l'hôpitalovo pravilo - unizg.hr

L'Hôpitalovo pravilo

. . . olak²ava ra£unanje limesa oblika

0

0i

∞∞

.

Teorem. (L'Hôpitalovo pravilo) Neka su zadane f , g : D ⊆ R→ R i c ∈ R ∪ {±∞}. Vrijedi

limx→c

f (x)

g(x)= lim

x→c

f ′(x)

g ′(x)

ako su zadovoljeni sljede¢i uvjeti:

(i) limx→c f (x) i limx→c g(x) su oba 0 ili oba beskona£ni.

(ii) Postoji limx→cf ′(x)g ′(x) ∈ R ∪ {±∞}.

Analogna tvrdnja vrijedi i za jednostrane limese.

Primjer 1(a)

Izra£unajmo pomo¢u L'Hôpitalova pravila limes

limx→0

sin x

x.

Imamo

limx→0

sin x

x=

(0

0

)

L'H= lim

x→0

(sin x)′

x ′

= limx→0

cos x

1

= limx→0

cos x

= cos 0

= 1.

Primjer 1(a)


limx→0

sin x

x.

Imamo

limx→0

sin x

x=

(0

0

)L'H= lim

x→0

(sin x)′

x ′

= limx→0

cos x

1

= limx→0

cos x

= cos 0

= 1.

Primjer 1(b)


limx→0

1− cos x

x2.

Imamolimx→0

1− cos x

x2=

(0

0

)

L'H= lim

x→0

(1− cos x)′

(x2)′

= limx→0

sin x

2x

=

(0

0

)L'H= lim

x→0

(sin x)′

(2x)′

= limx→0

cos x

2

=cos 0

2

=1

2.

Primjer 1(b)


limx→0

1− cos x

x2.

Imamolimx→0

1− cos x

x2=

(0

0

)L'H= lim

x→0

(1− cos x)′

(x2)′

= limx→0

sin x

2x

=

(0

0

)L'H= lim

x→0

(sin x)′

(2x)′

= limx→0

cos x

2

=cos 0

2

=1

2.

Primjer 1(b)


limx→0

1− cos x

x2.

Imamolimx→0

1− cos x

x2=

(0

0

)L'H= lim

x→0

(1− cos x)′

(x2)′

= limx→0

sin x

2x

=

(0

0

)

L'H= lim

x→0

(sin x)′

(2x)′

= limx→0

cos x

2

=cos 0

2

=1

2.

Primjer 1(b)


limx→0

1− cos x

x2.

Imamolimx→0

1− cos x

x2=

(0

0

)L'H= lim

x→0

(1− cos x)′

(x2)′

= limx→0

sin x

2x

=

(0

0

)L'H= lim

x→0

(sin x)′

(2x)′

= limx→0

cos x

2

=cos 0

2

=1

2.

Napomena

Jo² neki neodre�eni oblici mogu se svesti na 00ili ∞∞ . Primjerice, za neodre�eni oblik 0 · ∞ to

moºemo posti¢i tako da jedan od njegovih faktora shvatimo kao nazivnik nazivnika dvojnog

razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.

Primjer. Imamo limx→0+

x · ln x

= (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′(1x

)′= lim

x→0+

1x

− 1x2

= limx→0+

(−x) = 0.

Napomena



razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.


x · ln x = (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′(1x

)′= lim

x→0+

1x

− 1x2

= limx→0+

(−x) = 0.

Napomena



razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.


x · ln x = (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)

L'H= lim

x→0+

(ln x)′(1x

)′= lim

x→0+

1x

− 1x2

= limx→0+

(−x) = 0.

Napomena



razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.


x · ln x = (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′(1x

)′

= limx→0+

1x

− 1x2

= limx→0+

(−x) = 0.

Napomena



razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.


x · ln x = (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′(1x

)′= lim

x→0+

1x

− 1x2

= limx→0+

(−x) = 0.

Napomena



razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.


x · ln x = (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′(1x

)′= lim

x→0+

1x

− 1x2

= limx→0+

(−x)

= 0.

