Funksiya limiti

Bu nima va nega kerak

Harakatdagi jismning oniy (lahzalik) tezligini topish uchun vaqt oralig'i $\Delta t$ ni cheksiz kichraytirib nolga intiltiramiz ($\Delta t \to 0$). Bunda masofa o'zgarishi $\Delta s$ ham nolga intilib, $\left[\dfrac{0}{0}\right]$ ko'rinishidagi noaniqlik hosil bo'ladi.

Spidometr ko'rsatadigan aniq tezlik, egri chiziqqa o'tkazilgan urinmaning qiyaligi va ko'plab fizik jarayonlar aynan funksiya limiti tushunchasiga tayanadi.

Limit — bu matematik analiz (hosila va integral) eshigini ochuvchi asosiy tushunchadir. DTM imtihonlarida $\left[\dfrac{0}{0}\right]$ va $\left[\dfrac{\infty}{\infty}\right]$ noaniqliklarini ochish hamda birinchi ajoyib limitga doir masalalar muntazam tushadi.

Asosiy g'oya

1. Funksiya limiti ta'rifi

Intuitiv ta'rif: Agar argument $x$ biror $x_0$ songa cheksiz yaqinlashganda ($x \to x_0$), $f(x)$ funksiya qiymatlari tayin $A$ songa istalgancha yaqinlashsa, $A$ soni $f(x)$ ning $x \to x_0$ dagi limiti deyiladi: $\lim_{x \to x_0} f(x) = A$

Limit jarayonida $x$ soni $x_0$ ga intiladi, lekin $x_0$ ga teng bo'lishi shart emas. Masalan, $f(x) = \dfrac{x^2 - 1}{x - 1}$ funksiya $x = 1$ da aniqlanmagan, lekin uning $x \to 1$ dagi limiti $2$ ga teng.

2. Limitlarning asosiy xossalari

Agar $\lim_{x \to x_0} f(x) = A$ va $\lim_{x \to x_0} g(x) = B$ mavjud bo'lsa:

1. $\lim_{x \to x_0} [f(x) \pm g(x)] = A \pm B$
2. $\lim_{x \to x_0} [c \cdot f(x)] = c \cdot A$ ($c = \text{const}$)
3. $\lim_{x \to x_0} [f(x) \cdot g(x)] = A \cdot B$
4. $\lim_{x \to x_0} \dfrac{f(x)}{g(x)} = \dfrac{A}{B}$ ($B \neq 0$)

3. Noaniqliklarni ochish usullari

1. To'g'ridan-to'g'ri qo'yish: Maxraj nolga teng bo'lmasa, $x$ o'rniga $x_0$ qo'yiladi.
2. $\left[\dfrac{0}{0}\right]$ noaniqligi (ko'phadlar): Surat va maxrajni ko'paytuvchilarga ajratib, nolga aylanuvchi $(x - x_0)$ hadni qisqartiramiz.
3. $\left[\dfrac{0}{0}\right]$ noaniqligi (irratsional ifodalar): Surat va maxrajni qo'shmasiga ko'paytirib, ildizdan qutilamiz.
4. Birinchi ajoyib limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{\text{tg}\,x}{x} = 1$ Umumiy holatda: $\lim_{x \to 0} \dfrac{\sin(kx)}{mx} = \dfrac{k}{m}$.
5. $\left[\dfrac{\infty}{\infty}\right]$ noaniqligi ($x \to \infty$ da ratsional kasrlar):
   • Darajalar teng bo'lsa: limit bosh koeffitsiyentlar nisbatiga teng: $\lim_{x \to \infty} \dfrac{a_n x^n + \dots}{b_n x^n + \dots} = \dfrac{a_n}{b_n}$.
   • Maxraj darajasi katta bo'lsa: limit $0$ ga teng.
   • Surat darajasi katta bo'lsa: limit $\infty$ ga teng.

Yechilgan misollar

1-misol. Ratsional kasrda $\left[\dfrac{0}{0}\right]$ noaniqligini ochish

$\lim_{x \to 1} \dfrac{x^2 + 2x - 3}{x^2 - 1}$ limitni hisoblang.

Yechish: $x = 1$ da $\left[\dfrac{0}{0}\right]$ noaniqligi hosil bo'ladi. Surat va maxrajni ko'paytuvchilarga ajratamiz: $\lim_{x \to 1} \frac{(x - 1)(x + 3)}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{x + 3}{x + 1}.$ $x = 1$ ni qo'yamiz: $\frac{1 + 3}{1 + 1} = \frac{4}{2} = 2.$

Javob: $2$.

2-misol. Irratsional ifoda va qo'shmasiga ko'paytirish

$\lim_{x \to 7} \dfrac{\sqrt{x + 2} - 3}{x - 7}$ limitni hisoblang.

Yechish: Surat va maxrajni $(\sqrt{x + 2} + 3)$ ga ko'paytiramiz: $\lim_{x \to 7} \frac{(\sqrt{x + 2} - 3)(\sqrt{x + 2} + 3)}{(x - 7)(\sqrt{x + 2} + 3)} = \lim_{x \to 7} \frac{(x + 2) - 9}{(x - 7)(\sqrt{x + 2} + 3)}.$ $(x - 7) \text{ qisqaradi: } \lim_{x \to 7} \frac{1}{\sqrt{x + 2} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}.$

Javob: $\dfrac{1}{6}$.

3-misol. Cheksizlikdagi limit $\left[\dfrac{\infty}{\infty}\right]$

$\lim_{x \to \infty} \dfrac{6x^3 - 4x + 1}{2x^3 + 5x^2 - 3}$ limitni hisoblang.

