1. 应用洛必达法则求以下极限:
    1. `lim_(x to 0) (x "cot" x - 1)/x^2`;
    2. `lim_(x to 0) (1-"cot" x^2)/(x^2 sin x^2)`;
    3. `lim_(x to 0) (arcsin 2x - 2 arcsin x)/(x^2 sin x)`;
    4. `lim_(x to 0) (ln sin ax)/(ln sin bx)`;
    5. `lim_(x to 0) (1/x - 1/(e^x-1))`;
    6. `lim_(x to 0) (("cot"x)/x - "csc"^2 x)`;
    7. `lim_(x to a) (a^x - x^a)/(x-a)` (a > 0);
    8. `lim_(x to a) (x^alpha - a^alpha)/(x^beta - a^beta)`.
  2. 应用洛必达法则求以下极限:
    1. `lim_(x to 0) x^(x^2)`;
    2. `lim_(x to 0) [(1+x)^(1/x)/e]^(1/x)`;
    3. `lim_(x to 0^+) (ln{:1/x:})^(1/x)`;
    4. `lim_(x to pi/4) (tan x)^(tan 2x)`;
    5. `lim_(x to 0) (arcsin x / x)^(1/x^2)`;
    6. `lim_(x to +oo) (2/pi arctan x)^x`;
    7. `lim_(x to pi/4) tan2x tan(pi/4 - x)`;
    8. `lim_(x to oo) x(pi/4 - arctan {:x/(x+1):})`.
  3. 应用洛必达法则求下列数列的极限:
    1. `lim_(n to oo) n(root n a - 1)` (a > 0);
    2. `lim_(n to oo) n^2 (root n a - root (n+1) a)` (a > 0);
    3. `lim_(n to oo) ((n-1)/(n+x))^n`;
    4. `lim_(n to oo) (cos {:pi/n:} / cosh {:pi/n:})^(n^2)`;
    5. `lim_(n to oo) cos^n {:x/sqrt n:}`;
    6. `lim_(n to oo) tan^n (pi/4 + 1/n)`.
  4. 求 `mu` 使下述极限有非零的有限值: `lim_(x to 0) x^mu ((x-sin x)/x^3 - (1-cos x)/3x^2)`.
  5. 设 `f(x)` 有二阶导数. 证明: `lim_(h to 0) (f(x+h) + f(x-h) - 2f(x)) / h^2 = f''(x)`.
    使用洛必达法则, `lim_(h to 0) (f(x+h) + f(x-h) - 2f(x))/h^2`
    `= lim_(h to 0) (f'(x+h) - f'(x-h))/(2h)`
    `= lim_(h to 0) (f'(x+h) - f'(x))/(2h)` `+ lim_(h to 0) (f'(x) - f'(x-h))/(2h)`
    `= 1/2 f''(x) + 1/2 f''(x)`.
    注意二阶导数不存在时, 上述极限仍有可能存在. 如 `f(x) = |x|` 在 `x = 0` 处.
  6. 设 `f(x)` 在 `(a, +oo)` 上可导, 且 `lim_(x to +oo) [f'(x) + bf(x)] = c` (b ≠ 0). 证明: `lim_(x to +oo) f(x) = c/b`.
  7. 设函数 `f(x)` 在 `x = 0` 附近有二阶导数且 `f''(0) != 0`. 对充分接近于零的 `x != 0`, 由拉格朗日中值定理知存在 `theta = theta_x in (0, 1)` 使 `f(x) - f(0) = f'(theta x) x`. 证明: `lim_(x to 0) theta = 1/2`, 并以函数 `f(x) = arcsin x` 为例说明条件 `f''(0) != 0` 不能去掉.