1. 根据数列极限的四则运算和已经掌握的极限求以下极限:
    1. `lim_(n to oo) (3n^3 + 2n^2 - n - 2)/(2n^3 - 3n + 1)`;
    2. `lim_(n to oo) (2 root 3 n - sqrt n + 1) /(5 root 3 n + 3 sqrt n - 2)`;
    3. `lim_(n to oo) (3^n + (-2)^n)/(5^n + (-3)^n)`;
    4. `lim_(n to oo) (3^n n^2 a + 2^n n^3 b)/(3^n n^2 + 2^n n^3)`;
    5. `lim_(n to oo) (sqrt(n+1) - sqrt(n-1))`;
    6. `lim_(n to oo) root(3)(n^2) (root(3)(n+1) - root 3 n)`;
    7. `lim_(n to oo) (1/(1*2) + 1/(2*3) + cdots + 1/(n(n+1)))`;
    8. `lim_(n to oo) 1/n^2 [1 + 3 + cdots + (2n-1)]`;
    9. `lim_(n to oo) (1 + a + a^2 + cdots + a^n) /(1 + b + b^2 + cdots + b^n)` `\(|a| lt 1, |b| lt 1)`;
    10. `lim_(n to oo) (1-1/2^2)(1-1/3^2) cdots (1-1/n^2)`;
    11. `lim_(n to oo) (1+1/2)(1+1/4) cdots (1+1/2^n)`;
    12. `lim_(n to oo) (1/(1*2*3) + 1/(2*3*4) + cdots + 1/(n(n+1)(n+2)))`.
  2. 下述推理是否正确? `lim_(n to oo) (1/n^2 + 2/n^2 + cdots + n/n^2)`
    `= lim_(n to oo) 1/n^2 + lim_(n to oo) 2/n^2 + cdots + lim_(n to oo) n/n^2`
    `= 0 + 0 + cdots + 0 = 0`.
  3. 下述推理是否正确? `lim_(n to oo) (1 + 1/n)^n`
    `= lim_(n to oo) underbrace((1 + 1/n)(1 + 1/n) cdots (1 + 1/n))_(n 个)`
    `= lim_(n to oo) (1 + 1/n) lim_(n to oo) (1 + 1/n) cdots lim_(n to oo) (1 + 1/n)`
    `= 1 * 1 * cdots * 1 = 1`.
  4. 已知 `lim_(n to oo) (x_n - y_n) = 0` 且 `lim_(n to oo) x_n = a`, 求证 `lim_(n to oo) y_n = a`. 下述推理是否正确?

    因为 `0 = lim_(n to oo) (x_n - y_n) = lim_(n to oo) x_n - lim_(n to oo) y_n = a - lim_(n to oo) y_n`, 所以 `lim_(n to oo) y_n = a`.

  5. 运用两边夹法则求以下极限:
    1. `lim_(n to oo) sqrt(1 - 1/n)`;
    2. `lim_(n to oo) (sin n!)/sqrt n`;
    3. `lim_(n to oo) (1 - 1/root(n)(2)) e^(cos^2 n)`;
    4. `lim_(n to oo) root(n)(n log_2 n)`;
    5. `lim_(n to oo) (1/sqrt(n^2+1) + 1/sqrt(n^2+2) + cdots + 1/sqrt(n^2 + n))`;
    6. `lim_(n to oo) (1/sqrt(n^2+1) + 1/sqrt(n^2+2) + cdots + 1/sqrt((n+1)^2))`;
    7. `lim_(n to oo) 1/root(n)(n!)`;
    8. `lim_(n to oo) root(n^2)(n!)`;
    9. `lim_(n to oo) sum_(k=1)^n (1/root(k)(n^k+1) + 1/root(k)(n^k-1))`;
    10. `lim_(n to oo) sin(pi sqrt(n^2+1))`;
    11. `lim_(n to oo) (-1)^n sin(pi sqrt(n^2 + n))`;
    12. `lim_(n to oo) underbrace(sin sin cdots sin)_(n 次) x`.
