8 9 Qanday xulosaga keldingiz?
2-misol. Populatsiyadagi sichqonlar soni haftalar kechishi bilan
quyidagicha o‘zgaradi (8-rasm):
8-rasm. 3- va 6- hafta oralig‘ida sichqonlar soni o‘rtacha qanday o‘zgargan?
7 haftalik vaqt oralig‘da-chi?
Sichqonlar populatsiyasining o‘sish tezligi
( 240 110)ta sichqon
sichqon
43
(6 3)ta hafta
hafta
−
≈
−
, ya’ni 3- va 6- hafta oralig‘ida
sichqonlar soni haftasiga o‘rtacha 43 taga ko‘paygan.
Xuddi shunday 7 haftada
(315 50)ta sichqon
sichqon
38
(7 0)ta hafta
hafta
−
≈
−
.
7 hafta oralig‘ida sichqonlar soni haftasiga o‘rtacha 38 taga ko‘paygan.
▲
Umumiy holda:
x miqdor
a dan
b gacha o‘zgarganda
y=f (
x ) miqdor
o‘zgarishining
o‘rtacha tezligi ( )
( )
f b f a b a −
−
orttirmalar nisbatiga teng, bu yerda
f (
b )
– f (
a )
– funksiya orttirmasi,
b – a esa argument orttirmasi.
h =
b – a deb belgilasak, o‘rtacha tezlik
(
)
( )
f a h f a h +
−
ko‘rinishni oladi.
10 11 (
)
( )
f a h f a h +
−
kasr suratini
y = f (
x )
funksiyaning argumenti
x ning
h orttirmasiga mos keluvchi
orttirmasi deb atash qabul qilingan.
Kasrning o‘zi esa
ayirmali nisbat deb atashadi.
Mashqlar 6. Nuqtaning to‘g‘ri chiziq bo‘ylab yurgan yo‘li vaqtga qanday bog‘lan-
ganligi 9-rasmdagi grafikda tasvirlangan.
9-rasm. Nuqtaning
a) dastlabki 4 sekund;
b) so‘nggi 4 sekund;
c) 8 sekund mobaynidagi o‘rtacha tezligini toping.
7. 1) Dalaga turli miqdordagi (dozadagi) dori bilan ishlov berilganda 1 m
2
da mavjud bo‘lgan zararli hasharotlar sonining o‘zgarishi
10-rasmdagi
grafikda ko‘rsatilgan.
10 11 10-rasm. a) 1) doza 0 grammdan 10 grammgacha oshirilsa; 2) 4 grammdan
7 gramm gacha oshirilsa, 1 m
2
da mavjud bo‘lgan zararli hasharotlar soni-
ning o‘zgarishini toping.
b) doza 10 grammdan 14 grammgacha oshirilsa, qanday hodisa ro‘y
beradi?
2) Moddiy nuqtaning to‘g‘ri chiziq bo‘yicha harakat qonuni
s (
t ) ning
grafigi rasmda berilgan.
a) s(2), s(3), s(5), s(7) sonlar nechaga teng?
b) Qaysi oraliqlarda funksiya o‘suvchi?
c) Qaysi oraliqda funksiya kamayuvchi?
d) s(3)–s(1), s(5)–s(4), s(7)–s(6), s(8)–s(6) orttirmalarni hisoblang.
12 13 3–4 LIMIT HAQIDA TUSHUNCHA x ning qiymatlari 2 dan kichik bo‘lib, 2 ga yaqinlasha borganda
f (
x )
=x 2
funksiyaning qiymatlari jadvalini qaraylik:
x 1
1,9
1,99
1,999
1,9999
f (
x )
1
3,61
3,9601
≈ 3,996 00 ≈ 3,999 60
Jadvaldan ko‘rinib turibdiki,
x ning qiymatlari 2 ga qancha yaqin bo‘la versa
(
yaqinlashsa ),
f (
x ) funksiyaning mos qiymatlari ham 4 soniga yaqinlasha-
veradi.
Bunday holatda
x argument (o‘zgaruvchi) 2 ga
chapdan yaqinlash- ganda f (
x ) ning qiymatlari 4 soniga
yaqinlashadi deymiz.
