1. Masalani bayon qilish: Har bir ketma-ketlik $x_n$ limitini $n \to \infty$da hisoblash kerak. 3.1 \ $x_n = 2\left(\sqrt{n(n^2-2)} - \sqrt{n^3-1}\right)$. \textbf{Yeсhim:} \begin{enumerate} \item Ichki ildizlarni osonlashtiramiz: $\sqrt{n(n^2-2)} = \sqrt{n^3 - 2n}$, $\sqrt{n^3 - 1}$. \item Toq raqamlar o'xshash: $$x_n = 2\left(\sqrt{n^3 - 2n} - \sqrt{n^3 - 1}\right) = 2 \cdot \frac{(n^3 - 2n) - (n^3 - 1)}{\sqrt{n^3 - 2n} + \sqrt{n^3 - 1}} = 2 \cdot \frac{-2n + 1}{\sqrt{n^3 - 2n} + \sqrt{n^3 - 1}}.$$ \item Ildiz ostidagi asosiy daraja $n^{3/2}$, shuning uchun ayirma yaqin $\frac{-2n}{2n^{3/2}} = -\frac{1}{n^{1/2}} \to 0$. \item Natija: $$\lim_{n\to\infty} x_n = 0.$$ \end{enumerate} 3.2 \ $x_n = \frac{(n-1)^{10} + (n-1)^{15}}{3n^{15}}$. \begin{enumerate} \item Katta darajali ko'paytirgichlarni $n^{15}$ bo'yicha yozamiz: $$x_n = \frac{n^{10}(1 - \frac{1}{n})^{10} + n^{15}(1 - \frac{1}{n})^{15}}{3 n^{15}} = \frac{n^{10}(1 - \frac{1}{n})^{10}}{3 n^{15}} + \frac{n^{15}(1 - \frac{1}{n})^{15}}{3 n^{15}}.$$ \item Soddalashtiramiz: $$= \frac{(1 - \frac{1}{n})^{10}}{3 n^5} + \frac{(1 - \frac{1}{n})^{15}}{3}.$$ \item $n \to \infty$da $\frac{1}{n^5} \to 0$, va $\left(1 - \frac{1}{n}\right)^k \to 1$: $$\lim_{n\to\infty} x_n = 0 + \frac{1}{3} = \frac{1}{3}.$$ \end{enumerate} 3.3 \ $x_n = \sqrt{(n^2 + 2)(n^2 - 4)} - \sqrt{n^4 - 9}$. \begin{enumerate} \item Ichki ko'paytmani ochamiz: $$(n^2 + 2)(n^2 - 4) = n^4 - 4 n^2 + 2 n^2 - 8 = n^4 - 2 n^2 - 8.$$ \item Shuning uchun $x_n = \sqrt{n^4 - 2 n^2 - 8} - \sqrt{n^4 - 9}$. \item Ko'paytma ko'paytma argumentlar ayirmasini kasr ko'rinishida yozamiz: $$x_n = \frac{(n^4 - 2 n^2 - 8) - (n^4 - 9)}{\sqrt{n^4 - 2 n^2 - 8} + \sqrt{n^4 - 9}} = \frac{-2 n^2 + 1}{\sqrt{n^4 - 2 n^2 - 8} + \sqrt{n^4 - 9}}.$$ \item Dominant ildiz $n^2$, shuning uchun: $$x_n \approx \frac{-2 n^2}{2 n^2} = -1.$$ \item Limit: $$\lim_{n\to\infty} x_n = -1.$$ \end{enumerate} 3.4 \ $x_n = \frac{\sqrt{n^5 - 8} - n \sqrt{n^3}}{\sqrt{n}}$. \begin{enumerate} \item $n \sqrt{n^3} = n \cdot n^{3/2} = n^{5/2}$. \item Ildizlarni: $\sqrt{n^5 - 8} = n^{5/2} \sqrt{1 - \frac{8}{n^5}}$. \item Shunday qilib: $$x_n = \frac{n^{5/2} \sqrt{1 - \frac{8}{n^5}} - n^{5/2}}{n^{1/2}} = n^2 (\sqrt{1 - \frac{8}{n^5}} - 1).