Limes Funkcije

10

Lime Limes s funk funkci cije je

Zadatak 10.1 3

4x2  3x + 1 lim = x!1 2x2  2x + 3

4 4x2  3x + 1 : x2 lim = l i m x!1 2x2  2x + 3 : x2 x!1 2  40+0 = = 2: 20+0

x

2 x

+ +

1 x2

3

=

x2

Zadatak 10.2

x2 lim x!2 x

 4 = lim (x  2) (x + 2) = lim (x + 2) = 4: !2  2 !2 x  2 x

x

Zadatak 10.3

x2 lim x!5 x2

 7x + 10 = lim (x  5) (x  2) = lim x  2 = 5  2 = 3:  9x + 20 !5 (x  5) (x  4) !5 x  4 5  4 x

x

Zadatak 10.4

 lim !1

x

x



p  2 x

  10 x

p   10x x + x2  10x  p  2 lim = !1 1 x + x  10x p  2 2 x2  x  10x x2  x2 + 10x p  p  lim = lim = !1 x + x2  10x !1 x + x2  10x 10x :x 10 p  p  lim = lim 10 = !1 x + x2  10x : x !1 1 + x

=

x

=



x



x

=

p  2



x

x

x2

x

x

x

10 q  !1  1+

=

lim

x

x2

10 q  !1 1 + 1 

= lim x

10x

x2

=

1+

10 p 

10

=

10

=

x

10 = 5: 2

Zadatak 10.5

p  3

lim

!0

x

1 + x2  1 x2

p  3

= lim

!0

x

p 1 +  + p 1 +  p   + p 1 + 1+ p 1 +   1 p   + p 1 + + 1 =  1+

1 + x2  1 x2

3

= lim

!0 x2

x

3

3

x2

3

x2

x2

x2

2

3

2

3

x2 + 1

2

3

x2 + 1

3

3

x2

=

= lim

p 1 +  + p 1 +  p 1 +  + p 1 +  p 1 +  + p 1 +

!0 x2 

x

= lim

!0 x2

x

1 + x2  1

3

2

x2

3

x2

3

x2

3

x2

x2

3

2

x2

x2

= lim

!0 x2

x

3

2

x2

 = +1  = +1  = +1

1 1 = lim p  = : p  2 3 x!0 3 2 2 3 1+x + 1+x +1 Zadatak 10.6

p x  1 lim p  !1 x  1 3

x

=

2 4

x

!

smjena x = t6 1 = t

) !1

3 5 = lim! t

t2 1 t3

1 = 1

(t  1) (t + 1) t+1 2 = lim = : t!1 (t  1) (t2 + t + 1) t!1 t2 + t + 1 3

= lim Zadatak 10.7

3 lim p  !3 x + 1  2 x

x

= = = = =

p   3 x+1+2 p   lim p  = !3 x + 1  2 x+1+2 p  (x  3)  x+1+2 lim = p  2 !3 x + 1  22 p  (x  3)  x+1+2 lim = !3 x+14 p  (x  3)  x+1+2 lim = !3 x3 p  x

x

          lim +1+2 =4

x

x

x

x

!3

x

:

Zadatak 10.8

lim

!7

x

p  x3 x2  49

2

= =

p  p  x3 2+ x3  p  lim = !7 x2  49 2+ x3 p  2 22  x3 p  lim 2 = !7 (x  49) 2 + x  3 4  (x  3) li

x

x

2

 





7x p  = x!7 (x  7) (x + 7) 2 + x3  (x  7) p  = lim = x!7 (x  7) (x + 7) 2 + x3 1 p  = lim = x!7 (x + 7) 2 + x3 1 1 =  = : 14  4 56 = lim

 

 





Zadatak 10.9

x2

 16 lim p  !4 x + 5  3

x

p   16 x+5+3 p   lim p  = !4 x + 5  3 x+5+3 p  (x2  16) x + 5 + 3 lim = p  2 !4 2 x+5 3 p  (x  4) (x + 4) x + 5 + 3 lim = !4 x+59 p  (x  4) (x + 4) x + 5 + 3 lim = !4 x4 p  lim (x + 4) x + 5 + 3 = 8  6 = 48: !4 x2