Endi
x ning qiymatlari 2 dan katta bo‘lib, 2 ga yaqinlasha borganida
f (
x )
=x 2
funksiyaning qiymatlari jadvalini qaraylik:
x 3
2,1
2,01
2,001
2,0001
f (
x )
9
4,41
4,0401
≈ 4,004 00 ≈ 4,000 40
Bunday holatda
x argument 2 ga
o‘ngdan yaqinlashganda, f (
x ) funksiya
qiymatlari 4 soniga
yaqinlashadi deymiz.
Yuqoridagi ikki holatni umumlashtirib,
x argument 2 ga
yaqinlashganda, f (
x ) ning qiymatlari 4 soniga
yaqinlashadi deymiz va buni quyidagicha
yozamiz:
2
2
lim
4.
x x →
=
Bu yozuv shunday o‘qiladi:
x argument 2 ga yaqinlashganda
, f (
x ) =
x 2
funksiyaning
limiti 4 ga teng.
Umumiy holda
funksiya limiti tushunchasiga quyidagicha yondashi-
ladi:
x≠a bo‘lib, uning qiymatlari
a soniga yaqinlashsa
, f (
x ) ning mos
qiymatlari
A soniga
yaqinlashsin. Bu holda
A sonni
x a ga
yaqinlashganda f (
x ) funk siya ning
limiti deyiladi va bunday belgilanadi:
lim ( )
.
x a f x A →
=
Ayrim hollarda mazkur holatni
x ning qiymatlari
a ga
intilganda f (
x )
funksiya
A ga
intiladi, deymiz.
12 13 lim ( )
.
x a f x A →
=
yozuv o‘rniga
x → a da
f (
x )
→ A yozuv ham qo‘llaniladi.
Eslatma. x ning qiymati
a ga
intilganda x ≠ a sharti bajarilishining
muhimligini aytib o‘tish joiz.
Misol. x → 0 bo‘lganda
2
5
( )
x x f x x +
=
funksiyaning limitini toping.
x ≠ 0
sharti bajarilmasin, ya’ni
x= 0
bo‘lsin.
x =0 qiymatni
f (
x ) ga bevosita
qo‘yib ko‘rsak,
0
0
ko‘rinishdagi
aniqmaslikka ega bo‘lamiz.
Boshqa tomondan,
(5
)
( )
x x f x x +
=
bo‘lgani uchun bu funksiya ushbu
5
, agar
0 bo‘lsa
( )
aniqlanmagan, agar
0 bo‘lsa,
x x f x x +
≠
=
≠
=
ko‘rinishni oladi.
y=f (
x ) funksiyaning grafigi (0; 5) koordinatali nuqtasi “olib tashlangan”
y=x + 5 to‘g‘ri chiziq ko‘rinishida bo‘ladi (11-rasm):
11-rasm. (0; 5) koordinatali nuqta
y = f (
x ) funksiyaning
uzilish nuqtasi deyiladi.
Ko‘rinib turibdiki, bu nuqtadan farqli bo‘lgan nuqtalarda
x ning
qiymatlari 0 ga
yaqinlashganda f (
x ) funksiyaning mos qiymatlari 5 ga
yaqinlashadi, ya’ni uning
limiti mavjud:
2
0
5
lim
5
x x x x →
+
=
.
▲
14 15 Amalda, funksiya limitini topish uchun, lozim bo‘lsa, tegishli
soddalashtirishlarni bajarish maqsadga muvofiq.
1-misol. Limitlarni hisoblang:
a)
2
2
lim
x x →
; b)
2
0
3
lim
x x x x →
+
; c)
2
3
9
lim
3
x x x →
−
−
.
a)
x ning qiymatlari 2 ga
yaqinlashganda x 2
ning qiymatlari 4 ga
yaqin lashadi, ya’ni
2
2
lim
4
x x →
=
.
b)
x ≠ 0 bo‘lgani uchun
2
0
0
0
(
3)
3
lim
lim
lim (x 3) 3
x x x x x x x x x →
→
→
+
+
=
=
+
=
x .
c)
x ≠ 3 bo‘lgani uchun
2
3
3
3
(
3)(
3)
9
lim
lim
lim (
3) 6
3
3
x x x x x x x x x →
→
→
+
−
− =
=
+
=
−
−
.
