Limes funkcije

10 Limes funkcije

Zadatak 10.3 x 2 7x + 10

5) (x 4) x!5 x 4 5 4 Zadatak 10.4

x 2 2 10x x+ x 2 10x lim x

x 10x = lim

= 10x

x!1

x!1

1 x+ x 2

x 2 x 2 10x 2

x 2 x 2 + 10x

= lim

p 2 = lim p

= 10x

x!1

x+ x

10x

x!1 x+ x 2

10

10x

:x

= lim

10x :x x!1 x 2 10x =

p 2 = lim

x!1 x+ x

1+

10 10

= lim

= lim

x!1

1+ 10 x 2 1+ 1 x

10x

x!1

10 10

= 5:

1+ 1 0 2

Zadatak 10.5

1 1+x 2 1 1+x 2 + 1+x 2 +1 lim

1+x 2

2 x!0 = x x!0 x p 3

2 = lim

1+x 2 2 p

+ 3 1+x 2 +1

1+x 2 3 1 3

= lim

= 1+x 2 + 1+x 2 +1

x!0

x 3 x 3 x+1+2 lim p

= lim p

x!3

x+1 2 x!3 x+1 2 x+1+2

Zadatak 10.8 p

2 x 3 2 x 3 2+ x 3 lim

= lim

x!7

x 2 49 x!7 x 2

49 2+ x 3

= lim

x!7 (x 2

49) 2 + x 3

4 (x 3)

= lim

p (x 2 49) 2 + x 3

x!7 x!7

= lim p

x!4

x+5 3 x!4 x+5 3 x+5+3

(x

16) x+5+3

= lim

2 x!4 =

x+5

(x

4) (x + 4)

x+5+3

= lim

x!4

x+5 9

(x

4) (x + 4)

x+5+3

= lim

x!4

= lim (x + 4)

x + 5 + 3 = 8 6 = 48:

x!4

Zadatak 10.10

x 3 27 x 3 3 3 12 x+3 lim p

= lim p

x!3

12 x 3 x!3 12 x 3 12 x+3

(x

3 ) 12 x+3

= lim

x!3

12 2 x 3 2

(x

3) (x 2 + 3x + 9) 12 x+3

= lim

x!3

12 x 9

2 (x p 3) (x + 3x + 9) 12 x+3

= lim

x!3

(x

3) (x 2 + 3x + 9) 12 x+3

= lim

x!3

(x 3)

lim x + 3x + 9

12 x+3=

x!3

Zadatak 10.12 x 2 + 5x + 4 3x

x2 3x+7 8x 3 8x 3x 3 8x 3 x2 3x+7 = lim 1+ 2 =

x! +1 =e x2 =e 3x+7 =

3x+7 3x

x 2x 2 +x+1 x 2 +x+1 lim

2 2 x 2x +x+1 x +x 1

= lim 1+

x!+1

x 2 x+1 2x 2x

= lim 1+

x!+1

x+1

x2 2x 2x x 2x +1 2x

x2 x +1

= lim 1+

x!+1

x 2 x+1

x! =e lim =e +1 x2 x +1 =e 4 :

lim

2x

x +1 2x

4x2

x! +1 x2

Zadatak 10.14 sin 7x

x+4+2 lim p

x+4 2 x!0 x+4 2 x+4+2

p sin 3x x+4+2

p sin 3x x+4+2

p sin 3x x+4+2

= lim

x!0

p x sin 3x x+4+2

Zadatak 10.17 p

x+1 1 x+1 1 x+1+1 lim

x+1 1 x+1+1

= lim

x!0

sin 2x x+1+1 p

x+1 2 1 2

= lim

x!0 sin 2x x+1+1 x!0 sin 2x x+1+1

= lim

x!0 sin 2x x+1+1 x

= lim

x!0 sin 2x x+1+1 x

x+9+3 lim p

x+9 3 x!0 x+9 3 x+9+3

p sin 4x x+9+3

p sin 4x x+9+3

p sin 4x x+9+3

= lim

x!0

p x sin 4x x+9+3

= lim

x!0

4x sin 4x

= 4 lim

lim

x+9+3=

x!0 4x

x!0

5lim sin 5x 3lim sin 3x 5x 3x = x!0

4 x!0 2 x!0 2 2 2 Zadatak 10.21

Zadatak 10.22 x+1

lim x [ln (x + 1) ln (x 1)] = lim x ln =

x!+1

x!+1

x 1 x+1 x

= lim ln

x!+1

x 1 x+1 x

= ln lim

x!+1

3x+7 3x

3x+7 3x

9x x! lim x2 3x+7

+1 1 x+ 3 7 = ln e 24 x2 = ln e = 24 ln e = 24: Zadatak 10.24

x! lim

lim 2 3x ln (x + 1) + ln (x 4) ln x + 5x + 6 =

x!+1 x!+1

8x 10

x2 +5x+6 3x

= ln e = ln e +1 x2 +5x+6 = ln e 24 = 24: Zadatak 10.25

x! lim

+1 x2 +5x+6 3x

x x+3 +x+1 +1 x+3 +1 lim

x2

x +x+1

x x+3 +x+1 x 2 +1

x 1 +x+1 +1

(x+2) x2

x 1 +2

x 1 +x+1 +1

x! =e lim +1 x2 +3x+2 =e 1 = e: Zadatak 10.26

lim x 1 x2 +x+1

x2 +x+1

=e x! +1 +2 x +1

x t 1 lim

x 1 2 smjena

= 4 1 t= 1

x 2 =) x = 2 + t 5 = lim 1+ =

x!2

x!+1

2t

x ! 2 =) t ! 1

2t 1 1 2 1

= lim 1+

=e 2 :

x!+1

2t

Zadatak 10.27

smjena

lim = 6 1 =) e =t+1

t!0 ln(1+t)

=a

ln(1+t) =a

Zadatak 10.28 Izraµcunati lijevi i desni limes funkcije

3x x 2

f (x) = x 4

u taµcki x = 4:

3x 2 x 3 (4 ") (4 ") L(4) = lim

3x 2 x 2 smjena

3 (4 + ") (4 + ") D(4) = lim

Zadatak 10.29 Izraµcunati lijevi i desni limes funkcije

ln x 2 1 ln x 2 1

D(1) = lim

ln x 2 1 = ln lim

1= 4 x=1+" 5=

x!1 +

x!1

" > 0; " ! 0

2 = ln lim 2 (1 + ") 1 = ln lim 1 + 2" + " 1=

"!0

"!0

= ln lim 2 2" + " =

1:

"!0

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