββ2 (β2 β 3π₯ 3 β 2) [(β2β 3π₯ 3 ) 2 + 2(β2 β 3π₯ 3 ) + 4] β [1 β (β2π₯ + 3 3 ) + ( β2π₯ + 3 3 ) 2 ] (1 + β2π₯ + 3 3 ) β [1 β (β2π₯ + 3 3 ) + (β2π₯ + 3 3 ) 2 ] β [(β2β 3π₯ 3 ) 2 + 2(β2 β 3π₯ 3 ) + 4] =
25.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 25 = πππ π₯ββ2 [( β2β 3π₯ 3 ) 3 + 2( β2 β 3π₯ 3 ) 2 + 4( β2β 3π₯ 3 ) β 2( β2 β 3π₯ 3 ) 2 β 4( β2β 3π₯ 3 ) β 8] β [1 β ( β2π₯ + 3 3 ) + ( β2π₯ + 3 3 ) 2 ] [1 β ( β2π₯ + 3 3 ) + ( β2π₯ + 3 3 ) 2 + ( β2π₯ + 3 3 ) β ( β2π₯ + 3 3 ) 2 + ( β2π₯ + 3 3 ) 3 ] β [( β2β 3π₯ 3 ) 2 + 2( β2 β 3π₯ 3 ) + 4] = = πππ π₯ββ2 [(2 β 3π₯) + 2( β2 β 3π₯ 3 ) 2 + 4( β2β 3π₯ 3 ) β 2( β2 β 3π₯ 3 ) 2 β 4( β2β 3π₯ 3 ) β 8] β [1 β ( β2π₯ + 3 3 ) + ( β2π₯ + 3 3 ) 2 ] [1 β ( β2π₯ + 3 3 ) + ( β2π₯ + 3 3 ) 2 + ( β2π₯ + 3 3 ) β ( β2π₯ + 3 3 ) 2 + (2π₯ + 3)] β [( β2β 3π₯ 3 ) 2 + 2( β2 β 3π₯ 3 ) + 4] = = πππ π₯ββ2 [(2 β 3π₯) β 8] β [1 β ( β2π₯ + 3 3 ) + (β2π₯ + 3 3 ) 2 ] [1 + (2π₯ + 3)] β [(β2 β 3π₯ 3 ) 2 + 2(β2β 3π₯ 3 ) + 4] = = πππ π₯ββ2 (2 β 3π₯ β 8) β [1 β (β2π₯ + 3 3 ) + ( β2π₯ + 3 3 ) 2 ] (1 + 2π₯ + 3) β [(β2β 3π₯ 3 ) 2 + 2(β2 β 3π₯ 3 ) + 4] = = πππ π₯ββ2 (3π₯ β 6) β [1 β (β2π₯ + 3 3 ) + (β2π₯ + 3 3 ) 2 ] (2π₯ + 4) β [(β2 β 3π₯ 3 ) 2 + 2(β2 β 3π₯ 3 ) + 4] = = πππ π₯ββ2 3 β ( π₯ β 2) β [1 β (β2(β2) + 3 3 ) + (β2(β2)+ 3 3 ) 2 ] β2 β ( π₯ β 2) β [(β2 β 3(β2)3 ) 2 + 2 (β2 β 3(β2)3 ) + 4] = = πππ π₯ββ2 3 β [1 β (ββ4+ 3 3 ) + (ββ4 + 3 3 ) 2 ] β2 β [(β2 + 6 3 ) 2 + 2(β2+ 6 3 ) + 4] = πππ π₯ββ2 3 β [1 β (β1) + (β1)2] β2 β [(2)2 + 2(2) + 4] = = πππ π₯ββ2 3 β [1 + 1 + 1] β2 β [4 + 4 + 4] = β 9 24 = β 3 8 β΄ πππ πββπ β π β ππ π β π π + β ππ+ π π = β π π __________________________________________________________________________________________ 61. πππ π₯β2 β3π₯2 β 7π₯ + 1 3 + 1 β2π₯2 β 5π₯ + 3 3 β 1 = πππ π₯β2 β3 β (2)2 β 7 β (2) + 1 3 + 1 β2 β (2)2 β 5 β (2) + 3 3 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β2 β3π₯2 β 7π₯ + 1 3 + 1 β2π₯2 β 5π₯ + 3 3 β 1 = = πππ π₯β2 (β3π₯2 β 7π₯ + 1 3 + 1)β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 )(1) + (1)2 ] β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 )(1) + (1)2 ] (β2π₯2 β 5π₯ + 3 3 β 1)β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 )(1) + (1)2 ] β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 )(1) + (1)2 ] = = πππ π₯β2 (β3π₯2 β 7π₯ + 1 3 + 1)β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] (β2π₯2 β 5π₯ + 3 3 β 1)β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] =
