• Nenhum resultado encontrado

Limites Trigonométricos Resolvidos. Sete páginas e 34 limites resolvidos. sen x. = 1 logo = 1. lim = 4. lim. = n. lim. lim lim =? m m m.

N/A
N/A
Protected

Academic year: 2021

Share "Limites Trigonométricos Resolvidos. Sete páginas e 34 limites resolvidos. sen x. = 1 logo = 1. lim = 4. lim. = n. lim. lim lim =? m m m."

Copied!
7
0
0

Texto

(1)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

Usar o limite fundamental e alguns artifícios :

lim 1

0 = → x senx x

1.

x x x sen lim 0 →

= ? à

x x x sen lim 0 →

=

0

0

, é uma indeterminação.

x x x sen lim 0 →

=

x x x sen 1 lim 0 →

=

x x x sen lim 1 0 →

= 1 logo

x x x sen lim 0 →

= 1

2.

x x x 4 sen lim 0 →

= ? à

x x x 4 sen lim 0 →

=

0 0

à

x x x 4 4 sen . 4 lim 0 →

= 4.

y y y sen lim 0 →

=4.1= 4 logo

x x x 4 sen lim 0 →

=4

3.

x x x 2 5 sen lim 0 →

= ? à

x = x x 5 5 sen . 2 5 lim 0 → y = y y sen . 2 5 lim 0 2 5

logo

x x x 2 5 sen lim 0 →

=

2 5

4.

nx mx x sen lim 0 →

= ? à

nx mx x sen lim 0 →

=

mx mx n m x sen . lim 0 →

=

n m

.

y y y sen lim 0 →

=

n m

.1=

n m

logo

nx mx x sen lim 0 →

=

n m

5.

x x x sen2 3 sen lim 0 →

= ? à

x x x sen2 3 sen lim 0 →

=

→ = x x x x x sen2 3 sen lim 0 → = x x x x x 2 2 sen . 2 3 3 sen . 3 lim 0

.

2 3 2 2 sen lim 3 3 sen lim 0 0 = → → x x x x x x

.

.1 2 3 sen lim sen lim 0 0 = → → t t y y t y

=

2 3

logo

x x x sen2 3 sen lim 0 →

=

2 3

6.

sennx senmx x 0 lim →

= ? à

nx mx x sen sen lim 0 →

=

x nx x mx x sen sen lim 0 →

=

nx nx n mx mx m x sen . sen . lim 0 →

=

nx nx mx mx n m x sen sen . lim 0 →

=

n m

Logo

sennx senmx x 0 lim →

=

n m

7.

= → x tgx x 0 lim

? à

= → x tgx x 0 lim 0 0

à

= → x tgx x 0 lim = → x x x x cos sen lim 0 → x x = x x 1 . cos sen lim 0 x x x x cos 1 . sen lim 0 →

=

x x x x x cos 1 lim . sen lim 0 0 → →

= 1

Logo

x = tgx x 0 lim

1

8.

( )

1 1 lim 2 2 1 − − → a a tg a

= ? à

( )

1 1 lim 2 2 1 − − → a a tg a

=

0 0

à Fazendo

   → → − = 0 1 , 1 2 t x a t

à

( )

t t tg t 0 lim →

=1

logo

( )

1 1 lim 2 2 1 − − → a a tg a

=1

(2)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

9.

x x x x x sen2 3 sen lim 0 + − →

= ? à

x x x x x sen2 3 sen lim 0 + − →

=

0 0

à

( )

x x x x x f 2 sen 3 sen + − =

=

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

=

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

=

x x x x . 5 5 sen . 5 1 . 3 3 sen . 3 1 + −

à

0 lim → x x x x x . 5 5 sen . 5 1 . 3 3 sen . 3 1 + −

=

5 1 3 1 + −

=

6 2 −

=

3 1 −

logo

x x x x x sen2 3 sen lim 0 + − →

=

3 1 −

10.

3 0 sen lim x x tgx x − →

= ? à

0 3 sen lim x x tgx x − →

=

x x x x x x x 1 cos 1 . sen . cos 1 . sen lim 2 2 0 + →

=

2 1

( )

3sen x x tgx x f = −

=

3 sen cos sen x x x x

=

cos3 cos . sen sen x x x x x

=

(

)

x x x x cos . cos 1 . sen 3 −

=

x x x x x cos cos 1 . 1 . sen 2 −

=

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

=

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

=

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

Logo

3 0 sen lim x x tgx x − →

=

2 1

11.

