Категория:Задачник Кузнецова Пределы Задача 1

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В этой категории варианты Задачи 1, раздела "Пределы" в задачнике Кузнецова Л.А.

Варианты

Доказать, что $\lim_{n\to\infty} {a_{n}} = a$ (указать $N(\varepsilon)$ ).

Задача	Условие	Задача	Условие
1-1	$a_{n} = \frac {3n - 2} {2n - 1}, \ a = \frac {3} {2}$	1-2	$a_{n} = \frac {4n - 1} {2n + 1},\ a = 2$
1-3	$a_{n} = \frac {7n + 4} {2n + 1},\ a = \frac {7} {2}$	1-4	$a_{n} = \frac {2n - 5} {3n + 1},\ a = \frac {2} {3}$
1-5	$a_{n} = \frac {7n - 1} {n + 1},\ a = 7$	1-6	$a_{n} = \frac {4n^2 + 1} {3n^2 + 2},\ a = \frac {4} {3}$
1-7	$a_{n} = \frac {9 - n^3} {1 + 2n^3},\ a = - \frac {1} {2}$	1-8	$a_{n} = \frac {4n - 3} {2n + 1},\ a = 2$
1-9	$a_{n} = \frac {1 - 2n^2} {2 + 4n^2},\ a = - \frac {1} {2}$	1-10	$a_{n} = - \frac {5n} {n + 1},\ a = -5$
1-11	$a_{n} = \frac {n + 1} {1 - 2n},\ a = - \frac {1} {2}$	1-12	$a_{n} = \frac {2n + 1} {3n - 5},\ a = \frac {2} {3}$
1-13	$a_{n} = \frac {1 - 2n^2} {n^2 + 3},\ a = -2$	1-14	$a_{n} = \frac {3n^2} {2 - n^2},\ a = -3$
1-15	$a_{n} = \frac {n} {3n - 1},\ a = \frac {1} {3}$	1-16	$a_{n} = \frac {3n^3} {n^3 - 1},\ a = 3$
1-17	$a_{n} = \frac {4 + 2n} {1 -3n},\ a = - \frac {2} {3}$	1-18	$a_{n} = \frac {5n + 15} {6 - n},\ a = -5$
1-19	$a_{n} = \frac {3 - n^2} {4 + 2n^2},\ a = - \frac {1} {2}$	1-20	$a_{n} = \frac {2n - 1} {2 - 3n},\ a = - \frac {2} {3}$
1-21	$a_{n} = \frac {3n - 1} {5n + 1},\ a = \frac {3} {5}$	1-22	$a_{n} = \frac {4n - 3} {2n + 1},\ a = 2$
1-23	$a_{n} = \frac {1 - 2n^2} {2 + 4n^2},\ a = - \frac {1} {2}$	1-24	$a_{n} = \frac {5n + 1} {10n - 3},\ a = \frac {1} {2}$
1-25	$a_{n} = \frac {2 - 2n} {3 + 4n},\ a = - \frac {1} {2}$	1-26	$a_{n} = \frac {23 - 4n} {2 - n},\ a = 4$
1-27	$a_{n} = \frac {1 + 3n} {6 - n},\ a = -3$	1-28	$a_{n} = \frac {2n + 3} {n + 5},\ a = 2$
1-29	$a_{n} = \frac {3n^2 + 2} {4n^2 - 1},\ a = \frac {3} {4}$	1-30	$a_{n} = \frac {2 - 3n^2} {4 + 5n^2},\ a = - \frac {3} {5}$
1-31	$a_{n} = \frac {2n^3} {n^3 - 2},\ a = 2$