Napomena



razlomka:

0 · ∞ =01∞

ili 0 · ∞ =∞10

.


x · ln x = (0 · (−∞))

= limx→0+

ln x1x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′(1x

)′= lim

x→0+

1x

− 1x2

= limx→0+

(−x) = 0.

Primjer 2: L'Hôpitalovo pravilo ne pomaºe ba² svaki put!

Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)lim

x→+∞

(x + sin x)′

(x − sin x)′= lim

x→+∞

ne moºemo izra£unati vrijednost limesa na lijevoj strani. Ali moºemo sljede¢im ra£unom:

limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,

pri £emu u predzadnjoj jednakosti koristimo da je limx→+∞sin xx = 0: kako je za sve x ∈ R

−1 ≤ sin x ≤ 1, za sve x ∈ 〈0,+∞〉 imamo

−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x

pa je limx→+∞sin xx = 0 (ovdje primjenjujemo tzv. Teorem o sendvi£u).


Ra£unom

limx→+∞

x + sin x

x − sin x

=

(+∞+∞

)lim

x→+∞

(x + sin x)′


x→+∞


limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)

limx→+∞

(x + sin x)′


x→+∞


limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H= lim

x→+∞

(x + sin x)′

(x − sin x)′

= limx→+∞


limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H= lim

x→+∞

(x + sin x)′


x→+∞

1 + cos x

1− cos x


limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H= lim

x→+∞

(x + sin x)′


x→+∞

1 + cos x

1− cos x︸︷︷︸periodi£na

nekonstantnafunkcija

ne postoji


limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H

6= limx→+∞

(x + sin x)′


x→+∞

1 + cos x



ne postoji

ne moºemo izra£unati vrijednost limesa na lijevoj strani.

Ali moºemo sljede¢im ra£unom:

limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H

6= limx→+∞

(x + sin x)′


x→+∞

1 + cos x



ne postoji


limx→+∞

x + sin x

x − sin x

= limx→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H

6= limx→+∞

(x + sin x)′


x→+∞

1 + cos x



ne postoji


limx→+∞

x + sin x

x − sin x= lim

x→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H

6= limx→+∞

(x + sin x)′


x→+∞

1 + cos x



ne postoji


limx→+∞

x + sin x

x − sin x= lim

x→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,

pri £emu u predzadnjoj jednakosti koristimo da je limx→+∞sin xx = 0:

kako je za sve x ∈ R−1 ≤ sin x ≤ 1, za sve x ∈ 〈0,+∞〉 imamo

−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H

6= limx→+∞

(x + sin x)′


x→+∞

1 + cos x



ne postoji


limx→+∞

x + sin x

x − sin x= lim

x→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0

= 1,

pri £emu u predzadnjoj jednakosti koristimo da je limx→+∞sin xx = 0:

kako je za sve x ∈ R−1 ≤ sin x ≤ 1, za sve x ∈ 〈0,+∞〉 imamo

−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x



Ra£unom

limx→+∞

x + sin x

x − sin x=

(+∞+∞

)L'H

6= limx→+∞

(x + sin x)′


x→+∞

1 + cos x



ne postoji


limx→+∞

x + sin x

x − sin x= lim

x→+∞

x + sin x

x − sin x·

1x1x

= limx→+∞

1 + sin xx

1− sin xx

=1 + 0

1− 0= 1,



−1

x︸︷︷︸→0

kad x→+∞

≤ sin x

x≤ 1

x︸︷︷︸→0

kad x→+∞

Γ sin xx

Γ 1

x

Γ− 1

x


Zadatak 43(a)

Izra£unajte limes limx→0+

ln x

ctg x.

Rje²enje. Imamo

limx→0+

ln x

ctg x

=

(−∞+∞

)L'H= lim

x→0+

(ln x)′

(ctg x)′

= limx→0+

1x

− 1sin2 x

= limx→0+

− sin2 x

x

=

(0

0

)L'H= lim

x→0+

(− sin2 x

)′x ′

= limx→0+

−2 sin x · cos x

1

= −2 sin 0 · cos 0

= 0.

Zadatak 43(a)


ln x

ctg x.

