1. We are asked to find the limits as $n \to \infty$ for each sequence $x_n$ in problems 3.1 to 3.20. 2. For each, we simplify the expression and analyze the dominant terms for large $n$, then compute the limits: 3.1 $x_n = 2\left[\sqrt{n(n^2 - 2)} - \sqrt{n^3 - 1}\right] = 2\left[\sqrt{n^3 - 2n} - \sqrt{n^3 - 1}\right]$ - Factor inside radicals: leading term $\sqrt{n^3}$ dominates. - Rewrite: $2\left[n^{3/2}\sqrt{1 - \frac{2}{n^2}} - n^{3/2}\sqrt{1 - \frac{1}{n^3}}\right] = 2n^{3/2}\left[\sqrt{1 - \frac{2}{n^2}} - \sqrt{1 - \frac{1}{n^3}}\right]$ - Use binomial expansion for large $n$: $\sqrt{1 - a} \approx 1 - \frac{a}{2}$. - Approximate difference inside brackets $\approx 1 - \frac{1}{n^3\cdot 2} - \left(1 - \frac{2}{n^2 \cdot 2}\right) = \frac{1}{n^3 \cdot 2} - \frac{2}{n^2 \cdot 2} = -\frac{1}{n^2}$ (dominant term). - So $x_n \approx 2n^{3/2} \left(-\frac{1}{n^2}\right) = -2 n^{3/2 - 2} = -2 n^{-1/2} \to 0$. Answer: $\lim_{n \to \infty} x_n = 0$. 3.2 $x_n = \frac{(n-1)^{10} + (n-1)^{15}}{3 n^{15}}$ - Leading term in numerator is $(n-1)^{15} \approx n^{15}$ for large $n$. - Therefore: $x_n \approx \frac{n^{15}}{3 n^{15}} = \frac{1}{3}$. Answer: $\lim_{n \to \infty} x_n = \frac{1}{3}$. 3.3 $x_n = \sqrt{(n^2+2)(n^2-4)} - \sqrt{n^4 - 9}$ - Expand inside first root: $n^4 + 2 n^2 - 4 n^2 - 8 = n^4 - 2 n^2 - 8$. - So $x_n = \sqrt{n^4 - 2 n^2 - 8} - \sqrt{n^4 - 9}$. - Using $(a-b) = \frac{a^2 - b^2}{a+b}$ for difference of roots: - Numerator: $(n^4 - 2 n^2 - 8) - (n^4 - 9) = -2 n^2 + 1$. - Denominator: $\sqrt{n^4 - 2 n^2 - 8} + \sqrt{n^4 - 9} \approx 2 n^2$ for large $n$. - So $x_n \approx \frac{-2 n^2 + 1}{2 n^2} = -1 + \frac{1}{2 n^2} \to -1$. Answer: $\lim_{n \to \infty} x_n = -1$. 3.4 $x_n = \frac{\sqrt{n^5 - 8} - n \sqrt{n^3}}{\sqrt{n}}$ - $n \sqrt{n^3} = n \cdot n^{3/2} = n^{5/2}$. - Rewrite numerator $\sqrt{n^5 -8} - n^{5/2}$. - Approximate numerator: $\sqrt{n^5 - 8} = n^{5/2} \sqrt{1 - \frac{8}{n^5}} \approx n^{5/2} \left(1 - \frac{4}{n^5}\right) = n^{5/2} - \frac{4}{n^{5/2}}$. - So numerator $\approx n^{5/2} - \frac{4}{n^{5/2}} - n^{5/2} = - \frac{4}{n^{5/2}}$. - Divide by $\sqrt{n} = n^{1/2}$: $x_n \approx -\frac{4}{n^{5/2}} \cdot \frac{1}{n^{1/2}} = -\frac{4}{n^3} \to 0$. Answer: $\lim_{n \to \infty} x_n = 0$. 