26.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 26 = πππ π₯β2 [( β3π₯2 β 7π₯ + 1 3 ) 3 β (β3π₯2 β 7π₯ + 1 3 ) 2 + ( β3π₯2 β 7π₯ + 1 3 ) + (β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] [( β2π₯2 β 5π₯ + 3 3 ) 3 + (β2π₯2 β 5π₯ + 3 3 ) 2 + ( β2π₯2 β 5π₯ + 3 3 ) β (β2π₯2 β 5π₯ + 3 3 ) 2 β (β2π₯2 β 5π₯ + 3 3 ) β 1] β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] = = πππ π₯β2 [(3π₯2 β 7π₯ + 1) β (β3π₯2 β 7π₯ + 1 3 ) 2 + ( β3π₯2 β 7π₯ + 1 3 ) + (β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] [(2π₯2 β 5π₯ + 3) + (β2π₯2 β 5π₯ + 3 3 ) 2 + ( β2π₯2 β 5π₯ + 3 3 ) β (β2π₯2 β 5π₯ + 3 3 ) 2 β (β2π₯2 β 5π₯ + 3 3 ) β 1] β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] = = πππ π₯β2 [(3π₯2 β 7π₯ + 1) + 1] β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] [(2π₯2 β 5π₯ + 3) β 1] β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] = = πππ π₯β2 (3π₯2 β 7π₯ + 2) β [(β2π₯2 β 5π₯ + 3 3 ) 2 + ( β2π₯2 β 5π₯ + 3 3 ) + 1] (2π₯2 β 5π₯ + 2) β [(β3π₯2 β 7π₯ + 1 3 ) 2 β ( β3π₯2 β 7π₯ + 1 3 ) + 1] = = πππ π₯β2 ( π₯ β 2) β (3π₯ β 1) β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] ( π₯ β 2) β (2π₯ β 1) β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] = = πππ π₯β2 (3π₯ β 1) β [(β2π₯2 β 5π₯ + 3 3 ) 2 + (β2π₯2 β 5π₯ + 3 3 ) + 1] (2π₯ β 1) β [(β3π₯2 β 7π₯ + 1 3 ) 2 β (β3π₯2 β 7π₯ + 1 3 ) + 1] = = πππ π₯β2 (3 β 2 β 1) β [(β2β 22 β 5 β 2 + 3 3 ) 2 + (β2 β 22 β 5 β 2 + 3 3 ) + 1] (2 β 2 β 1) β [(β3β 22 β 7 β 2 + 1 3 ) 2 β (β3 β 22 β 7 β 2 + 1 3 ) + 1] = = πππ π₯β2 5 β [(β8β 10 + 3 3 ) 2 + (β8 β 10 + 3 3 ) + 1] 3 β [(β12β 14 + 1 3 ) 2 β (β12β 14 + 1 3 ) + 1] = πππ π₯β2 5 β [(1)2 + 1 + 1] 3 β [(β1)2 β (β1)+ 1] = = πππ π₯β2 5 β [1 + 1 + 1] 3 β [1 + 1 + 1] = 15 9 = 5 3 β΄ πππ πβπ βππ π β ππ + π π + π βππ π β ππ + π π β π = π π __________________________________________________________________________________________ 62. πππ π₯β1 β5π₯ + 4 β 3 β π₯ β 2 3 + 1 = πππ π₯β1 β5 β 1 + 4 β 3 β1β 2 3 + 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 β5π₯ + 4 β 3 β π₯ β 2 3 + 1 = πππ π₯β1 (β5π₯ + 4 β 3)(β5π₯ + 4 + 3) [(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 )(1)+ (1)2 ] (β π₯ β 2 3 + 1) [(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 )(1) + (1)2](β5π₯ + 4 + 3) = = πππ π₯β1 (β5π₯ + 4 β 3)(β5π₯ + 4 + 3)[(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] (β π₯ β 2 3 + 1) [(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1](β5π₯ + 4 + 3) =