Yechish: Surat va maxrajni $x^3$ ga bo'lamiz: $\lim_{x \to \infty} \frac{6 - \dfrac{4}{x^2} + \dfrac{1}{x^3}}{2 + \dfrac{5}{x} - \dfrac{3}{x^3}} = \frac{6 - 0 + 0}{2 + 0 - 0} = \frac{6}{2} = 3.$

Javob: $3$.

4-misol. Birinchi ajoyib limitni qo'llash

$\lim_{x \to 0} \dfrac{1 - \cos(2x)}{x^2}$ limitni hisoblang.

Yechish: $1 - \cos(2x) = 2\sin^2 x$ formulasidan foydalanamiz: $\lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2 \cdot \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 2 \cdot 1^2 = 2.$

Javob: $2$.

Mashq

1. $\lim_{x \to 2} (3x^2 - 5x + 4)$ ni hisoblang.
2. $\lim_{x \to -3} \dfrac{x^2 - 9}{x + 3}$ ni hisoblang.
3. $\lim_{x \to 4} \dfrac{x^2 - 6x + 8}{x - 4}$ ni hisoblang.
4. $\lim_{x \to 1} \dfrac{x^3 - 1}{x - 1}$ ni hisoblang.
5. $\lim_{x \to 0} \dfrac{\sqrt{9 + x} - 3}{x}$ ni hisoblang.
6. $\lim_{x \to \infty} \dfrac{8x^2 + 3x - 5}{2x^2 - 7}$ ni hisoblang.
7. $\lim_{x \to \infty} \dfrac{5x + 4}{3x^2 - 1}$ ni hisoblang.
8. $\lim_{x \to 0} \dfrac{\sin(8x)}{4x}$ ni hisoblang.
9. $\lim_{x \to 0} \dfrac{\text{tg}(6x)}{\sin(2x)}$ ni hisoblang.
10. $\lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}$ ni hisoblang.

Javoblar va yechim yo'llari

1. To'g'ridan-to'g'ri qo'yamiz: $3(2^2) - 5(2) + 4 = 12 - 10 + 4 = 6$. Javob: $6$.

2. $\lim_{x \to -3} \dfrac{(x - 3)(x + 3)}{x + 3} = \lim_{x \to -3} (x - 3) = -6$. Javob: $-6$.

3. $\lim_{x \to 4} \dfrac{(x - 4)(x - 2)}{x - 4} = 4 - 2 = 2$. Javob: $2$.

4. $x^3 - 1 = (x - 1)(x^2 + x + 1) \implies 1^2 + 1 + 1 = 3$. Javob: $3$.

5. Qo'shmasiga ko'paytiramiz: $\dfrac{x}{x(\sqrt{9 + x} + 3)} = \dfrac{1}{\sqrt{9 + x} + 3} \to \dfrac{1}{6}$. Javob: $\dfrac{1}{6}$.

6. Darajalar teng ($2$), koeffitsiyentlar nisbati: $\dfrac{8}{2} = 4$. Javob: $4$.

7. Maxraj darajasi ($2$) katta bo'lgani sababli limit $0$ ga teng. Javob: $0$.

8. $\lim_{x \to 0} \dfrac{\sin(8x)}{4x} = \dfrac{8}{4} = 2$. Javob: $2$.

9. $\lim_{x \to 0} \left(\dfrac{\text{tg}(6x)}{6x} \cdot \dfrac{2x}{\sin(2x)} \cdot \dfrac{6}{2}\right) = 1 \cdot 1 \cdot 3 = 3$. Javob: $3$.

10. $x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3) \implies \lim_{x \to 9} \dfrac{1}{\sqrt{x} + 3} = \dfrac{1}{6}$. Javob: $\dfrac{1}{6}$.

Keng tarqalgan xatolar

1. $\left[\dfrac{0}{0}\right]$ noaniqligini ko'rib, "limit mavjud emas" deb yanglishish. ✗ $\lim_{x \to 2} \dfrac{x - 2}{x^2 - 4} = 0$ deb xulosa qilish. ✓ $\left[\dfrac{0}{0}\right]$ noaniqligi ifodani soddalashtirish kerakligini bildiradi: $\dfrac{x - 2}{(x - 2)(x + 2)} = \dfrac{1}{x + 2} \to \dfrac{1}{4}$.

2. Cheksizlikdagi limitda darajalarni hisobga olmaslik. ✗ $\lim_{x \to \infty} \dfrac{3x + 1}{x^2 + 5} = 3$ deb koeffitsiyentlarni bo'lish. ✓ Maxraj darajasi katta bo'lsa, limit nolga teng ($0$). Koeffitsiyentlar faqat darajalar teng bo'lgandagina olinadi.

3. Birinchi ajoyib limitda koeffitsiyentni unutish. ✗ $\lim_{x \to 0} \dfrac{\sin(5x)}{x} = 1$ deb yozish. ✓ Argument $5x$ bo'lgani uchun: $\lim_{x \to 0} 5 \cdot \dfrac{\sin(5x)}{5x} = 5 \cdot 1 = 5$.

4. Qo'shmasiga ko'paytirganda maxrajni unutish. ✗ Suratni qo'shmaga ko'paytirib, maxrajga yozmasdan tashlab ketish. ✓ Kasr qiymati saqlanishi uchun surat ham, maxraj ham bir xil ifodaga ko'paytiriladi.

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Keyingi qadam

Limit tushunchasi va noaniqliklarni ochish usullarini to'liq egalladik. Keyingi darsda ushbu poydevor asosida hosila ta'rifi va uning geometrik ma'nosi bilan tanishamiz.