ListaExercícios#01

Prévia do material em texto

UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
 
LISTA DE EXERCÍCIOS DE LIMITES # 01 
 
 
Resolva os limites abaixo: 
 
 
1) 
4
2
5x
23x
lim
5x












 2) 










 x1
xx
lim
2
1x
 = 3 
 
 
3) 3
38x
x1
lim
21x












 4) 
2
1
x
1
1xx
1
lim
0x











 
 
 
5) 
24
5
4x
3xx3
lim 2
2
2x












 6) 
16
1
4x
x
1
4
1
lim
4x
















 
 
 7) 
4
5
524x
3x2
lim
21x












 8) 










 xx
1
x
1
lim
2
0x
 = 1 
 
 
9) 
4
1
xx
5x2
lim 2
1x












 10) 
10
xx
x5x5
lim
2
1x












 
 
 
11) 
3
33
1x 43
1
1x
x21x
lim 











 12) 








 1x2x
3xx
lim 2
2
1x
 
 
 
 
 
 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
 
RESOLUÇÃO DA LISTA DE EXERCÍCIOS DE LIMITES # 01 
 
 
 
1) 










 5x
23x
lim
5x
 = 













 23x
23x
5x
23x
lim
5x
 
 
    
   









 23x5x
23x
lim
22
5x
 = 
   









 23x5x
23x
lim
5x
 
 
 
  











 23x5x
5x
lim
5x   4
2
22
1
23x
1
lim
5x










 
 
 
2) 










 x1
xx
lim
2
1x
 = 













 x1
x1
x1
xx
lim
2
1x
 =   
  










22
2
1x x1
x1xx
lim
 
 
 










 x1
xxxxx
lim
22
1x
 =    










 x1
x1xx1x
lim
2
1x
 
 
  










 x1
x1x)x1)(x1(x
lim
1x
 =  










 x1
xx)x1()x1(
lim
1x
 
 
 
 xx)x1(lim
1x


 = (1 + 1)(1) + 1 = 3 
 
3) 











 38x
x1
lim
21x 












 38x
38x
38x
x1
lim
2
2
21x
 
 
  
  










 2
2
2
2
1x
)3(8x
38x)x1(
lim
 =  










 1x
38x)x1(
lim 2
2
1x
 
 
 
3
1x
38x
lim
2
1x












 
 
 











 )1x)(1x(
38x)1x(
lim
2
1x
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
4) 









 x
1
1xx
1
lim
0x
 = 

















1
1x
1
x
1
lim
0x
 = 
















 1x
1x1
x
1
lim
0x
 
 
 



















 1x1
1x1
1x
1x1
x
1
lim
0x
 = 
   















 1x11x
)1x(1
x
1
lim
0x
 
 
 
   















 1x11x
x
x
1
lim
0x
 = 
   2
1
1x11x
1
lim
0x












 
 
5) 










 4x
3xx3
lim
2
2
2x
 = 













 3xx3
3xx3
4x
3xx3
lim
2
2
2
2
2x
 
 
 
 




























 33xx4x
3xx)3(
lim
22
2
22
2x
= 
 




















 33xx4x
3xx9
lim
22
2
2x
 
 
 
 




















 33xx4x
6xx
lim
22
2
2x
=
  24
5
33xx)2x)(2x(
)3x)(2x(
lim
22x












 
 
6) 














 4x
x
1
4
1
lim
4x
 = 














 4x
x4
4x
lim
4x
= 
16
1
x4
1
lim
4x







 
 
7) 










 524x
3x2
lim
21x
 = 













 3x2
3x2
524x
3x2
lim
21x
 
 
 
 
 




















 3x2524x
3x)2(
lim
2
22
1x
 = 
 




















 3x2524x
)3x(4
lim
21x
 
 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
 
 























 524x
524x
3x2524x
)1x(
lim
2
2
21x
 
 
  








































3x2524x
524x)1x(
lim
2
2
2
2
1x =   




















 3x22524x
524x)1x(
lim
2
2
1x
 
 
  




















 3x2)1x)(1x(
524x)1x(
lim
2
1x
 =   4
5
3x2)1x(
524x
lim
2
1x






















 
 
8) 










 xx
1
x
1
lim
2
0x
 = 














 1x
1
1
x
1
lim
0x
 = 














 1x
1x1
x
1
lim
0x
 
 
 












 1x
x
x
1
lim
0x
 = 






 1x
1
lim
0x
 = 1 
 
 
9) 










 xx
5x2
lim
2
1x
= 












 5x2
5x2
xx
5x2
lim
2
1x
 
 
  
  









 5x2xx
5x)2(
lim
2
22
1x
= 
  









 5x2xx
)5x(4
lim
2
1x
 
 
 
 









 5x2)1x(x
1x
lim
1x
= 
 









 5x2)1x(x
)1x(
lim
1x
 
 
 
  4
1
5x2x
1
lim
1x












 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
f(x) x 
 – – + + + – – 
– 1 1 2 
10) 










 xx
x5x5
lim
2
1x
= 













 xx
xx
xx
x5x5
lim
2
1x
=   
  









 2
2
2
1x xx
xxx5x5
lim
 
 
   











2
2
1x xx
xxx5x5
lim
=  










 )1x(x
xx)1x(x5
lim
1x
=  
10
)1
xx5
lim
1x












 
 
11) 










 1x
x21x
lim
33
1x
=    
    













23323
2332333
1x x2x2)1x(1x
x2x2)1x(1x
1x
x21x
lim
 
    
    













23323
2332333
1x x2x2)1x(1x
x2x2)1x(1x
1x
x21x
lim
 
 
 
   
   




















 23323
3333
1x x2x2)1x(1x)1x(
x21x
lim
 
 
   




















 233231x x2x2)1x(1x)1x(
1x
lim 
 
   




















 233231x x2x2)1x(1x)1x(
)1x(
lim 
 
   




















 233231x x2x2)1x(1x
1
lim
= 
    323323 43
1
2)2)(2(2
1




















 
 
12) 
fparaalsindoestudooFazemos
0
3
1x2x
3xx
lim 2
2
1x









 
 
 








 1x2x
3xx
lim 2
2
1x
 
 
  

)x(flim
1x
 e 
  

)x(flim
`1x