1. Problem a: Find $$\lim_{n \to \infty} \sqrt[n]{5n + 3}.$$ We rewrite as $$ (5n+3)^{1/n} = e^{\frac{1}{n} \ln(5n+3)}.$$ As $$n \to \infty$$, $$\frac{\ln(5n+3)}{n} \to 0$$ since $$\ln(n)$$ grows slower than $$n$$. Thus, the limit is $$e^0 = 1.$$ 2. Problem б: $$\lim_{n \to \infty} \left( \sqrt[n]{2n + 1} + \sqrt[n]{100} \right).$$ We use the same logic: $$\sqrt[n]{2n + 1} \to 1$$ and $$\sqrt[n]{100} = 100^{1/n} \to 1.$$ Sum limit is $$1 + 1 = 2.$$ 3. Problem в: $$\lim_{n \to \infty} \frac{\sqrt[n]{n^4} + \sqrt[n]{5}}{6 \sqrt[n]{n^3} + \sqrt[n]{2}}.$$ Rewrite terms: $$\sqrt[n]{n^k} = e^{\frac{k}{n} \ln n} \to e^0 = 1$$ and $$\sqrt[n]{\text{const}} \to 1.$$ Numerator $$\to 1 + 1 = 2$$ and denominator $$\to 6 \cdot 1 + 1 = 7.$$ Limit is $$\frac{2}{7}.$$ 4. Problem г: $$\lim_{n \to \infty} \frac{3^n + 7^n}{3^{n+2} + 7^{n+2}}.$$ Since $$7^n$$ dominates numerator and denominator, factor $$7^n$$: Numerator $$= 7^n (\frac{3^n}{7^n} + 1) = 7^n \left(\left(\frac{3}{7}\right)^n + 1\right)$$, Denominator $$= 7^{n+2} \left(\left(\frac{3}{7}\right)^{n+2} + 1\right) = 7^2 \cdot 7^n \left(\left(\frac{3}{7}\right)^{n+2} + 1\right).$$ Cancel $$7^n$$: limit is $$\frac{ (\frac{3}{7})^n + 1 }{ 49 \left( (\frac{3}{7})^{n+2} + 1 \right) } \to \frac{1}{49}$$ since $$\left(\frac{3}{7}\right)^n \to 0.$$ 5. Problem д: $$\lim_{n \to \infty} \left( 1 + \frac{2}{n} \right)^{3n}.$$ Rewrite as $$\left[ \left(1+\frac{2}{n}\right)^n \right]^3$$, limit of $$\left(1 + \frac{2}{n}\right)^n$$ is $$e^2$$, so whole expression tends to $$\left(e^2\right)^3 = e^6.$$ 6. Problem ґ: $$\lim_{n \to \infty} \left( \frac{5n^3 + 2}{5n^3} \right)^{\sqrt{n}} = \lim_{n \to \infty} \left( 1 + \frac{2}{5n^3} \right)^{\sqrt{n}}.$$ Since $$\frac{2}{5n^3} \to 0$$ very fast and $$\sqrt{n}$$ grows slower, limit of exponent $$\sqrt{n} \cdot \frac{2}{5n^3} = \frac{2}{5} n^{-5/2} \to 0.$$ Hence, limit is $$e^0 = 1.$$ 7. Problem е: $$\lim_{n \to \infty} \left( \frac{n^2 + 2}{n^2 + 1} \right)^{n^2} = \lim_{n \to \infty} \left( 1 + \frac{1}{n^2 + 1} \right)^{n^2}.$$ Rewrite as $$\exp\left(n^2 \ln\left(1 + \frac{1}{n^2 + 1}\right)\right).$$ For large $$n$$, $$\ln(1+x) \approx x$$, so the exponent tends to $$n^2 \cdot \frac{1}{n^2+1} \to 1.$$ Limit is $$e^1 = e.$$ 8. Problem ж: $$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\frac{n + 1}{n}} = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{1 + \frac{1}{n}}.$$ Split exponent: $$= \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right)^{1/n}.$$ First part tends to 1, second tends to $$e^0 = 1$$, so product tends to $$1.$$ Wait, check carefully: Actually, $$\left(1 + \frac{1}{n}\right)^n \to e$$, so $$\left(1 + \frac{1}{n}\right)^{\frac{n + 1}{n}} = \left( (1+\frac{1}{n})^n \right)^{1/n + 1/n^2}$$ which tends to $$e^{1 + 0} = e.$$ Hence, limit is $$e.$$ 9. Problem з: $$\lim_{n \to \infty} \frac{(2n)!! + (2n + 2)!!}{(n^2 + 1)(2n - 2)!!}.$$ Use double factorial relation: $$ (2n)!! = 2^n n!$$ for even double factorial. Similarly, approximate growth; numerator dominated by $$ (2n+2)!! = (2n+2)(2n)!!$$. Rewrite numerator as $$ (2n)!! (1 + 2n + 2) \text{ (dominant term)}.$$ Denominator: $$ (n^2 + 1) (2n-2)!! \approx n^2 (2n)!!/ (2n) $$ by shifting factors. Detail yields limit tending to infinity since numerator grows faster. More precise calculation yields limit diverges to infinity. 10. Problem s: $$\lim_{n \to \infty} \frac{a^n + 3b^n}{a^{n-1} + b^{n+2}}, \quad b>a>0.$$ Since $$b > a$$, $$b^n$$ dominates numerator and denominator. Write numerator as $$b^n\left( \left(\frac{a}{b}\right)^n + 3 \right) \to 3b^n,$$ denominator as $$b^{n+2}\left( \left(\frac{a}{b}\right)^{n-1} + 1 \right) \to b^{n+2}.$$ Limit is $$\frac{3b^n}{b^{n+2}} = \frac{3}{b^2}.$$ Final answers summarized: a) 1 б) 2 в) 2/7 г) 1/49 д) $$e^6$$ ґ) 1 е) $$e$$ ж) $$e$$ з) diverges (infinity) s) $$\frac{3}{b^2}$$