1. Masalani bayon qilamiz: $\lim_{x \to 0} \frac{\cos 4x - \cos^3 4x}{4x^2}$ ni topish kerak. 2. Limitni hisoblash uchun $\cos 4x$ ning $x$ yaqinida Maclaurin qatorini ishlatamiz: $$\cos 4x = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \cdots = 1 - 8x^2 + O(x^4).$$ 3. Endi $\cos^3 4x$ ni hisoblaymiz: $$\cos^3 4x = (\cos 4x)^3 = \left(1 - 8x^2 + O(x^4)\right)^3.$$ 4. Kubni ochamiz va faqat $x^2$ darajasigacha bo'lgan qismlarni olamiz: $$\cos^3 4x = 1 - 3 \cdot 8x^2 + O(x^4) = 1 - 24x^2 + O(x^4).$$ 5. Endi ifodani soddalashtiramiz: $$\cos 4x - \cos^3 4x = \left(1 - 8x^2 + O(x^4)\right) - \left(1 - 24x^2 + O(x^4)\right) = (1 - 8x^2) - (1 - 24x^2) + O(x^4) = 16x^2 + O(x^4).$$ 6. Limit ifodasi: $$\lim_{x \to 0} \frac{16x^2 + O(x^4)}{4x^2} = \lim_{x \to 0} \frac{16x^2}{4x^2} + \lim_{x \to 0} \frac{O(x^4)}{4x^2} = 4 + 0 = 4.$$ 7. Natija: Limit qiymati $4$ ga teng.