Rje²enje. Imamo

limx→0+

ln x

ctg x=

(−∞+∞

)

L'H= lim

x→0+

(ln x)′

(ctg x)′

= limx→0+

1x

− 1sin2 x

= limx→0+

− sin2 x

x

=

(0

0

)L'H= lim

x→0+

(− sin2 x

)′x ′

= limx→0+

−2 sin x · cos x

1

= −2 sin 0 · cos 0

= 0.

Zadatak 43(a)


ln x

ctg x.

Rje²enje. Imamo

limx→0+

ln x

ctg x=

(−∞+∞

)L'H= lim

x→0+

(ln x)′

(ctg x)′

= limx→0+

1x

− 1sin2 x

= limx→0+

− sin2 x

x

=

(0

0

)L'H= lim

x→0+

(− sin2 x

)′x ′

= limx→0+

−2 sin x · cos x

1

= −2 sin 0 · cos 0

= 0.

Zadatak 43(a)


ln x

ctg x.

Rje²enje. Imamo

limx→0+

ln x

ctg x=

(−∞+∞

)L'H= lim

x→0+

(ln x)′

(ctg x)′

= limx→0+

1x

− 1sin2 x

= limx→0+

− sin2 x

x

=

(0

0

)

L'H= lim

x→0+

(− sin2 x

)′x ′

= limx→0+

−2 sin x · cos x

1

= −2 sin 0 · cos 0

= 0.

Zadatak 43(a)


ln x

ctg x.

Rje²enje. Imamo

limx→0+

ln x

ctg x=

(−∞+∞

)L'H= lim

x→0+

(ln x)′

(ctg x)′

= limx→0+

1x

− 1sin2 x

= limx→0+

− sin2 x

x

=

(0

0

)L'H= lim

x→0+

(− sin2 x

)′x ′

= limx→0+

−2 sin x · cos x

1

= −2 sin 0 · cos 0

= 0.

Zadatak 43(b)

Izra£unajte limes limx→1

x3 − 2x2 − x + 2

x3 − 7x + 6.

Rje²enje. Imamo

limx→1

x3 − 2x2 − x + 2

x3 − 7x + 6=

(0

0

)

L'H= lim

x→1

(x3 − 2x2 − x + 2

)′(x3 − 7x + 6)′

= limx→1

3x2 − 4x − 1

3x2 − 7

=3 · 12 − 4 · 1− 1

3 · 12 − 7

=1

2.

Zadatak 43(b)


x3 − 2x2 − x + 2

x3 − 7x + 6.

Rje²enje. Imamo

limx→1

x3 − 2x2 − x + 2

x3 − 7x + 6=

(0

0

)L'H= lim

x→1

(x3 − 2x2 − x + 2

)′(x3 − 7x + 6)′

= limx→1

3x2 − 4x − 1

3x2 − 7

=3 · 12 − 4 · 1− 1

3 · 12 − 7

=1

2.

Zadatak 43(c)

Izra£unajte limes limx→+∞

ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)

L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′

= limx→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)

L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′

= limx→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)

L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′

= limx→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)

L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′

= limx→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)

L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′

= limx→+∞

ex

120

= +∞.

Zadatak 43(c)


ex

x5.

Rje²enje. Imamo

limx→+∞

ex

x5=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(x5)′= lim

x→+∞

ex

5x4

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(5x4)′= lim

x→+∞

ex

20x3

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(20x3)′= lim

x→+∞

ex

60x2

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(60x2)′= lim

x→+∞

ex

120x

=

(+∞+∞

)L'H= lim

x→+∞

(ex)′

(120x)′= lim

x→+∞

ex

120

= +∞.

Napomena

Potpuno analogno kao u zadatku 43(c) pokaºe se da vrijedi:

limx→+∞

ex

p(x)=

{+∞ za svaki polinom p s vode¢im koe�cijentom > 0

−∞ za svaki polinom p s vode¢im koe�cijentom < 0

limx→+∞

p(x)

ex= 0 za svaki polinom p.

Zadatak 43(d)


(1− cos x) · ctg x .