$$ \item $\sqrt{1 + x} \approx 1 + \frac{x}{2}$ (kichik $x$ uchun), shunday qilib: $$\sqrt{1 - \frac{8}{n^5}} - 1 \approx -\frac{4}{n^5}.$$ \item Natija: $$x_n \approx n^2 \cdot \left(-\frac{4}{n^5}\right) = -\frac{4}{n^3} \to 0.$$ \item Limit: $$\lim_{n\to\infty} x_n = 0.$$ \end{enumerate} 3.5 \ $x_n = \sqrt{n^2 - 3 n + 2} - n$. \begin{enumerate} \item $\sqrt{n^2 - 3 n + 2} = n \sqrt{1 - \frac{3}{n} + \frac{2}{n^2}}$. \item $ \sqrt{1 + x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}$, shu bilan: $$\sqrt{1 - \frac{3}{n} + \frac{2}{n^2}} \approx 1 - \frac{3}{2n} + \frac{2}{2 n^2} = 1 - \frac{3}{2n} + O\left(\frac{1}{n^2}\right).$$ \item Shunda: $$x_n = n \left(1 - \frac{3}{2n} + O(\frac{1}{n^2})\right) - n = - \frac{3}{2} + O\left(\frac{1}{n}\right).$$ \item Limit: $$\lim_{n\to\infty} x_n = -\frac{3}{2}.$$ \end{enumerate} 3.6 \ $x_n = n + \sqrt[3]{4 - n^2}$. \begin{enumerate} \item $\sqrt[3]{4 - n^2} = \sqrt[3]{- n^2 (1 - \frac{4}{n^2})} = - n^{2/3} \sqrt[3]{1 - \frac{4}{n^2}}$. \item Limitda $n \to \infty$, shuning uchun: $$x_n = n - n^{2/3} \to \infty.$$ \item Limit mavjud emas (yoki $+\infty$). \end{enumerate} 3.7 \ $x_n = \sqrt{n + 2} - \sqrt{n^2 - 2 n + 3}$. \begin{enumerate} \item Avval tanaffuslar sonlarini taqqoslaymiz (ikkala ildizni ham oson usul bilan yozish qiyin, shuning uchun o'simta-asos metoddan foydalanamiz). \item Asosiy ildiz: $$\sqrt{n^2 - 2n + 3} = n \sqrt{1 - \frac{2}{n} + \frac{3}{n^2}} \approx n - 1 + O\left(\frac{1}{n}\right).$$ \item $\sqrt{n + 2} \approx \sqrt{n}$ va $n - 1 >> \sqrt{n}$. \item Shuning uchun $x_n \approx \sqrt{n} - (n - 1) \to -\infty$. \item Limit yo'q (cheksiz minusga ketadi). \end{enumerate} 3.8 \ $x_n = \sqrt{(n+2)(n+1)} - \sqrt{(n-1)(n+3)}$. \begin{enumerate} \item Izohlamalar: $$(n+2)(n+1) = n^2 + 3 n + 2,$$ $$(n-1)(n+3) = n^2 + 2 n - 3.$$ \item $x_n = \sqrt{n^2 + 3 n + 2} - \sqrt{n^2 + 2 n - 3}$. \item Farqni kasrga aylantiramiz: $$x_n = \frac{(n^2 + 3 n + 2) - (n^2 + 2 n - 3)}{\sqrt{n^2 + 3 n + 2} + \sqrt{n^2 + 2 n - 3}} = \frac{n + 5}{\sqrt{n^2 + 3 n + 2} + \sqrt{n^2 + 2 n - 3}}.