Vypočítejte limitu posloupnosti:

\( \lim_{n \to \infty} \frac{5 - 2n}{3n - 7} \) lim (5 - 2n) / (3n - 7) as n->infinity

\( \lim_{n \to \infty} \frac{n^2 + 4}{n + 4} \) lim (n^2 + 4) / (n + 5) as n->infinity

\( \lim_{n \to \infty} \frac{n^2}{n^3 + 5} \) lim (n^2) / (n^3 + 5) as n->infinity

\( \lim_{n \to \infty} \frac{(2n - 1)(n^2 - n + 2)}{(n + 1)(2n + 1)(3n + 1)} \) limit of ( (2n - 1)(n^2 - n + 2)) / ((n + 1)(2n + 1)(3n + 1)) as n=+oo

\( \lim_{n \to \infty} \frac{n!}{(n + 1)! - n!} \) lim n! / ((n + 1)! - n!) as n->infinity

\( \lim_{n \to \infty} n^{\sin \frac{1}{n}} \) lim (n^(sin (1/n))) as n->infinity

\( \lim_{n \to \infty} \frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}} \) lim ((-2)^n + 3^n) / ((-2)^(n+1) + 3^(n+1)) as n->+oo

\(\lim_{n \to \infty} \frac{2n + \sin n}{3n - 1} \) lim (2n + sin n) / (3n - 1) as n->infinity

\( \lim_{n \to \infty} \frac{\sqrt[3]{n^2} \sin n!}{n + 1} \) lim ( cuberoot (n^2) sin n! ) / (n + 1) as n->+oo

\( \lim_{n \to \infty} ( \sqrt{n^2 + 2n} - n ) \) lim sqrt(n^2 + 2n) - n as n->infinity

\( \lim_{n \to \infty} ( \sqrt{n + 1} - \sqrt{n} ) \) lim sqrt(n + 1) - sqrt(n) as n->infinity Answer: 0 resp. lze použít racionalizace jako v 31

\( \lim_{n \to \infty} \frac{1 + a + a^2 + \dots + a^n}{1 + b + b^2 + \dots + b^n}, |a| < 1, |b| < 1 \)

\( \lim_{n \to \infty} ( \frac{1}{n^2} + \frac{2}{n^2} + \dots + \frac{n-1}{n^2} ) \)

\( \lim_{n \to \infty} \sqrt{2} \sqrt[4]{2} \sqrt[8]{2} \dots \sqrt[2^n]{2} \)

\( \lim_{n \to \infty} ( \frac{n + 4}{n + 3})^n \) lim ((n + 4)/(n + 3))^n as n->infinity

\( \lim_{n \to \infty} (\frac{n - 3}{n})^n \) lim ((n - 3)/n )^n as n->infinity upravíme na: lim (1 - 3/n)^n as n->+oo