1. Masala: Limit topish masalasi berilgan. Limitni topish uchun quyidagi qoidalar va formulalardan foydalanamiz. 2. Limit formulasi: Agar $\lim_{x \to a} f(x) = L$ bo'lsa, demak $f(x)$ $x$ $a$ ga yaqinlashganda $L$ ga yaqinlashadi. 3. Limitni topishda asosiy qoidalar: - To'g'ridan-to'g'ri qiymatni qo'yish. - Agar to'g'ridan-to'g'ri qo'yishda $\frac{0}{0}$ ko'rinish chiqsa, algebraik soddalashtirish yoki L'Hôpital qoidasi qo'llaniladi. - Limitni chap va o'ngdan tekshirish. 4. Misol 1: $\lim_{x \to 2} (3x + 1)$ - To'g'ridan-to'g'ri qo'yamiz: $3(2) + 1 = 7$ - Javob: 7 5. Misol 2: $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ - To'g'ridan-to'g'ri qo'yishda $\frac{0}{0}$ ko'rinish. - Soddalashtiramiz: $\frac{(x-1)(x+1)}{x-1} = x + 1$ - Endi $x \to 1$ qo'yamiz: $1 + 1 = 2$ - Javob: 2 6. Misol 3: $\lim_{x \to 0} \frac{\sin x}{x}$ - Bu limitning qiymati $1$ ekanligi ma'lum. - Javob: 1 7. Misol 4: $\lim_{x \to \infty} \frac{2x^2 + 3}{x^2 - 1}$ - Darajasi teng bo'lgan polinomlar limitida koeffitsientlar nisbatini olamiz. - Javob: $\frac{2}{1} = 2$ 8. Misol 5: $\lim_{x \to 0} \frac{e^x - 1}{x}$ - Bu limit $1$ ga teng. - Javob: 1 9. Misol 6: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ - $\frac{(x-3)(x+3)}{x-3} = x + 3$ - $x \to 3$ qo'yamiz: $3 + 3 = 6$ - Javob: 6 10. Misol 7: $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ - Bu limit $\frac{1}{2}$ ga teng. - Javob: $\frac{1}{2}$ 11. Misol 8: $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ - $\frac{(x-1)(x^2 + x + 1)}{x-1} = x^2 + x + 1$ - $x \to 1$ qo'yamiz: $1 + 1 + 1 = 3$ - Javob: 3 12. Misol 9: $\lim_{x \to 0} \frac{\ln(1+x)}{x}$ - Bu limit $1$ ga teng. - Javob: 1 13. Misol 10: $\lim_{x \to \infty} \frac{3x + 1}{2x + 5}$ - Darajasi teng bo'lgan polinomlar koeffitsientlari nisbatini olamiz: $\frac{3}{2}$ - Javob: $\frac{3}{2}$ 14. Misol 11: $\lim_{x \to 0} \frac{\tan x}{x}$ - Bu limit $1$ ga teng. - Javob: 1 15. Misol 12: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ - $\frac{(x-2)(x+2)}{x-2} = x + 2$ - $x \to 2$ qo'yamiz: $2 + 2 = 4$ - Javob: 4 16. Misol 13: $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$ - $\lim_{x \to 0} \frac{e^{2x} - 1}{x} = 2$ - Javob: 2 17. Misol 14: $\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$ - Soddalashtirish uchun konjugat ko'paytiramiz: - $\frac{\sqrt{x} - 1}{x - 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{x - 1}{(x - 1)(\sqrt{x} + 1)} = \frac{1}{\sqrt{x} + 1}$ - $x \to 1$ qo'yamiz: $\frac{1}{1 + 1} = \frac{1}{2}$ - Javob: $\frac{1}{2}$ 18. Misol 15: $\lim_{x \to 0} \frac{\sin 3x}{x}$ - $\lim_{x \to 0} \frac{\sin 3x}{x} = 3$ - Javob: 3 Har bir misolda limitni topish uchun to'g'ridan-to'g'ri qiymat qo'yish, algebraik soddalashtirish va maxsus limitlar qoidasidan foydalanildi.