$$ \item Dominat ildiz taxminan $2 n$: $$x_n \approx \frac{n}{2 n} = \frac{1}{2}.$$ \item Limit: $$\lim_{n\to\infty} x_n = \frac{1}{2}.$$ \end{enumerate} 3.9 \ $x_n = n^2 \left(\sqrt{n(n^4 - 1)} - \sqrt{n^5 - 8}\right)$. \begin{enumerate} \item Ichki ildizlar: $$\sqrt{n(n^4 - 1)} = \sqrt{n^5 - n}, \quad \sqrt{n^5 - 8}.$$ \item Farqni kasrga aylantiramiz: $$x_n = n^2 \cdot \frac{(n^5 - n) - (n^5 - 8)}{\sqrt{n^5 - n} + \sqrt{n^5 - 8}} = n^2 \cdot \frac{- n + 8}{\sqrt{n^5 - n} + \sqrt{n^5 - 8}}.$$ \item Denominat taxminan $2 n^{5/2}$: $$x_n \approx n^2 \cdot \frac{8 - n}{2 n^{5/2}} = \frac{n^2 (8-n)}{2 n^{5/2}} = \frac{8 n^2 - n^3}{2 n^{5/2}} = \frac{8 n^2}{2 n^{5/2}} - \frac{n^3}{2 n^{5/2}} = 4 n^{-(1/2)} - \frac{n^{(3 - 5/2)}}{2} = 4 n^{-1/2} - \frac{n^{1/2}}{2}.$$ \item Limitda $n^{1/2} \to \infty$, demak $x_n \to -\infty$. \end{enumerate} 3.10 \ $x_n = \frac{n^3 - 3 n^2 + 4}{n^3 - 5 n^2}$. \begin{enumerate} \item Katta darajali $n^3$ bo'yicha qisqartiramiz: $$x_n = \frac{n^3 (1 - \frac{3}{n} + \frac{4}{n^3})}{n^3 (1 - \frac{5}{n})} = \frac{1 - \frac{3}{n} + \frac{4}{n^3}}{1 - \frac{5}{n}}.$$ \item $n \to \infty$ uchun: $$\lim_{n\to\infty} x_n = \frac{1 - 0 + 0}{1 - 0} = 1.$$ \end{enumerate} 3.11 \ $x_n = \sqrt{n^2 + 3 n - 2} - \sqrt{n^2 - 3}$. \begin{enumerate} \item Farqni kasrga aylantiramiz: $$x_n = \frac{(n^2 + 3 n - 2) - (n^2 - 3)}{\sqrt{n^2 + 3 n - 2} + \sqrt{n^2 - 3}} = \frac{3 n + 1}{\sqrt{n^2 + 3 n - 2} + \sqrt{n^2 - 3}}.$$ \item Ildiz taxminan $2 n$: $$x_n \approx \frac{3 n}{2 n} = \frac{3}{2}.$$ \item Limit: $$\lim_{n\to\infty} x_n = \frac{3}{2}.$$ \end{enumerate} 3.12 \ $x_n = \sqrt{n\left(\sqrt{n+2} - \sqrt{n-3}\right)}$. \begin{enumerate} \item Ichki ildizdagi farq: $$\sqrt{n+2} - \sqrt{n-3} = \frac{(n+2) - (n-3)}{\sqrt{n+2} + \sqrt{n-3}} = \frac{5}{\sqrt{n+2} + \sqrt{n-3}}.$$ \item Yuqoridagi ildiz taxminan $2 \sqrt{n}$: $$\sqrt{n+2} + \sqrt{n-3} \approx 2 \sqrt{n}.$$ \item Shunday qilib $$x_n = \sqrt{n \cdot \frac{5}{2 \sqrt{n}}} = \sqrt{\frac{5n}{2 \sqrt{n}}} = \sqrt{\frac{5 \sqrt{n}}{2}} = \sqrt{\frac{5}{2}} n^{1/4}.