3.5 $x_n = \sqrt{n^2 - 3 n + 2} - n$ - Rewrite radical: $\sqrt{n^2 - 3n + 2} = n \sqrt{1 - \frac{3}{n} + \frac{2}{n^2}}$. - For large $n$, $\sqrt{1 - \frac{3}{n} + \frac{2}{n^2}} \approx 1 - \frac{3}{2n}$ (using binomial expansion). - So $x_n \approx n \left(1 - \frac{3}{2 n}\right) - n = n - \frac{3}{2} - n = -\frac{3}{2}$. Answer: $\lim_{n \to \infty} x_n = -\frac{3}{2}$. 3.6 $x_n = n + \sqrt[3]{4 - n^2}$ - For large $n$, $\sqrt[3]{4 - n^2} \approx \sqrt[3]{-n^2} = - n^{2/3}$. - Thus $x_n \approx n - n^{2/3}$. - Since $n$ grows faster than $n^{2/3}$, $x_n \to \infty$. Answer: $\lim_{n \to \infty} x_n = +\infty$. 3.7 $x_n = \sqrt{n+2} - \sqrt{n^2 - 2 n + 3}$ - For large $n$, $\sqrt{n+2} \approx \sqrt{n} \to \infty$ slowly. - $\sqrt{n^2 - 2 n + 3} = n \sqrt{1 - \frac{2}{n} + \frac{3}{n^2}} \approx n - 1$. - So $x_n \approx \sqrt{n} - (n -1) \to - \infty$ as $n$ grows. Answer: $\lim_{n \to \infty} x_n = -\infty$. 3.8 $x_n = \sqrt{(n+2)(n+1)} - \sqrt{(n-1)(n+3)}$ - Multiply out inside roots: $\sqrt{n^2 + 3 n + 2} - \sqrt{n^2 + 2 n -3}$. - Difference: $\frac{(n^2 + 3 n + 2) - (n^2 + 2 n - 3)}{\sqrt{n^2 + 3 n + 2} + \sqrt{n^2 + 2 n -3}} = \frac{n + 5}{\text{denominator}}$. - Denominator dominated by $2 n$ for large $n$. - So $x_n \approx \frac{n + 5}{2 n} \to \frac{1}{2}$. Answer: $\lim_{n \to \infty} x_n = \frac{1}{2}$. 3.9 $x_n = n^2[ \sqrt{n(n^4 -1)} - \sqrt{n^5 -8} ]$ - Inside first root $n(n^4 -1) = n^5 - n$. - Difference: $\sqrt{n^5 - n} - \sqrt{n^5 - 8}$. - Use formula for difference: $\frac{(n^5 - n) - (n^5 - 8)}{\sqrt{n^5 - n} + \sqrt{n^5 - 8}} = \frac{-n + 8}{\text{denominator}}$. - Denominator $\approx 2 n^{5/2}$. - So bracket approx $\approx \frac{8 - n}{2 n^{5/2}} = -\frac{n - 8}{2 n^{5/2}}$. - Multiply by $n^2$: $x_n \approx n^2 \cdot \left(- \frac{n - 8}{2 n^{5/2}} \right) = - \frac{n^3 - 8 n^2}{2 n^{5/2}} = - \frac{n^{3} - 8 n^2}{2 n^{5/2}}$ - Simplify powers: $n^{3} / n^{5/2} = n^{1/2}$, $n^{2} / n^{5/2} = n^{-1/2}$. - So $x_n \approx -\frac{n^{1/2} - 8 n^{-1/2}}{2} \to -\infty$ as $n^{1/2} \to \infty$. Answer: $\lim_{n \to \infty} x_n = -\infty$. 3.10 $x_n = \frac{n^3 - 3 n^2 +4}{n^3 - 5 n^2}$ - Leading terms dominate: $x_n \approx \frac{n^3}{n^3} = 1$. Answer: $\lim_{n \to \infty} x_n = 1$. 