=

x

=



x

=





x

=

x

=



 



x

 



Zadatak 10.10

x3

 27 lim p  !3 12  x  3

x

= = = = = = =

p   33 12  x + 3 p   lim p  = !3 12  x  3 12  x + 3 p  (x3  33 ) 12  x + 3 lim = p  2 !3 12  x  32 p  (x  3) (x2 + 3x + 9) 12  x + 3 lim !3 12  x  9p  (x  3) (x2 + 3x + 9) 12  x + 3 lim !3 3  x p  (x  3) (x2 + 3x + 9) 12  x + 3 lim !3  (xp  3)  lim x2 + 3x + 9 12  x + 3 = !3 27  6 = 162: x3

x



x







  

x

x

x

x







 =  =  =

Zadatak 10.11

lim

!+1

x



x+1 x 1





x

 lim 1 + ! 1  lim 1 + ! 1 

= = =



+

x

lim

!+1

x

= e

2

1+

2x

lim

x!+1

x1

x

 1 = x+1x+1 x1

+

x



x+1 x 1

1

x



x1

2



2

x1

x

=

x

=

= e2 :

Zadatak 10.12

lim

!+1

x



x2 + 5x + 4 x2 3 x + 7



3x



= = =

x

x

+

+

x2 + 5x + 4 x2 + 3x x2 3x + 7

+

x

lim

x!+1 x

lim

1



8x  3 x2  3x + 7

8x3 2 3x+7

3x

=e

lim

=



 

7

3x



=

3x

= x

2 x 3 +7 8x3

2 24x 9x 2 3x+7



8x3 x2 3x+7

3x

=

=

x!+1 x

9 24 x

x!+1 1

= e





8x  3 = lim 1 + 2 x!+1 x  3x + 7 = e

3x

x2 + 5x + 4 x2 3x + 7

 lim 1 + ! 1  lim 1 + ! 1  lim 1 + ! 1 

3+ 7 x2

x

= e24 :

Zadatak 10.13

lim

!+1

x



x2 + x + 1 x2 x + 1



2x



= = = =

x

x

+

+

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

+

x

lim

!+1

x

= e

2x

x2 + x + 1 x2 x + 1

 lim 1 + ! 1  lim 1 + ! 1  lim 1 + ! 1  lim

x!+1 x

1+

2x 2 x+1



2x x2

x 2x

x2

2x

1

=

 

 +1 

x+1 =e



lim

x!+1 x

1



2x

=

2x

= 2 x

x

+1

2x

2 4x 2 x+1



2x x2 x+1

= e4 :

2x

=

Zadatak 10.14

lim

sin7x

!0 x

x

sin7x sin7x  7 = 7  lim = 7  1 = 7: x!0 x!0 7x 7x

= lim

Zadatak 10.15 sin6x sin6x  6 = 6x sin6x 6x 6x lim = lim sin3x = lim sin3 x!0 sin3x x!0  3x x!0 3x x  3 3x

=

x 6 lim sin6 6x

!0

x

x 3 lim sin3 3x

=

!0

x

6 = 2: 3

Zadatak 10.16

p 

sin3x sin3x x+4+2 p   lim p  = lim p  = x!0 x!0 x+42 x+42 x+4+2 p  sin3x x + 4 + 2 = lim p  = 2 x!0 x + 4  22 = = = =

    p  + 4 + 2 sin3 lim = !  + 4 4 p  + 4 + 2 sin3 lim = ! p  + 4 + 2 sin3 3= lim ! 3   p  sin3  lim 3  lim +4+2 = 0

x

x

x

x x

x

x

x x

0

x

x

0

x

x

!0 3x

!0

x

x

= 3  1  4 = 12:

x

Zadatak 10.17

lim

!0

x

p 

x+1 1 sin2x



= lim

p 

!0

x

= lim





  p sin2+ 1 +1 1 + 1 p  + 1 + 1 = lim ! sin2 !0

x

=

p    p x + 1 + 1 = x+1+1 p  p  x+11 p  x + 1 + 1 = x+1 1 sin2x x

2

x

x

0

x

x

x

2



x+1

1  =  p  ! sin2 +1+1 p  + 1 + 1 = lim ! sin2 p  + 1 + 1 = lim ! 2  1  = p  lim ! 2 +1+1 1  = p  1  lim 2 +1+1