$$ \item $n^{1/4} \to \infty$, limit mavjud emas. \end{enumerate} 3.13 \ $x_n = \sqrt{n(n+5)} - n$. \begin{enumerate} \item $\sqrt{n(n+5)} = \sqrt{n^2 + 5 n} = n \sqrt{1 + \frac{5}{n}}$. \item $\sqrt{1 + x} \approx 1 + \frac{x}{2}$ uchun, $$x_n = n \left(1 + \frac{5}{2 n} + O\left(\frac{1}{n^2}\right) \right) - n = \frac{5}{2} + O\left(\frac{1}{n}\right).$$ \item Limit: $$\lim_{n \to \infty} x_n = \frac{5}{2}.$$ \end{enumerate} 3.14 \ $x_n = \sqrt{n^3 + 8(\sqrt{n^3 + 2} - \sqrt{n^3 - 1})}$. \begin{enumerate} \item Farqni ichki ildizlarning chiziqli yaqinlashuvi yordamida hisoblaymiz: $$\sqrt{n^3 + 2} - \sqrt{n^3 - 1} = \frac{3}{\sqrt{n^3 + 2} + \sqrt{n^3 - 1}} \approx \frac{3}{2 n^{3/2}}.$$ \item Shunday qilib: $$x_n = \sqrt{n^3 + 8 \cdot \frac{3}{2 n^{3/2}}} = \sqrt{n^3 + \frac{12}{n^{3/2}}} = n^{3/2} \sqrt{1 + \frac{12}{n^{9/2}}}.$$ \item Ahamiyatga ega bo'lmagan kichik qo'shimcha limitda $1$ga teng: $$ \Rightarrow x_n \approx n^{3/2}.$$ \item Limit $n \to \infty$: mavjud emas, cheksizga ketadi. \end{enumerate} 3.15 \ $x_n = \frac{\sqrt{(n^2+1)(n^2+2)} - \sqrt{n^4+2}}{2 \sqrt{n^3}}$. \begin{enumerate} \item Ichki ildizlarni ko'paytma va yig'indiga yozamiz: $$\sqrt{(n^2+1)(n^2+2)} = \sqrt{n^4 + 3 n^2 + 2}.$$ \item Farqni kasr shaklida yozamiz: $$\frac{\sqrt{n^4 + 3 n^2 + 2} - \sqrt{n^4 + 2}}{2 \sqrt{n^3}} = \frac{\frac{(n^4 + 3 n^2 + 2) - (n^4 + 2)}{\sqrt{n^4 + 3 n^2 + 2} + \sqrt{n^4 + 2}}}{2 \sqrt{n^3}} = \frac{3 n^2}{(\sqrt{n^4 + 3 n^2 + 2} + \sqrt{n^4 + 2}) 2 \sqrt{n^3}}.$$ \item Yuqoridagi ildizlar $\approx 2 n^2$, pastda esa $2 n^{3/2}$, shuning uchun: $$x_n \approx \frac{3 n^2}{2 n^2 \cdot 2 n^{3/2}} = \frac{3 n^2}{4 n^{7/2}} = \frac{3}{4 n^{3/2}} \to 0.$$ \item Limit: $$\lim_{n\to\infty} x_n = 0.$$ \end{enumerate} 3.16 \ $x_n = n - \sqrt{n(n-1)}$. \begin{enumerate} \item $\sqrt{n(n-1)} = \sqrt{n^2 - n} = n \sqrt{1 - \frac{1}{n}} \approx n \left(1 - \frac{1}{2 n}\right) = n - \frac{1}{2}.$ \item Shunday qilib: $$x_n \approx n - \left(n - \frac{1}{2}\right) = \frac{1}{2}.$$ \item Limit: $$\lim_{n\to\infty} x_n = \frac{1}{2}.$$ \end{enumerate} 3.17 \ $x_n = n - \sqrt[3]{n^3 - 5}$. \begin{enumerate} \item $\sqrt[3]{n^3 - 5} = n \sqrt[3]{1 - \frac{5}{n^3}} \approx n \left(1 - \frac{5}{3 n^3} \right) = n - \frac{5}{3 n^2}.