Úlohy
Úloha
Určete limitu \(\lim_{n \to \infty} {{1 - n} \over n^2} =\)
- \(= \lim_{n \to \infty} ({1 \over n^2} - {n \over n^2}) = \lim_{n \to \infty} {1 \over n^2} - \lim_{n \to \infty} {1 \over n} = \)
- \(= 0 - 0 = 0\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{n!} \over {(n+1)!-n!}} =\)
- \(= \lim_{n \to \infty} {{n!} \over {n!(n+1-1)}} = \lim_{n \to \infty} {1 \over n} = \)
- \(= 0\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{n + 4} \over {n^2 - 5}} =\)
- \(= \lim_{n \to \infty} {{{n \over n^2} + {4 \over n^2}} \over {{n^2 \over n^2} - {5 \over n^2}}} = \lim_{n \to \infty} {{{1 \over n} + {4 \over n^2}} \over {1 - {5 \over n^2}}} = \)
- \(= {0 \over 1} = 0\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{n^3 + 5n^2 - 2} \over {2n^3 - n}} =\)
- \(= \lim_{n \to \infty} {{{n^3 \over n^3} + {{5n^2} \over n^3} - {2 \over n^3}} \over {{{2n^3} \over n^3} - {n \over n^3}}} = \lim_{n \to \infty} {{1 + {5 \over n} - {2 \over n^3}} \over {2 - {1 \over n^2}}} = \)
- \(= {{1 + 0 - 0} \over {2 - 0}} = {1 \over 2}\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{6n^4 + n} \over {9n^3 - n^2}} =\)
- \(= \lim_{n \to \infty} {{{{6n^n} \over n^4} + {n \over n^4}} \over {{{9n^3} \over n^4} - {n^2 \over n^4}}} = \lim_{n \to \infty} {{6 + {1 \over n^3}} \over {{9 \over n} - {1 \over n^2}}} = \)
- \(= {{6 + 0} \over {0 - 0}} = \infty\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{-n + 8} \over {-2n^4 - 4n^2 + 3}} =\)
- \(= \lim_{n \to \infty} {{-{n \over n^4} + {8 \over n^4}} \over {{{{-2n^4} \over n^4} - {{4n^2} \over n^4} + {3 \over n^4}}}} = \lim_{n \to \infty} {{-{1 \over n^3} + {8 \over n^4}} \over {-2 - {4 \over n^2} + {3 \over n^4}}} = \)
- \(= {{0 + 0} \over {-2 - 0 + 0}} = 0\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{(n + 1)!} \over {n! - (n + 1)!}} =\)
- \(= \lim_{n \to \infty} {{n!(n+1)!} \over {n!(1 - (n + 1))}} = \lim_{n \to \infty} {{n + 1} \over {-n}} = \)
- \(= -1 - 0 = -1\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{1 + 2 + 3 + \ldots + n} \over {n^2}} =\)
- \(= \lim_{n \to \infty} {{{n \over 2}(1 + n)} \over {n^2}} = \lim_{n \to \infty} {1 \over 2} \cdot {{1 + n} \over {n}} = \)
- \(= {1 \over 2}(0 + 1) = {1 \over 2}\)
Úloha
Určete limitu \(\lim_{n \to \infty} {{1 + 3 + 5 + \ldots + 2n - 1} \over {2 + 4 + 6 + \ldots + 2n}} =\)
- \(= \lim_{n \to \infty} {{{n \over 2}(1 + 2n - 1)} \over {{n \over 2}(2 + 2n)}} = \lim_{n \to \infty} {{n} \over {1 + n}} = \)
- \(= \lim_{n \to \infty} {{{n \over n}} \over {{1 \over n} + {n \over n}}} = \lim_{n \to \infty} {{1} \over {{1 \over n} + 1}} = \)
- \(= {1 \over {0 + 1}} = 1\)