3 0 sen 1 1 lim x x tgx x + − + →

=? à

x tgx x x tgx x 1 1 sen 1 . sen lim 3 0 + + + − →

=

x tgx x x x x x x x 1 1 sen 1 . cos 1 1 . sen . cos 1 . sen lim 2 2 0 + + + + →

=

2 1 . 2 1 . 1 1 . 1 1 . 1

=

4 1

( )

1 31 x senx tgx x f = + − +

=

x tgx x x tgx sen 1 1 1 . sen 1 1 3 + + + − − +

=

x tgx x x tgx sen 1 1 1 . sen 3 + + + − 3 0 sen 1 1 lim x x tgx x + − + →

=

4 1

12.

a x a x a x − − → sen sen lim

= ? à

a x a x a x − − → sen sen lim

=

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

=

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

=

a cos

Logo

a x a x a x − − → sen sen lim

= cosa

(3)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

13.

(

)

a x a x a sen sen lim 0 − + →

= ? à

(

)

a x a x a sen sen lim 0 − + →

=

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

=

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

=

x cos

Logo

(

)

a x a x a sen sen lim 0 − + →

=cosx

14.

(

)

a x a x a cos cos lim 0 − + →

= ? à

(

)

a x a x a cos cos lim 0 − + →

=

a x a x x a x a       − −       + + − → 2 sen . 2 sen 2 lim 0

=

      −       −       + − → 2 . 2 2 sen . 2 2 sen . 2 lim 0 a a a x a

=

      −       −       + − → 2 2 sen . 2 2 sen lim 0 a a a x a

=

x sen −

Logo

(

)

a x a x a cos cos lim 0 − + →

=-senx

15.

a x a x a x − − → sec sec lim

= ? à

a x a x a x − − → sec sec lim

=

a x a x a x − − → cos 1 cos 1 lim

=

a x a x x a a x − − → cos . cos cos cos lim

=

(

x a

)

x a x a a x .cos .cos cos cos lim − − →

=

(

x a

)

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

=

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

=

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

=

a a a cos . cos 1 . 1 . 1 sen

=

a a a cos 1 . cos sen

=

tga sec

.

a

Logo

a x a x a x − − → sec sec

lim

=

tga sec

.

a

16.

x x x 1 sec lim 2 0 − →

= ? à

x x x 1 sec lim 2 0 − →

=

(

x

)

x x x x cos 1 1 . cos 1 . sen 1 lim 2 2 0 + − →

=

2

( )

x x x f cos 1 1 2 − =

=

x x x cos 1 cos 2 −

=

(

x

)

x x cos 1 . 1 cos . 2 − −

=

(

)

(

)

(

x

)

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

=

x 1 1 cos 1 1 2 −

=

sen x 1 1 1 2

(4)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

17.

tgx gx x − − → 1 cot 1 lim 4 π

= ? à

tgx gx x − − → 1 cot 1 lim 4 π

=

tgx tgx x − − → 1 1 1 lim 4 π

=

tgx tgx tgx x − − → 1 1 lim 4 π

=

tgx tgx tgx x − − − → 1 ) 1 .( 1 lim 4 π

=

x tgx 1 lim 4 − →π

= 1

Logo

tgx gx x − − → 1 cot 1 lim 4 π

= -1

18.

x x x 2 3 0 sen cos 1 lim − →

= ? à

x x x 2 3 0 sen cos 1 lim − →

=

(

)

(

)

x x x x x 2 2 0 1 cos cos cos 1 . cos 1 lim − + + − →

=

(

)

(

)

(

x

)(

x

)

x x x x 1 cos .1 cos cos cos 1 . cos 1 lim 2 0 − + + + − →

=

x x x x 1 cos cos cos 1 lim 2 0 + + + →

=

2 3

Logo

x x x 2 3 0 sen cos 1 lim − →

=

2 3

19.

x x x 1 2.cos 3 sen lim 3 − →π

= ? à

x x x 1 2.cos 3 sen lim 3 − →π

=

(

)

1 cos . 2 1 . sen lim 3 x x x + − →π

=

− 3

( )

x x x f cos . 2 1 3 sen − =

=

(

)

x x x cos . 2 1 2 sen − +

=

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

=

(

)

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

=

(

)

[

]

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

=

[

]

x x x cos . 2 1 1 cos 4 . sen 2 − −

=

(

)(

)

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

=

(

)

1 cos . 2 1 . senx + x

20.

tgx x x x − − → 1 cos sen lim 4 π

= ? à

tgx x x x − − → 1 cos sen lim 4 π

=

x

(

x

)

cos lim 4 − →π

=

2 2 −

( )

tgx x x x f − − = 1 cos sen

=

x x x x cos sen 1 cos sen − −

=

x x x x cos sen 1 cos sen − −

=

x x x x x cos sen cos cos sen − −

=

(

)

x x x x x cos cos sen . 1 cos sen − − −

=

x x x x x sen cos cos . 1 cos sen − − −

=

−cosx

21.

lim

(

3

)

.cossec( )

3 x x

x→ − π

= ? à

xlim→3

(

3−x

)

.cossec(πx)