27.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 27 = πππ π₯β1 [(β5π₯ + 4) 2 β (3)2 ][(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] ( β π₯ β 2 3 + 1)[(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] (β5π₯ + 4 + 3) = = πππ π₯β1 (5π₯ + 4 β 9)[(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] [(β π₯ β 2 3 ) 3 β(β π₯ β 2 3 ) 2 + ( β π₯ β 2 3 ) + (β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1](β5π₯ + 4 + 3) = = πππ π₯β1 (5π₯ β 5)[(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] [(β π₯ β 2 3 ) 3 β (β π₯ β 2 3 ) 2 + (β π₯ β 2 3 ) + ( β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1](β5π₯ + 4 + 3) = = πππ π₯β1 (5π₯ β 5)[(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] [ π₯ β 2 + 1](β5π₯ + 4 + 3) = πππ π₯β1 5( π₯ β 1) [(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] ( π₯ β 1)(β5π₯ + 4 + 3) = = πππ π₯β1 5 β [(β π₯ β 2 3 ) 2 β (β π₯ β 2 3 ) + 1] (β5π₯ + 4 + 3) = πππ π₯β1 5 β [(β1β 2 3 ) 2 β (β1 β 2 3 ) + 1] (β5 β 1 + 4 + 3) = = πππ π₯β1 5 β [(ββ1 3 ) 2 β (ββ1 3 ) + 1] (3 + 3) = πππ π₯β1 5 β [1 + 1 + 1] (3 + 3) = 15 6 = 5 2 β΄ πππ πβπ β ππ+ π β π β πβ π π + π = π π __________________________________________________________________________________________ 63. πππ π₯β2 β5π₯ β 2 3 β 2 β π₯ β 1 β 1 = πππ π₯β2 β5 β 2 β 2 3 β 2 β2 β 1 β 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β2 β5π₯ β 2 3 β 2 β π₯ β 1 β 1 = πππ π₯β2 ( β5π₯ β 23 β 2) [(β5π₯ β 23 ) 2 + (β5π₯ β 23 )(2) + (2)2 ](β π₯ β 1 + 1) (β π₯ β 1 β 1)(β π₯ β 1 + 1) [(β5π₯ β 2 3 ) 2 + ( β5π₯ β 2 3 )(2)+ (2)2] = = πππ π₯β2 ( β5π₯ β 2 3 β 2) β [(β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] (β π₯ β 1 + 1) (β π₯ β 1 β 1)(β π₯ β 1 + 1) β [( β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] = = πππ π₯β2 [(β5π₯ β 2 3 ) 3 + 2(β5π₯ β 2 3 ) 2 + 4(β5π₯ β 2 3 ) β 2(β5π₯ β 2 3 ) 2 β 4(β5π₯ β 2 3 ) β 8](β π₯ β 1 + 1) [(β π₯ β 1) 2 β (1)2] β [(β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] = = πππ π₯β2 [(5π₯ β 2) + 2(β5π₯ β 2 3 ) 2 + 4(β5π₯ β 2 3 ) β 2(β5π₯ β 2 3 ) 2 β 4(β5π₯ β 2 3 ) β 8](β π₯ β 1 + 1) [ π₯ β 1 β 1] β [(β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] =
