Exercícios de Cálculo Diferencial e Integral

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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 
1) 






 x3
)x2(tg
lim
0x
 
2) 2x
x x
1
1lim









 
3) 1x
1x
)e(lim
1x



 
4) x3
x x
2
1lim 







 
5) 
 )xln()1x(ln(xlim
x


 
6) 






1x7x3xlim
22
x
 
7)  








 1x
xln
lim
3
1x
 
8) 










 3x
9x
lim
4
81x
 
9) 





 
 x3
)x21ln(
lim
0x
 
10) 







 
 x2
e1x
lim
x2
0x
 
11) 










 )1xln(
93
lim
2x
0x
 
12) 










 )x(sen)x2(sen
32
lim
xx
0x
 
 LIMITES FUNDAMENTAIS 
 
 
1
x
)x(sen
lim
0x







 
 
 
)aln(
x
1a
lim
x
0x








 

 
 
 
  ex1lim x
1
0x


 
 
 
e
x
1
1lim
x
x








 
 
 LIMITES NOTÁVEIS 
 
 
0
x
)xln(
lim
x







 
 
 
1
x
1xln(
lim
0x





 

 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
y 
x 
x = 1 
y = 2 
1 
–1 1 
2 
 
13) Observe o gráfico abaixo e responda às questões a seguir. 
 
 
 
 
 
 
 
a) 
 )x(flim
1x 
 = 0 
b) 
 )x(flim
1x 
 = 0 
c) 
 )x(flim
1x 
 = 0 
d) 
 )x(flim
0x
 = 1 
e) 
 )x(flim
1x 
 = 2 
f) 
 )x(flim
1x 
 = + ∞ 
g) 
 )x(flim
1x
 = 

 
h) 
 )x(flim
x 
 = 2 
i) 
 )x(flim
x 
 = – ∞ 
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 
1) 






 x3
)x2(tg
lim
0x
 = 












 x3
)x2cos(
)x2(sen
lim
0x
 = 






 ))x2(cos(x3
)x2(sen
lim
0x
 = 







 )x2cos(
1
x
)x2(sen
lim
3
1
0x
 
 






 x
)x2(sen
lim
3
1
0x






 )x2cos(
1
lim
0x
 = 
1
1
x
)x2(sen
2
2
lim
3
1
0x








 
 
3
211
3
2
1
x2
)x2(sen
lim
3
2
0x







 
2) 2x
x x
1
1lim









 = 








x
x x
1
1lim
2
x x
1
1lim 







 = 
e1
x
1
1lim
x
x








 
3) 














1x
1x
1x
1x
)e(lim
1x
 












1x
1x)1x(
)e(lim
1x
 
2
1x
ee)e(
)11(1x
lim 


 
 
4) x3
x x
2
1lim 







 → Fazendo 
0xquando
2
x
x
2



 
 Reescrevendo o limite em função da variável 

 temos: 
  










2
3
1lim
0
 =  










6
1lim
0
 =  
6
0
1
1lim






















=   6
6
0
e1
1
lim
























 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
5) 
 ))xln()1x(ln(xlim
x


 = 

















 
 x
1x
lnxlim
x
 = 













 

x
x x
1x
lnlim
 
 
















x
x x
1
x
x
lnlim
 = 
















x
x x
1
1lnlim
 = 
























x
x x
1
1ln lim
 = ln(e) = 1 
6) 






1x7x3x 22
x
lim
 = + ∞ – ∞ → Indeterminação 
 















 1x7x3x
1x7x3x
1x7x3x
22
22
22
x
lim
 
 










 1x7x3x
1x7x3x
22
22
x
lim
 = 










 1x7x3x
6x3
22x
lim
 
 










 22x xx
x3
lim
 = 









 2x x2
x3
lim
 = 







 
 x2
x3
lim
x
 = 
2
3
x2
x3
lim
x





 

 
7)  








 1x
xln 3
1x
lim
 =  






 1x
xln3
lim
1x
 = 













)xln(
1x
1
3 lim
1x
 = 




















1x
1
)xln(3 lim
1x
 
 Fazendo x – 1 = 
 
 x = 

 + 1 → quando x → 1 

 

 → 0 
 Reescrevendo o limite em função da variável 

 temos: 
 



















1
)1ln(lim3
0
 = 



























1
)1(limln3
0
 = 3[ln(e)] = 3 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
 
8) 










 3x
9x
4
81x
lim
 → Fazendo 
4 x
 

 x = 

4 → quando x → 81 

 

→ 3 
 Reescrevendo o limite em função da nova variável 

 temos: 
 










 3
9
lim
4
3
 = 










 3
92
3
lim
 = 








 3
)3)(3(
lim
3
 = 
 3lim
3


 = 6 
9) 





 
 x3
)x21ln(
lim
0x
 = 








)x21ln(
x3
1
lim
0x
 =  












x3
1
)x21ln(lim
0x
 
  






















x3
1
x21limln
0x
 → Fazendo 2x = β 



2
x
 quando x → 0 

 
0
 
  



























32
1limln
0x
 = 
 































3
2
0x
1
1limln
 =  
3
2
0x
1
1limln



























= 
3
2]eln[
3
2
 
10) 







 
 x2
e1x
lim
x2
0x
= 










 x2
x
x2
1e
lim
x2
0x
 = 










 2
1
x2
1e
lim
x2
0x
 
 







 
 x2
1e
lim
x2
0x
 + 






 2
1
lim
0x
 → Fazendo 2x = β → quando x → 0 

 
0
 
 











1e
lim
0
 + 
2
1
 → ln(e) + 
2
1
 = 1 + 
2
1
 = 
2
3
 
UNIVERSIDADE FEDERAL DO RECÔNCAVO DA BAHIA 
CENTRO DE CIÊNCIAS EXATAS E TECNOLÓGICAS 
 
 
CET 007 – Cálculo Diferencial e Integral 
Profa. Ruth Exalta da Silva 
 
11) 










 )1xln(
93
lim
2x
0x
 = 










 )1xln(
939
lim
x
0x
 =  










 )1xln(
139
lim
x
0x
 
 Dividindo numerador e denominador por x e utilizando as propriedades temos: 
 



























x
)1xln(
x
13
lim9
x
0x
 = 





















 







 


x
)1xln(
lim
x
13
lim
9
0x
x
0x
 = 






























 


)1xln(
x
1
lim
x
13
lim
9
0x
x
0x
 
 






































 


x
1
0x
x
0x
)1xln(lim
x
13
lim
9
 = 


















































 


x
1
0x
x
0x
)1x(limln
x
13
lim
9
 = 
)eln(
)3ln(
9
= 9[ln(3)] 
12)










 )x(sen)x2(sen
32
lim
xx
0x
 → Dividindo numerador e desominador por x temos: 
 















x
)x(sen)x2(sen
x
1132
lim
xx
0x
 = 

















x
)x(sen
x
)x2(sen
x
13
x
12
lim
xx
0x
→ Utilizando as propriedades 
 



















 








 


x
)x(sen
lim
x
)x2(sen
lim
x
13
lim
x
12
lim
0x0x
x
0x
x
0x
 = 

























 








 


x
)x(sen
lim
x
)x2(sen
lim2
x
13
lim
x
12
lim
0x0x
x
0x
x
0x
 
 
1)1(2
)3ln()2ln(


 = ln(2) – ln(3) = 
 
3
2ln