$ \item $x_n = n - \left(n - \frac{5}{3 n^2}\right) = \frac{5}{3 n^2} \to 0.$ \item Limit: $$\lim_{n\to\infty} x_n = 0.$$ \end{enumerate} 3.18 \ $x_n = \sqrt[5]{n} \left( \sqrt[5]{\sqrt[5]{n^2}} - \sqrt[5]{n(n+2)} \right)$. \begin{enumerate} \item Ildizlarni yozamiz: $$\sqrt[5]{\sqrt[5]{n^2}} = \sqrt[5]{n^{2/5}} = n^{2/25},$$ $$\sqrt[5]{n(n+2)} = \sqrt[5]{n^2 + 2 n} = (n^2 (1 + \frac{2}{n}))^{1/5} = n^{2/5} (1 + \frac{2}{n})^{1/5}.$$ \item $x_n = n^{1/5} (n^{2/25} - n^{2/5} (1 + \frac{2}{n})^{1/5}) = n^{1/5 + 2/25} - n^{1/5 + 2/5} (1 + \frac{2}{n})^{1/5}$. \item $1/5 = 5/25$, shuning uchun: $$x_n = n^{7/25} - n^{3/5} (1 + O(\frac{1}{n})) = n^{0.28} - n^{0.6} (1 + o(1)) \to -\infty.$$ \item Limit mavjud emas. \end{enumerate} 3.19 \ $x_n = \sqrt{n+2} (\sqrt{n+3} - \sqrt{n-4})$. \begin{enumerate} \item Ichki ildiz farqini kasrga o'tkazamiz: $$\sqrt{n+3} - \sqrt{n-4} = \frac{7}{\sqrt{n+3} + \sqrt{n-4}} \approx \frac{7}{2 \sqrt{n}}.$$ \item Shunday qilib $$x_n = \sqrt{n+2} \cdot \frac{7}{2 \sqrt{n}} \approx \frac{7}{2} \cdot \sqrt{\frac{n+2}{n}} = \frac{7}{2} \cdot \sqrt{1 + \frac{2}{n}} \to \frac{7}{2}.$$ \item Limit: $$\lim_{n\to\infty} x_n = \frac{7}{2}.$$ \end{enumerate} 3.20 \ $x_n = \frac{\sqrt{n^2 + 1} - \sqrt{n^2 + n}}{n - \sqrt{n^2 - n}}$. \begin{enumerate} \item Farqni chiqaramiz: $$\sqrt{n^2 + 1} - \sqrt{n^2 + n} = \frac{(n^2 + 1) - (n^2 + n)}{\sqrt{n^2 + 1} + \sqrt{n^2 + n}} = \frac{1 - n}{\sqrt{n^2 + 1} + \sqrt{n^2 + n}} = \frac{1 - n}{2 n + o(n)}.$$ \item Pastga ham shunday amalni qilamiz: $$n - \sqrt{n^2 - n} = \frac{n^2 - (n^2 - n)}{n + \sqrt{n^2 - n}} = \frac{n}{2 n + o(n)}.$$ \item Shunday qilib, $x_n$ ga tayanamiz: $$x_n = \frac{\frac{1 - n}{2 n + o(n)}}{\frac{n}{2 n + o(n)}} = \frac{1 - n}{n} = \frac{1}{n} - 1 \to -1.$$ \item Limit: $$\lim_{n \to \infty} x_n = -1.$$ \end{enumerate} Javoblar: 3.1: 0 3.2: 1/3 3.3: -1 3.4: 0 3.5: -3/2 3.6: +\infty (limit mavjud emas) 3.7: -\infty (limit mavjud emas) 3.8: 1/2 3.9: -\infty (limit mavjud emas) 3.10: 1 3.11: 3/2 3.12: +\infty (limit mavjud emas) 3.13: 5/2 3.14: +\infty (limit mavjud emas) 3.15: 0 3.16: 1/2 3.17: 0 3.18: -\infty (limit mavjud emas) 3.19: 7/2 3.20: -1