3.11 $x_n = \sqrt{n^2 + 3 n - 2} - \sqrt{n^2 - 3}$ - Use difference formula: $\frac{(n^2 + 3 n - 2) - (n^2 - 3)}{\sqrt{n^2 + 3 n - 2} + \sqrt{n^2 -3}} = \frac{3 n + 1}{\text{denominator}}$. - Denominator approx $2 n$. - So $x_n \approx \frac{3 n + 1}{2 n} = \frac{3}{2} + \frac{1}{2 n} \to \frac{3}{2}$. Answer: $\lim_{n \to \infty} x_n = \frac{3}{2}$. 3.12 $x_n = \sqrt{n (\sqrt{n + 2} - \sqrt{n - 3})}$ - Inner difference: $\sqrt{n + 2} - \sqrt{n - 3} = \frac{(n + 2) - (n - 3)}{\sqrt{n + 2} + \sqrt{n - 3}} = \frac{5}{2 \sqrt{n}}$ approx. - So $x_n \approx \sqrt{n \cdot \frac{5}{2 \sqrt{n}}} = \sqrt{\frac{5 n}{2 n^{1/2}}} = \sqrt{\frac{5}{2} n^{1/2}} = \sqrt{\frac{5}{2}} n^{1/4} \to \infty$. Answer: $\lim_{n \to \infty} x_n = +\infty$. 3.13 $x_n = \sqrt{n + 5} - n$ - For large $n$, $\sqrt{n + 5} \approx \sqrt{n}$ grows slower than $n$. - So $x_n \approx \sqrt{n} - n \to -\infty$. Answer: $\lim_{n \to \infty} x_n = -\infty$. 3.14 $x_n = \sqrt{n^3 + 8(\sqrt{n^3 + 2} - \sqrt{n^3 + 1})}$ - Simplify inner difference: $\sqrt{n^3 + 2} - \sqrt{n^3 + 1} = \frac{(n^3 + 2) - (n^3 + 1)}{\sqrt{n^3 + 2} + \sqrt{n^3 + 1}} = \frac{1}{2 n^{3/2}}$ - Multiply by 8: $\frac{8}{2 n^{3/2}} = \frac{4}{n^{3/2}}$ - So expression inside first root: $n^3 + \frac{4}{n^{3/2}}$ - Dominated by $n^3$, so $\sqrt{n^3 + \frac{4}{n^{3/2}}} \approx \sqrt{n^3} = n^{3/2}$. Answer: $\lim_{n \to \infty} x_n = +\infty$. 3.15 $x_n = \frac{\sqrt{(n^2 + 1)(n^2 + 2)} - \sqrt{n^4 + 2}}{2 \sqrt{n^3}}$ - Expand inside first root: $n^4 + 2 n^2 + n^2 + 2 = n^4 + 3 n^2 + 2$. - Difference: $\frac{(n^4 + 3 n^2 + 2) - (n^4 + 2)}{\sqrt{n^4 + 3 n^2 + 2} + \sqrt{n^4 + 2}} = \frac{3 n^2}{\text{denominator}}$. - Denominator approx $2 n^2$. - So numerator of formula approx $\frac{3 n^2}{2 n^2} = \frac{3}{2}$. - Therefore numerator $\approx \frac{3}{2}$. - Divide by $2 \sqrt{n^3} = 2 n^{3/2}$: $x_n \approx \frac{3/2}{2 n^{3/2}} = \frac{3}{4 n^{3/2}} \to 0$. Answer: $\lim_{n \to \infty} x_n = 0$. 3.16 $x_n = n - \sqrt{n (n - 1)}$ - Rewrite $\sqrt{n (n - 1)} = n \sqrt{1 - \frac{1}{n}} \approx n (1 - \frac{1}{2 n}) = n - \frac{1}{2}$. - So $x_n \approx n - (n - \frac{1}{2}) = \frac{1}{2}$. Answer: $\lim_{n \to \infty} x_n = \frac{1}{2}$. 