Repaso 1 Calculo I 18102021 PDF

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REPASO 1 CALCULO I 18102021

PUNTO VII # 6

lim

√ 2+ x− √2 ∗√ 2+ x +√ 2 x √2+ x + √ 2

x →0

¿ lim x→ 0

2

2

( √ 2+ x ) −( √ 2 ) 2+ x−2 =lim x →0 x ( √ 2+ x+ √ 2 ) x → 0 x ( √ 2+x+ √ 2 )

=lim

1 1 x 1 = = =lim x ( √2+ x + √ 2 ) x→ 0 ( √2+ x+ √2 ) ( √ 2 + 0 +√ 2) 2 √ 2

Recordar a2 −b2=(a+ b)( a−b)

PUNTO VII #9

lim −x x→0 1 lim 4 ( 1 )− ( x +4 ) ( 1 ) lim 4− x − 4 lim −x 1 ∗1 − x→ 0 4 ( x+4 ) x → 0 4 (x + 4 ) 4 ( x+4 ) 4 ( x +4 ) x +4 4 x →0 lim = = = =¿ = x x x x x x→ 0 1 lim −1 x→ 0

∗1 lim −1 4 ( x+ 4 ) −1 −1 x→ 0 = = = 1 4 ( x+4 ) 4 (0+ 4 ) 16

PUNTO IV #12

lim cosx=cos x→

5π 3

( 53π )= 21

PUNTO IV #13

lim ( ln 3 x + e x ) =ln 3( 1 )+ e1 =ln 3+e x→ 1

PUNTO V, 1-F

SI

lim x→ c

lim ( 3 g ( x ) ) 3 lim g ( x ) 3 lim g ( x ) 3 g( x ) x→ c x→c x→c = = = 3 3 3 3 √ 3 f ( x )−2 g ( x) lim √ 3 f ( x )−2 g ( x ) lim (3 f ( x ) −2 g ( x ) ) lim 3 f ( x )−lim 2 g ( x )

√

x→ c

3 lim g(x ) ¿

x→c

√(3 lim f ( x )−2 lim g(x )) 3

x →c

x→ c

3(−2 ) −6 =3 √3 (5 )−2(−2) √ 19

=3

x→c

PUNTO VI #9

lim x →0

sen 3 t sen 3 t ∗3 lim ∗3 2t 3t 3 3 sen 3 t 3 = x→0 = (1) = = lim 2 x→ 0 3 t 3 2 2 2

√(

x →c

x →c

)

PUNTO VI #8

senx 1 senx cosx− senx cosx −senx − 1 cosx cosx cosx cosx =lim =lim =lim lim x→ 0 senx−cosx x →0 senx−cosx x→ 0 senx−cosx x → 0 senx−cosx 1 1−

lim cosx − senx x→ 0

¿

cosx senx−cosx

lim −1 ∗1

x →0

cosx = 1

∗1

lim−1 =

x →0

cosx

=

−1 =−1 cos 0...