28.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 28 = πππ π₯β2 (5π₯ β 10)(β π₯ β 1 + 1) ( π₯ β 2) β [(β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] = πππ π₯β2 5( π₯ β 2)(β π₯ β 1 + 1) ( π₯ β 2) β [(β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] = = πππ π₯β2 5(β π₯ β 1 + 1) [(β5π₯ β 2 3 ) 2 + 2(β5π₯ β 2 3 ) + 4] = πππ π₯β2 5(β2 β 1 + 1) [(β5β 2 β 2 3 ) 2 + 2(β5 β 2 β 2 3 ) + 4] = = πππ π₯β2 5(1 + 1) [(β10 β 2 3 ) 2 + 2(β10β 2 3 ) + 4] = = πππ π₯β2 10 [(β8 3 ) 2 + 2(β8 3 )+ 4] = = πππ π₯β2 10 [4 + 4 + 4] = 10 12 = 5 6 β΄ πππ πβπ β ππ β π π β π β π β π β π = π π __________________________________________________________________________________________ 64. πππ π₯β1 β3π₯3 β 5π₯ + 6 β 2 βπ₯2 β 3π₯ + 1 3 + 1 = πππ π₯β2 β3 β (1)3 β 5 β (1) + 6 β 2 β(1)2 β 3 β (1)+ 1 3 + 1 = 0 0 (π‘ππππ π’ππ πππππ‘ππππππçãπ) πππ π₯β1 β3π₯3 β 5π₯ + 6 β 2 βπ₯2 β 3π₯ + 1 3 + 1 = = πππ π₯β1 (β3π₯3 β 5π₯ + 6 β 2)(β3π₯3 β 5π₯ + 6 + 2)[(βπ₯2 β 3π₯ + 1 3 ) 2 β ( βπ₯2 β 3π₯ + 1 3 )(1) + (1)2 ] (βπ₯2 β 3π₯ + 1 3 + 1)[(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 )(1) + (1)2](β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 (β3π₯3 β 5π₯ + 6 β 2)(β3π₯3 β 5π₯ + 6 + 2)[(βπ₯2 β 3π₯ + 1 3 ) 2 β ( βπ₯2 β 3π₯ + 1 3 ) + 1] (βπ₯2 β 3π₯ + 1 3 + 1)[(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] (β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 [(β3π₯3 β 5π₯ + 6) 2 β (2)2 ][(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] [(βπ₯2 β 3π₯ + 1 3 ) 3 β (βπ₯2 β 3π₯ + 1 3 ) 3 + (βπ₯2 β 3π₯ + 1 3 ) + (βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1](β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 [(3π₯3 β 5π₯ + 6) β 4] [(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] [(βπ₯2 β 3π₯ + 1 3 ) 3 + 1](β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 (3π₯3 β 5π₯ + 2)[(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] [ π₯2 β 3π₯ + 1 + 1](β3π₯3 β 5π₯ + 6 + 2) =
29.
Limite - ExercΓcios
Resolvidos - Lista 1 Professor Alan - MatemΓ‘tico www.calculandocerto.com.br calculandocerto@gmail.com Calculando Certo PΓ‘gina 29 = πππ π₯β1 (3π₯3 β 5π₯ + 2)[(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] ( π₯2 β 3π₯ + 2)(β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 ( π₯ β 1)(3π₯2 + 3π₯ β 2)[(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] ( π₯ β 1)( π₯ β 2)(β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 (3π₯2 + 3π₯ β 2)[(βπ₯2 β 3π₯ + 1 3 ) 2 β (βπ₯2 β 3π₯ + 1 3 ) + 1] ( π₯ β 2)(β3π₯3 β 5π₯ + 6 + 2) = = πππ π₯β1 (3 β 12 + 3 β 1 β 2)[(β12 β 3 β 1 + 1 3 ) 2 β (β12 β 3 β 1 + 1 3 ) + 1] (1 β 2)(β3β 13 β 5 β 1 + 6 + 2) = = πππ π₯β1 4 β [1 + 1 + 1] β1 β (2 + 2) = β 12 4 = β3 β΄ πππ πβπ βππ π β ππ + π β π βπ π β ππ + π π + π = βπ
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