3.17 $x_n = n - \sqrt[3]{n^3 - 5}$ - Approximate cube root: $\sqrt[3]{n^3 - 5} = n \sqrt[3]{1 - \frac{5}{n^3}} \approx n (1 - \frac{5}{3 n^3}) = n - \frac{5}{3 n^2}$. - So $x_n \approx n - \left(n - \frac{5}{3 n^2}\right) = \frac{5}{3 n^2} \to 0$. Answer: $\lim_{n \to \infty} x_n = 0$. 3.18 $x_n = \sqrt[5]{n} \left( \sqrt[5]{n^2} - \sqrt[5]{n (n + 2)} \right)$ - Rewrite: $x_n = n^{1/5} \left(n^{2/5} - n^{1/5}(n + 2)^{1/5}\right) = n^{1/5} n^{2/5} - n^{1/5} n^{1/5} (n + 2)^{1/5} = n^{3/5} - n^{2/5} (n + 2)^{1/5}$. - Factor $n^{2/5}$: $x_n = n^{2/5} (n^{1/5} - (n + 2)^{1/5}) = n^{2/5} \left( n^{1/5} - (n + 2)^{1/5} \right)$ - Use binomial approx for difference: $(n + 2)^{1/5} = n^{1/5} (1 + \frac{2}{n})^{1/5} \approx n^{1/5} \left(1 + \frac{2}{5 n} \right)$ - So difference inside parentheses: $n^{1/5} - n^{1/5} - n^{1/5} \frac{2}{5 n} = - n^{1/5} \frac{2}{5 n} = - \frac{2}{5} n^{1/5 - 1} = - \frac{2}{5} n^{-4/5}$ - Multiply by $n^{2/5}$: $x_n = n^{2/5} \cdot \left(- \frac{2}{5} n^{-4/5} \right) = - \frac{2}{5} n^{-2/5} \to 0$. Answer: $\lim_{n \to \infty} x_n = 0$. 3.19 $x_n = \sqrt{n + 2} \left( \sqrt{n + 3} - \sqrt{n - 4} \right)$ - Difference: $\frac{(n + 3) - (n - 4)}{\sqrt{n + 3} + \sqrt{n - 4}} = \frac{7}{\sqrt{n + 3} + \sqrt{n - 4}} \approx \frac{7}{2 \sqrt{n}}$ - So $x_n \approx \sqrt{n} \cdot \frac{7}{2 \sqrt{n}} = \frac{7}{2}$. Answer: $\lim_{n \to \infty} x_n = \frac{7}{2}$. 3.20 $x_n = \frac{\sqrt{n^2 + 1} - \sqrt{n^2 + n}}{n - \sqrt{n^2 - n}}$ - Numerator difference: $\frac{(n^2 + 1) - (n^2 + n)}{\sqrt{n^2 +1} + \sqrt{n^2 + n}} = \frac{1 - n}{\text{denominator}}$. - Denominator approx $2 n$. - So numerator approx $\frac{1 - n}{2 n} = \frac{1}{2 n} - \frac{1}{2} \approx - \frac{1}{2}$. - Denominator: $n - \sqrt{n^2 - n} = n - n \sqrt{1 - \frac{1}{n}} \approx n - n \left(1 - \frac{1}{2 n} \right) = n - (n - \frac{1}{2}) = \frac{1}{2}$. - Therefore $x_n \approx \frac{-\frac{1}{2}}{\frac{1}{2}} = -1$. Answer: $\lim_{n \to \infty} x_n = -1$. Final summary: 3.1: 0 3.2: 1/3 3.3: -1 3.4: 0 3.5: -3/2 3.6: +\infty 3.7: -\infty 3.8: 1/2 3.9: -\infty 3.10: 1 3.11: 3/2 3.12: +\infty 3.13: -\infty 3.14: +\infty 3.15: 0 3.16: 1/2 3.17: 0 3.18: 0 3.19: 7/2 3.20: -1