1. 应用数列极限的 `epsilon-N` 定义证明:
    `lim_(n to oo) n/(2n-1) = 1/2`.
    并对 `epsilon = 0.1, 0.01, 0.001, 0.0001`, 写出使定义条件成立的 N 的具体数值.
  2. 根据数列极限的 `epsilon-N` 定义证明以下极限:
    1. `lim_(n to oo) (2 sqrt n - 1)/(3 sqrt n + 1) = 2/3`;
    2. `lim_(n to oo) (4n + (-1)^n sqrt n)/(5n+2) = 4/5`;
    3. `lim_(n to oo) 2^n/(n!) = 0`;
    4. `lim_(n to oo) (n!)/n^n = 0`;
    5. `lim_(n to oo) n/a^n = 0 \ (a gt 1)`;
    6. `lim_(n to oo) n^2 q^n = 0 \ (0 lt q lt 1)`.
  3. 下面关于 `lim_(n to oo) n^2/2^n = 0` 的证明是否正确: `AA epsilon gt 0`, 为使 `n^2/2^n lt epsilon`, 只要 `n^2 lt 2^n epsilon`. 取对数得 `2 ln n lt n ln 2 + ln epsilon`. 因为 `ln n ge 0`, 故从 `2 ln n lt n ln 2 + ln epsilon` 推得 `0 lt n ln 2 + ln epsilon`, 解得 `n gt - (ln epsilon)/(ln 2) = (ln (1/epsilon))/(ln 2)`. 所以令 `N = [(ln (1/epsilon))/(ln 2)] + 1` (不妨设 `0 lt epsilon lt 1` 以便 `N ge 1`). 则当 `n gt N` 时就有 `n^2/2^n lt epsilon`, 所以 `lim_(n to oo) n^2/2^n = 0`.
  4. 设 `x_n le a le y_n`, `n = 1, 2, cdots`, 且 `lim_(n to oo) (y_n - x_n) = 0`, 证明: `lim_(n to oo) x_n = lim_(n to oo) y_n = a`.
  5. 设存在常数 `0 lt lambda lt 1` 使得 `|x_(n+1)| le lambda |x_n|`, `n = 1, 2, cdots`, 证明: `lim_(n to oo) x_n = 0`.
  6. 已知 `lim_(n to oo) x_n = a`, 证明:
    1. `lim_(n to oo) |x_n| = |a|`;
    2. `lim_(n to oo) sqrt(x_n) = sqrt(a)` (假定 `a ge 0`, `x_n ge 0`, `n = 1, 2, cdots`);
    3. `lim_(n to oo) root(3)(x_n) = root(3)(a)`.
  7. 证明: `lim_(n to oo) x_n = a` 的充要条件是 `lim_(n to oo) x_(2n) = a` 且 `lim_(n to oo) x_(2n-1) = a`.
  8. 已知 `|b| lt a`, 证明: `lim_(n to oo) root(n)(a^n+b^n) = a`.
  9. 已知 `lim_(n to oo) x_n = a`, 证明: `lim_(n to oo) ([n x_n])/n = a`.
  10. 已知 `x_n != 0`, `n = 1, 2, cdots`, 且 `lim_(n to oo) |x_(n+1)/x_n| = l lt 1`, 证明: `lim_(n to oo) x_n = 0`.
  11. 已知 `lim_(n to oo) |x_n| = l lt 1`, 证明: `lim_(n to oo) x_n^n = 0`.
  12. 已知 `lim_(n to oo) x_n = a`, 证明: `lim_(n to oo) (x_1 + 2x_2 + cdots + n x_n)/(1 + 2 + cdots + n) = a`.
  13. 已知 `lim_(n to oo) (x_(n+1) - x_n) = a`, 证明: `lim_(n to oo) x_n/n = a`.
  14. 用通俗易懂的语言描述满足下列各个条件的数列:
    1. `EE N in NN`, `AA epsilon gt 0`, `AA n gt N`, 有 `|x_n - a| lt epsilon`;
    2. `AA epsilon gt 0`, `EE N in NN`, `EE n gt N`, 使 `|x_n - a| lt epsilon`;
    3. `AA epsilon gt 0`, `AA N in NN`, `AA n gt N`, 有 `|x_n - a| lt epsilon`;
    4. `AA epsilon gt 0`, `AA N in NN`, `EE n gt N`, 使 `|x_n - a| lt epsilon`.
    1. 用逻辑语言写出实数 a 是非空有上界的实数集合 S 的上确界的条件;
    2. 用逻辑语言写出数列 `{x_n}` 不收敛 (即没有极限) 的条件;
    3. 写出第 14 题各个条件的否定条件, 并用通俗易懂的语言描述满足这样 条件的数列.