Rje²enje. Imamo

limx→0

(1− cos x) · ctg x =

(0 · ∞︸︷︷︸

+∞ za x→0+−∞ za x→0−

)

= limx→0

(1− cos x) · 1

tg x

= limx→0

1− cos x

tg x

=

(0

0

)L'H= lim

x→0

(1− cos x)′

(tg x)′

= limx→0

sin x1

cos2 x

= limx→0

sin x · cos2 x

= sin 0 · cos2 0

= 0.

Zadatak 43(d)


(1− cos x) · ctg x .

Rje²enje. Imamo

limx→0

(1− cos x) · ctg x =

(0 · ∞︸︷︷︸

+∞ za x→0+−∞ za x→0−

)= lim

x→0(1− cos x) · 1

tg x

= limx→0

1− cos x

tg x

=

(0

0

)L'H= lim

x→0

(1− cos x)′

(tg x)′

= limx→0

sin x1

cos2 x

= limx→0

sin x · cos2 x

= sin 0 · cos2 0

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞))

= limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)

L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′

= limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)

L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − ·2 ln x · 1

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x − x · 2 ln x · 1x1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x −�x · 2 ln x · 1�x

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x −�x · 2 ln x · 1�x

1

= limx→1+

(− ln2 x − 2 ln x

)

= − ln2 1− 2 ln 1

= 0.

Zadatak 43(e)


ln x · ln(x − 1).

Rje²enje. Imamo

limx→1+

ln x · ln(x − 1) = (0 · (−∞)) = limx→1+

ln(x − 1)1

ln x

=

(−∞+∞

)L'H= lim

x→1+

(ln(x − 1))′(1

ln x

)′ = limx→1+

1x−1

− 1ln2 x· 1x

= limx→1+

−x · ln2 x

x − 1

=

(0

0

)L'H= lim

x→1+

(−x · ln2 x

)′(x − 1)′

= limx→1+

− ln2 x −�x · 2 ln x · 1�x

1

= limx→1+

(− ln2 x − 2 ln x

)= − ln2 1− 2 ln 1

= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)

= limx→+∞

(e ln x

) 1

x= lim

x→+∞e

ln xx = e limx→+∞

ln xx = e0 = 1,

pri £emu predzadnja jednakost vrijedi jer je

limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x

= limx→+∞

eln xx = e limx→+∞

ln xx = e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e

ln xx

= e limx→+∞ln xx = e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx

= e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx

= e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)

L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx

= e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′

= limx→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx

= e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1

= limx→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx

= e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x

= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx

= e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx = e0

= 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(a)


x1

x .

Rje²enje. Imamo

limx→+∞

x1

x =((+∞)0

)= lim

x→+∞

(e ln x

) 1

x= lim

x→+∞e


ln xx = e0 = 1,


limx→+∞

ln x

x=

(+∞+∞

)L'H= lim

x→+∞

(ln x)′

x ′= lim

x→+∞

1x

1= lim

x→+∞

1

x= 0.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e

3 ln x4+ln x = e limx→0+

3 ln x4+ln x = e3,

pri £emu zadnja jednakost vrijedi jer je

limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

1= lim

x→0+3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x

= limx→0+

e3 ln x4+ln x = e limx→0+

3 ln x4+ln x = e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

1= lim

x→0+3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e

3 ln x4+ln x

= e limx→0+3 ln x4+ln x = e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

1= lim

x→0+3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

1= lim

x→0+3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)

L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

1= lim

x→0+3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′

= limx→0+

3

1= lim

x→0+3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3x1x

= limx→0+

3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

�x1

�x

= limx→0+

3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

�x1

�x

= limx→0+

3

= 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x

= e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

�x1

�x

= limx→0+

3 = 3.

Zadatak 44(b)


x3

4+ln x .

Rje²enje. Imamo

limx→0+

x3

4+ln x =(00)

= limx→0+

(e ln x

) 3

4+ln x= lim

x→0+e


3 ln x4+ln x = e3,


limx→0+

3 ln x

4 + ln x=

(−∞−∞

)L'H= lim

x→0+

(3 ln x)′

(4 + ln x)′= lim

x→0+

3

�x1

�x

= limx→0+

3 = 3.

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