=

0.∞

( ) (

x 3 x

)

.cossec( x) f = − π

=

(

) ( )

x x π sen 1 . 3−

=

(

x

)

x π π− − sen 3

=

(

x

)

x π π − − 3 sen 3

=

(

)

(

x

)

x − − 3 . 3 sen . 1 π π π π

=

(

)

(

x

)

x π π π π π − − 3 3 sen . 1

à

lim

(

3

)

.cossec( ) 3 x x x→ − π

=

(

)

(

x

)

x x π π π π π − − → 3 3 sen . 1 lim 3

=

π

1

22.

lim

.

sen(

1

)

x

x

x→∝

= ? à

) 1 sen( . lim x x x→∝

=

∞.0 x x x 1 1 sen lim       →∝

=

1 sen lim 0 = → t t t

à Fazendo

  → +∞ → = 0 1 t x x t

(5)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

23.

1

sen

.

3

sen

.

2

1

sen

sen

.

2

lim

2 2 6

+

+

x

x

x

x

x π

= ? à

1 sen . 3 sen . 2 1 sen sen . 2 lim 2 2 6 − + − + → x x x x x π

=

x x x 1 sen sen 1 lim 6− + + →π

=

6 sen 1 6 sen 1 π π + − +

=

2 1 1 2 1 1 + − +

=

3

à

( )

1 sen . 3 sen . 2 1 sen sen . 2 2 2 + − − + = x x x x x f

=

(

)

(

sen 1

)

. 2 1 sen 1 sen . 2 1 sen −       +       x x x x

=

(

)

(

sen

1

)

1

sen

+

x

x

=

x x sen 1 sen 1 + − +

24.

(

)

      − → 1 . 2 lim 1 x tg x x π

= ? à

(

)

      − → 1 . 2 lim 1 x tg x x π

=

0

.

à

( ) (

)

      − = 2 . 1 x tg x x f π

=

(

)

      − − 2 2 cot . 1 x g π πx

=

(

)

      − − 2 2 1 x tg x π π

=

(

)

      − − 2 2 2 . 1 . 2 x tg x π π π π

=

( )

x x tg −       − 1 . 2 2 2 2 π π ππ

=

      −       − 2 2 2 2 2 x x tg π π π ππ

à

(

)

      − → 1 . 2 lim 1 x tg x x π

=

      −       − → 2 2 2 2 2 lim 1 x x tg x π π π ππ

=

( )

t t tg t 0 lim 2 → π

=

π

2

Fazendo uma mudança de variável,

temos :

   → → − = 0 1 2 t x x x t π π

25.

( )

x x x senπ 1 lim 2 1 − →

= ? à

( )

x x x senπ 1 lim 2 1 − →

=

(

)

(

x

)

x x x π π π π π − − + → .sen 1 lim 1

=

π

2

( )

x x x f π sen 1− 2 =

=

(

(

)(

)

)

x x x π π − + − sen 1 . 1

=

(

)

(

x

)

x x − − + 1 sen 1 π π

=

(

)

(

x

)

x x − − + 1 . sen . 1 π π π π

=

(

)

(

x

)

x x π π π π π − − + sen . 1

26.

 −

g

x

g

x

x

cot

2

.

cot

2

lim

0

π

= ? à

 −

g

x

g

x

x

cot

2

.

cot

2

lim

0

π

=

.

0

( )

      − = g x g x x f 2 cot . 2 cot π

=

cotg2x.tgx

=

x

tg

tgx

2

=

x tg tgx tgx 2 1 2 −

=

tgx x tg tgx . 2 1 . 2 −

=

2 1−tg2x       − → g x g x x cot 2 .cot 2 lim 0 π

=

2 1 lim 2 0 x tg x − →

=

2 1

(6)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

BriotxRuffini :

1

0

0

...

0

-1

1

1

1

...

1

1

1

1

1

...

1

0

28.

x x x x x cos sen 1 2 cos 2 sen lim 4 − − − →π

= ? à

x x x x x cos sen 1 2 cos 2 sen lim 4 − − − →π

=

x

(

x

)

cos . 2 lim 4 − →π

=

2.cos4 π

=

2 2 . 2 −

=

2

( )

x x x x x f sen cos 1 2 cos 2 sen − − − =

=

(

)

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

=

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

=

x x x x x sen cos cos 2 cos . sen . 2 2 − −

=

(

)

x x x x x sen cos sen cos . cos . 2 − − −

=

−2.cosx

29.

(

)

1

1

2

1

sen

lim

1

x

x

x

= ? à

(

)

1 1 2 1 sen lim 1 − − − → x x x

=

(

)

(

)

1 1 1 2 . 1 1 sen . 2 1 lim 1 + − − − → x x x x

= 1

( )

( )

1 1 2 1 sen − − − = x x x f

=

(

)

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

=

(

)

1 1 1 2 . 1 1 2 1 sen − + − − − x x x

=

(

(

)

)

1 1 1 2 . 1 . 2 1 sen − + − − x x x

=

(

)

(

)

1 1 1 2 . 1 1 sen . 2 1 − + − − x x x

30.