= lim = = = =

x

0

x

x x

x

0

x

x

x

x

0 sin2x 2x

x

0 sin2x 2x

x

sin2x 0 2x

x

!

x

=

x

x

1 1 1  = : 1 22 4

Zadatak 10.18

p 

sin4x sin4x x+9+3 p   lim p  = lim p  = x!0 x!0 x+93 x+93 x+9+3 p  sin4x x + 9 + 3 = lim p  = 2 x!0 2 x+9 3 = = = =

    p  + 9 + 3 sin4 lim = !  + 9 9 p  + 9 + 3 sin4 lim = ! p  + 9 + 3 sin4 4= lim ! 4   p  sin4  lim 4  lim +9+3 =

x

x

x

0

x

x

x x

x

x

x x

0

x

0

x

!0 4x

x

= 4  1  6 = 24:

!0

x

x

Zadatak 10.19 sin5x x  5x  sin3  3x sin5x  sin3x 5x 3x lim = lim = sin x x!0 x!0 sin x  x x x x 5 sin5 5x

= lim



!0

x

= lim

x

x 5 sin5 5x

x

x

 3 sin3 3



=

x

x

=

x

x 5lim sin5 5 x 0 x

!

sin3  3lim !0 3

lim

!0

x

x

x

x

=

x

sin x

sin x

!0

x

=

 3 sin3 3

=

sin x x

53 =2 1

Zadatak 10.20

lim

!0

x

1  cos x x2

= lim

!0

x

2sin2 x2

x

2

2sin x2  sin x2

= lim

x x

!0



x

=

sin x2  sin x2 = 2  lim x x x!0  4 = 2 2 sin x2 sin x2 2 1 1  = lim x  lim x =  1  1 = : x!0 4 x!0 2 2 2 2

Zadatak 10.21 2

2

2

2sin2 x2 2sin x2  sin x2 1  cos x2 lim = lim 2 = lim x2 x2 = sin x2 x!0 x2 sin x2 x!0 x sin x2 x!0    4 2 2 x2 2

sin x2 2  lim = 4 x!0 x22 =

2

 lim !0 x

sin x2 x2

2



1

lim sinx2x

=

2

!0

x

1 : 2

Zadatak 10.22

x

lim x  [ln(x + 1)  ln (x  1)] =

!+1

=

lim x  ln

!+1

x

 lim ln ! 1 

x

+



x+1 x 1





x+1 x 1

x

x

x

=

   = +1

 ln lim 1 + ! 1  ln lim 1 + ! 1 

=

+

x

=

+

x

= ln lim

1+

!+1

x

2x

lim

= ln e

x!+1



x+1 x 1

x

 1 = x+1x+1 x1

x



2

x

1



x1

2

2



x1

x

= =

= ln e2 = 2 ln e = 2:

x1

Zadatak 10.23

lim 3x  ln x2 + 5x + 4

 

!+1

x

= = = = =

lim 3x  ln

!+1

x



x2

  ln 

x2 + 5x + 4 x2 3x + 7



x2 x2

x x

3x

x2 x2

x x

3x



 3x + 7

 =

=

 + 5 + 4 lim ln ! 1 3 +7 =  + 5 + 4 ln lim ! 1 3 +7 =  + 5 + 4  + 3  7 ln lim 1 + ! 1  8  3 3 + 7 ln lim 1 + ! 1 3 +7 =  8 3   

x

+

+

x

x2

+

x

x2 x

x x2

x2

+

1+

lim

8x3

!+1

x

x

x2

2 x!+1 x 3x+7

= ln e

lim

8x3

 3x + 7

3x

2 3x+7

lim

3x

=

24x2 9x

2 x!+1 x 3x+7

= ln e

8x3

x2 3x+7

=

9 24 x

x!+1 1

= ln e

=

x

x

= ln lim

3x

3x

x

x

x

3+ 7 x2

x

= ln e24 = 24ln e = 24:

Zadatak 10.24

lim 3x  ln (x + 1) + ln (x  4)  ln x2 + 5x + 6

!+1

x





 =

= = = = = =

 ( + 1) (  4)  lim 3  ln = ! 1 +5 +6   3  4 lim 3  ln = ! 1 +5 +6   3  4 lim ln = ! 1 +5 +6  3 4  1 = ln lim 1 + ! 1 +5 +6   3  4   5  6 ln lim 1 + ! 1  8  10+5 + 6 ln lim 1 + = ! 1 +5 +6  8  10   

x

x

x

x x2

x

+

x2 x2

x

+

x2 x2

+

x x

3x

x x

+

x

x x

x x

x2

x x2

+

x

3x

x2 x2

x2 x

x2

+

x

x

= ln lim

1+

!+1

x

= ln e

x



2

+5x+6 8x10

8x10

3x



x2 +5x+6



=

x2 + 5 x + 6

8x10 2 +5x+6

lim

x!+1 x

=

3x

x

x

3x

x

3x

24x

lim

= ln e



x!+1

2 x 30

= ln e24 = 24:

x2 +5x+6

Zadatak 10.25 x

lim

!+1

x



x+3 x+2



2

2

+x+1

x

x+1

= = = =

 lim 1 + ! 1  lim 1 + ! 1  lim 1 + ! 1  +

x

+

x

+

x

lim

1+

!+1

x

= e

lim

1

x!+1

x+2



x

x+3 x+2



+x+1

x+1

1

= 2 +x+1

x

x+3 x x+2



2

x

 1 +2  1

x

 2

x+1

=

+x+1

x+1

= (x+2)



1 x+2



x

2 +x+1 x+1

x+2

2 +x+1

lim

x

2 +x+1

2 x!+1 x +3x+2

x+1

=e

= = e1 = e:

Zadatak 10.26

x

  lim !2 2

x

1 x2

=

=

2 4

smjena 1 t = x2 = x = 2+ x 2 = t

lim

!



1+

) ) !1

1

2t



1 2

1

= e2

3   1 5 = !lim1 1 + 2 = t

1 t

x

+

t

Zadatak 10.27

2 66 4 =)

3 77 5

smjena ax t=e 1 = eax = t + 1 = = ax = ln (1 + t) = x = a1 ln (1 + t) x 0 = t 0 t t 1 = lim 1 = a lim = a lim ln(1+t) = t!0 t!0 ln (1 + t) t!0 ln (1 + t) a

eax 1 lim x!0 x





)

!

) !

)



= a

1 t) lim ln(1+ t

!0

=a

t

= a

1

lim ln (1 + t)



1 = lim 1t  ln (1 + t)

!0

t

= a

1 t

!0

t

= a

t

1 ln lim (1 + t) t

1 = a: ln e

1

=

t

!0

Zadatak 10.28 Izra µ  cunati lijevi i desni limes funkcije 

3x  x2 f (x) = x4 u ta µ  cki  x = 4: 2

3x  x L(4) = lim x!4 x4 

= lim

!0

"

= D(4) =

2 =4

smjena x=4 " " > 0; " 0

 !

12  3"  16 + 8"  "2

"

 4 lim   5 + ! 2 3  lim =4 ! 4

"

x

0



4+

"

x x

"

x

= lim = lim



!0 !0

"

"

 4

"  5  "

"

2

"

"

0

=

"

"

12 + 3"  16  8"  "2

"

"

 "2 + 5"  4 = lim = !0 " 4 = 0  5 + lim = +1: !0 "

smjena x = 4+" " > 0; " 0

2

3 5 = lim! 3 (4 4 ) (44 )

!

3 5 = lim! 3 (4 +4 +) (44+ ) "

"

0

 "2  5"  4 = lim = !0

"

"

4 = 0  5  lim = 1:

!0 "

"

cunati lijevi i desni limes funkcije  Zadatak 10.29 Izra µ 

"

"

2

=

u ta µ  cki  x =

1:

2 3   1 = ln lim   1 = 4 = 1  5 = lim ln ! ! 0 !0     ln lim (1  )  1 = ln lim 1 + 2 +  1 = !  !  ln lim 2 + = 1 ! 2 3   1 = ln lim   1 = 4 = 1 + 5 = lim ln ! ! 0 !0     ln lim (1 + )  1 = ln lim 1 + 2 +  1 = !  !  ln lim 2 + = 1 smjena

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