  6. 已知 `lim_(n to oo) x_n = a`, `lim_(n to oo) y_n = b`. 证明:
    1. `lim_(n to oo) min{x_n, y_n} = min{a, b}`;
    2. `lim_(n to oo) max{x_n, y_n} = max{a, b}`;
  7. 设 `a_1, a_2, cdots, a_m` 都是正实数. 求 `lim_(n to oo) root(n)(a_1^n + a_2^n + cdots + a_m^n)`.
  8. 已知 `x_n gt 0`, `n = 1, 2, cdots`, 且 `lim_(n to oo) x_n = a gt 0`. 证明: `lim_(n to oo) root(n)(x_n) = 1`.
  9. 已知 `lim_(n to oo) x_n = a`, 证明:
    1. 对任意 `b gt 0` 有 `lim_(n to oo) b^(x_n) = b^a`;
    2. 对任意 `b gt 0` 有 `lim_(n to oo) log_b x_n = log_b a` (假定 `x_n gt 0`, `n = 1, 2, cdots`, 且 `a gt 0`);
    3. 对任意 `b gt 0` 有 `lim_(n to oo) x_n^b = a^b` (假定 `x_n gt 0`, `n = 1, 2, cdots`, 且 `a gt 0`);
    4. `lim_(n to oo) sin x_n = sin a`, `lim_(n to oo) cos x_n = cos a`;
    5. 当 `-1 le x_n le 1` `(n = 1, 2, cdots)` 时, `lim_(n to oo) arcsin x_n = arcsin a`, `lim_(n to oo) arccos x_n = arccos a`.
  10. 已知 `x_n gt 0`, `n = 1, 2, cdots`, 且 `lim_(n to oo) x_(n+1)/x_n = a`. 证明: `lim_(n to oo) root(n)(x_n) = a`.
  11. 求下列极限:
    1. `lim_(n to oo) sum_(k=1)^n 1/(k(k+m))` (m 是正整数);
    2. `lim_(n to oo) sum_(k=1)^n (2^k k)/((k+2)!)`;
    3. `lim_(n to oo) sum_(k=1)^n (k^3 + 6k^2 + 11k + 5)/((k+3)!)`;
    4. `lim_(n to oo) 1/n^3 [1*3 + 2*4 + cdots + n(n+2)]`
    5. `lim_(n to oo) ((2^3-1)/(2^3+1) (3^3-1)/(3^3+1) cdots (n^3-1)/(n^3+1))`.
  12. 证明: 极限 `lim_(n to oo) sin n` 不存在.
  13. 已知 `lim_(n to oo) x_n = a`, 证明:
    1. `lim_(n to oo) (x_n + 2x_(n-1) + cdots + nx_1) /(n(n+1)) = a/2`;
    2. `lim_(n to oo) 1/2^n [1 + (n;1) x_1 + (n;2) x_2 + cdots + (n;n-1) x_(n-1) + x_n] = a`;
    3. `lim_(n to oo) (x_n + lambda x_(n-1) + lambda^2 x_(n-2) + cdots + lambda^(n-1) x_1) = a/(1-lambda)` (其中 `0 lt lambda lt 1` 为常数).
  14. 设 `lim_(n to oo) x_n = a`. 又设 `{p_n}` 是正数列, 满足
    `lim_(n to oo) p_n/(p_1 + p_2 + cdots + p_n) = 0`.
    证明: `lim_(n to oo) (p_1 x_n + p_2 x_(n-1) + cdots + p_n x_1) /(p_1 + p_2 + cdots + p_n) = a`.
  15. 已知 `lim_(n to oo) x_n = a` 且 `lim_(n to oo) y_n = b`, 证明:
    `lim_(n to oo) (x_1 y_n + x_2 y_(n-1) + cdots + x_(n-1) y_2 + x_n y_1)/n = ab`.