3 cos . 2 1 lim 3 π π − − → x x x

= ? à

3 cos . 2 1 lim 3 π π − − → x x x

=

                        + → 2 3 2 3 sen . 2 3 sen . 2 lim 3 x x x x π π π π

=

. 2 3 3 sen . 2        π +π

=

. 2 3 2 sen . 2         π

=

. 3 sen . 2      π

=

3 2 3 . 2 =

( )

3 cos . 2 1 π − − = x x x f

=

3 cos 2 1 . 2 π −       x x

=

3 cos 3 cos . 2 π π −       x x

=

( )

        −                 + − 2 3 . 2 . 1 2 3 sen . 2 3 sen 2 . 2 x x x π π π

=

                        + 2 3 2 3 sen . 2 3 sen . 2 x x x π π π

=

                        + 2 3 2 3 sen . 2 3 sen . 2 x x x π π π

31.

x x x x .sen 2 cos 1 lim 0 − →

= ? à

x x x x .sen 2 cos 1 lim 0 − →

=

x x x sen . 2 lim 3 π

= 2

(7)

Limites Trigonométricos Resolvidos

Sete páginas e 34 limites resolvidos

( )

x x x x f sen . 2 cos 1− =

=

(

)

x

x

x

sen

.

sen

2

1

1

2

=

x x x sen . sen 2 1 1− + 2

=

x

x

x

sen

.

sen

.

2

2

=

x x sen . 2

32.

x x x x 1 sen 1 sen lim 0 + − − →

= ? à

x x x x 1 sen 1 sen lim 0 + − − →

=

x x x x x 2.sen sen 1 sen 1 lim 0 − + + →

=

2

.

1

1

1

+

=1

( )

x x x x f sen 1 sen 1+ − − =

=

(

(

)

)

x x x x x sen 1 sen 1 sen 1 sen 1 . − − + − + +

=

(

)

x x x x x sen 1 sen 1 sen 1 sen 1 . + − + − + +

=

(

)

x x x x sen . 2 sen 1 sen 1 . + + −

=

x x x x sen . 2 sen 1 sen 1+ + −

=

1 . 2 1 1+

= 1

33.

x x x x cos sen 2 cos lim 0 − →

=

1 sen cos lim 0 x x x + →

=

2 2 2 2 +

=

2

( )

x x x x f sen cos 2 cos − =

=

(

(

)(

)

)

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

=

(

)

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

=

(

)

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

=

(

)

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

=

1 sen cosx+ x

=

2 2 2 2 +

= 2

34.

3 sen . 2 3 lim 3 π π − − → x x x

= ? à

3 sen . 2 3 lim 3 π π − − → x x x

=

3 sen 2 3 . 2 lim 3 π π −         − → x x x

=

3 sen 3 sen . 2 lim 3 π π π −       x x x

=

3 2 3 cos . 2 3 sen . 2 lim 3 π π π π −                         +             − → x x x x

=

3 3 2 3 3 cos . 2 3 3 sen . 2 lim 3 π π π π −                         +             − → x x x x

=

(

)

3 3 . 1 6 3 cos . 6 3 sen . 2 lim 3 x x x x − −           +       − → π π π π

35. ?

Referências

Documentos relacionados

Os direitos políticos, apesar de sua importância, não são absolutos. Por conta disso, a Constituição Federal veda que sejam cassados, mas admite a sua suspensão e perda nos

Considerando tal contexto, o Curso Superior de Tecnologia em Mídias Sociais Digitais do Centro Universitário Belas Artes de São Paulo visa desenvolver e aprimorar o profissional

Início das operações no Brasil, com a produção de papel não revestido na fábrica de Mogi Guaçu, no estado de São Paulo.. Início das operações no Brasil, com a produção

pelo Banco Central do Brasil que tenham clientes pessoas físicas ou pessoas jurídicas classificadas como microempresas na forma da legislação própria devem instituir

A aplicação de aparas de couro wet blue como carga em materiais cerâmicos é citada pela literatura como alternativa para inertização de resíduos provenientes da indústria de

Tendo sempre em vista o respeito aos direitos humanos, buscamos expor o mínimo possível a intimidade da personagem Ana Giselle, dando ênfase ao seu histórico de lutas e à forma como

 Fase III: avaliação da eficácia da vacina, baseada na demonstração da eficácia contra a doença da dengue, virologicamente confirmada, resultante da infecção por

Cerâmicas tradicionais como telhas, blocos e revestimento cerâmicos usualmente apresentam grande variação de composição devido ao largo intervalo de composições das