▲
Mashqlar Limitni hisoblang (
8–11 ):
8. a)
3
1
4
a lim (
4) b lim (5 2 ) c lim (3
1)
x x x x x x →
→−
→
+
−
−
3
1
4
a lim (
4) b lim (5 2 ) c lim (3
1)
x x x x x x →
→−
→
+
−
−
; b)
3
1
4
a lim (
4) b lim (5 2 ) c lim (3
1)
x x x x x x →
→−
→
+
−
−
; c)
3
1
4
a lim (
4) b lim (5 2 ) c lim (3
1)
x x x x x x →
→−
→
+
−
−
d)
2
2
2
2
0
0
d lim (5
3
2) e lim (1
) f lim (
5)
x h x x x h h x →
→
→
−
+
−
+
; e)
2
2
2
2
0
0
d lim (5
3
2) e lim (1
) f lim (
5)
x h x x x h h x →
→
→
−
+
−
+
; f)
2
2
2
2
0
0
d lim (5
3
2) e lim (1
) f lim (
5)
x h x x x h h x →
→
→
−
+
−
+
.
9. a)
5
2
0
a lim5 b lim7 c lim ,
x h x c →
→
→
; b)
5
2
0
a lim5 b lim7 c lim ,
x h x c →
→
→
; c)
5
2
0
a lim5 b lim7 c lim ,
x h x c →
→
→
c – o‘zgarmas son.
10. a)
2
2
1
2
0
0
3
5
1
a lim
b
lim
c lim
d lim
1
x h x x x x h h x x x h x x →
→
→
→
−
+
−
+
; b)
2
2
1
2
0
0
3
5
1
a lim
b
lim
c lim
d lim
1
x h x x x x h h x x x h x x →
→
→
→
−
+
−
+
; c)
2
2
1
2
0
0
3
5
1
a lim
b
lim
c lim
d lim
1
x h x x x x h h x x x h x x →
→
→
→
−
+
−
+
; d)
2
2
1
2
0
0
3
5
1
a lim
b
lim
c lim
d lim
1
x h x x x x h h x x x h x x →
→
→
→
−
+
−
+
.
11. a)
2
2
2
0
0
0
3
5
2
a lim
b
lim
c lim
x x x x x x x x x x x x →
→
→
−
−
−
; b)
2
2
2
0
0
0
3
5
2
a lim
b
lim
c lim
x x x x x x x x x x x x →
→
→
−
−
−
; c)
2
2
2
0
0
0
3
5
2
a lim
b
lim
c lim
x x x x x x x x x x x x →
→
→
−
−
−
.
d)
2
2
3
0
0
0
2
6
3
4
8
d lim
e
lim
f lim
h h h h h h h h h h h h →
→
→
+
−
−
; e)
2
2
3
0
0
0
2
6
3
4
8
d lim
e
lim
f lim
h h h h h h h h h h h h →
→
→
+
−
−
; f)
2
2
3
0
0
0
2
6
3
4
8
d lim
e
lim
f lim
h h h h h h h h h h h h →
→
→
+
−
−
;
g)
2
2
2
1
2
3
2
6
g lim
h
lim
i lim
1
2
3
x x x x x x x x x x x x →
→
→
−
−
− −
−
−
−
; h)
2
2
2
1
2
3
2
6
g lim
h
lim
i lim
1
2
3
x x x x x x x x x x x x →
→
→
−
−
− −
−
−
−
; i)
2
2
2
1
2
3
2
6
g lim
h
lim
i lim
1
2
3
x x x x x x x x x x x x →
→
→
−
−
− −
−
−
−
.
3
1
4
a lim (
4) b lim (5 2 ) c lim (3
1)
x x x x x x →
→−
→
+
−
−
3
1
4
a lim (
4) b lim (5 2 ) c lim (3
1)
x x x x x x →
→−
→
+
−
−
2
2
2
2
0
0
d lim (5
3
2) e lim (1
) f lim (
5)
x h x x x h h x →
→
→
−
+
−
+
2
2
2
2
0
0
d lim (5
3
2) e lim (1
) f lim (
5)
x h x x x h h x →
→
→
−
+
−
+
2
2
2
2
0
0
d lim (5
3
2) e lim (1
) f lim (
5)
x h x x x h h x →
→
→
−
+
−
+
5
2
0
a lim5 b lim7 c lim ,
x h x c →
→
→
5
2
0
a lim5 b lim7 c lim ,
x h x c →
→
→
5
2
0
a lim5 b lim7 c lim ,
x h x c →
→
→