Medicion Y Manipulacion En El Laboratorio De Quimica

Limites Trigonométricos Resolvidos
Sete páginas e 34 limites resolvidos Usar o limite fundamental e alguns artifícios :
senx =1 x→0 x lim

0 x x lim = , é uma indeterminação. =? à x →0 sen x x → 0 sen x 0 x 1 1 x lim = lim = = 1 logo lim =1 sen x x →0 sen x x →0 sen x x → 0 sen x lim x→0 x x sen 4 x sen 4 x sen 4 x 0 sen y à lim 4. = 4. lim = ? à lim = =4.1= 4 2. lim x →0 y →0 x→0 x→0 0 4x yx x sen 4 x lim =4 x→0 x sen 5 x 5 sen 5 x 5 sen y 5 sen 5 x = ? à lim . logo lim 3. lim = lim . = x →0 2 x x →0 2 y →0 2 x →0 2 x y 5x 2 1. lim 4. lim
sen mx = x →0 nx sen 3 x x →0 sen 2 x 3 2

logo

=

5 2

? à lim

5. lim

=? à logo

6. lim

x→0

senmx = sennx

? à

sen mx m = x →0 nx n sen y sen 3 x sen 3 x sen 3 x lim 3. lim sen 3 x 3 y →0 y 3 x →0 3 x 3x = . lim = limx = lim = . = .1 = sen t sen 2 x 2 x →0 sen 2 x x → 0 sen 2 x x→0 sen 2 x 2 lim lim 2. x→0 2 x t →0 t x 2x sen 3 x 3 lim = x →0 sen 2 x 2 sen mx sen mx sen mx m. sen mx x mx = lim m . mx = m lim = lim = lim Logo sen nx x →0 sen nx x →0 n sen nx x → 0 sen nx x →0 n n. nx nx x

sen mx m sen mx = lim . x →0 x→0 n nx mx

=

sen y m . lim n y →0 y

=

m m .1= n n

logo lim

senmx m = x →0 sennx n lim

7.

8.

sen x 0 tgx tgx tgx sen x 1 lim = ? à lim = = lim cos x = lim . = à lim x→ 0 x x→ 0 x x→ 0 x x→ 0 x → 0 cos x x 0 x tgx sen x 1 sen x 1 =1 lim . = lim . lim = 1 Logo lim x→ 0 x→ 0 x → 0 cos x x→ 0 x x cos x x x → 1 0 tg (t ) tg a 2 − 1 tg a 2 − 1 = ? à lim 2 = lim à Fazendo t = a 2 − 1,  à lim =1 2 a →1 a − 1 a →1 a − 1 t →0 t t →0 0 

(

)

(

)

logo limtg a 2 − 1 a2 −1

(

a →1

) =1

1

Limites Trigonométricos Resolvidos
Sete páginas e 34 limites resolvidos 9. lim
x − sen 3 x x + sen 2 x

x →0

= ? à lim

x →0

x − sen 3 x x + sen 2 x

=

0 0

à f (x ) =

x − sen 3 x x + sen 2 x

=

 sen 3 x  x.1 −  x   =  sen 5 x  x.1 +  x  

sen 3 x   sen 3 x sen 3 x x.1 − 3.  1 − 3. 1 − 3. 3. x   3. x3.x = 1 − 3 = −2 = − 1 logo = à lim sen 5 x sen 5 x x →0 sen 5 x  1+ 5 6 3  1 + 5. 1 + 5. x.1 + 5.  5. x 5. x 5. x   x − sen 3 x 1 lim =− x →0 x + sen 2 x 3 1 sen x 1 sen 2 x 1 tgx − sen x tgx − sen x 10. lim = ? à lim = lim = . . . 3 3 2 x →0 x →0 x→0 x cos x x 1 + cos x 2 x x sen x − sen x. cos x sen x − sen x tgx − sen x cos x sen x.(1 − cos x ) sen x 1 1 − cos x cos x . . f (x ) = = = == 3 x x 2 cos x x 3 . cos x x3 x3 x sen x 1 1 − cos x 1 + cos x . . . x x 2 cos x 1 + cos x tgx − sen x 1 Logo lim = x →0 2 x3

=

=

sen x 1 1 − cos 2 x 1 . . . 2 x cos x 1 + cos x x

=

1 sen x 1 sen 2 x . . . 2 x cos x x 1 + cos x

11. lim
lim

1 + tgx − 1 + sen x x
3

x →0

=? à

lim

tgx − sen x x
3

x →0

.

1 1 + tgx + 1 + sen x 1 4 tgx − sen x x3

=

senx 1 sen 2 x 1 1 . . . . x →0 x cos x x 2 1 + cos x 1 + tgx + 1 + sen x f (x ) = lim 1 + tgx − 1 + senx x3
3

= 1. . . . =
1

1 1 1 1 1 1 2 2

=
1 4

1 + tgx − 1 − sen x x3

.

1 + tgx + 1 + sen x

=

.

1 1 + tgx + 1 + sen x

1 + tgx − 1 + sen x x

x →0

=

12.

sen x − sen a lim x→a x−a

=? à

sen x − sen a lim x→a x−a

=

x−a x+a 2 sen . cos  2   2 = lim x→a x−a 2.   2 

x − a . cos x + a    )  2  2 . lim x→a 1 x−a 2.   2  2 sen(

= cos a

Logo lim

x→a

sen x − sen a x−a

= cosa

2

Limites Trigonométricos Resolvidos
Sete páginas e 34 limites resolvidos 13. lim
sen ( x + a ) − sen x a

a →0

= ? à lim

a →0

sen ( x + a ) − sen x a

=

 x+a−x  x+a+ x 2 sen  . cos  2 2     . lima→a 1  x−a 2.   2  sen ( x + a ) − sen x a →0 a

=

a  2x + a  2 sen  . cos  2   2  . = cos x lim a→a 1 a 2.  2

Logo lim

=cosx

14.

x+a+ x x−a− x − 2 sen . sen  cos( x + a ) − cos x cos( x + a ) − cos x 2 2     lim = ? à lim = lim a →0 a→0 a→0 a a a  2x + a  −a −a − 2. sen sen . sen   2x + a   2   2   2  . = − sen x Logo...