函数极限的运算法则
设函数极限\(\underset{x \to x_0}{\lim}f(x)\)与\(\underset{x \to x_0}{\lim}g(x)\)存在，那么存在关系： $$ \begin{equation}\label{f1} \lim_{x \to x_0}\left[f(x) ± g(x)\right] = \lim_{x \to x_0}f(x) ± \lim_{x \to x_0}g(x) \end{equation}$$ $$ \begin{equation}\label{f2} \lim_{x \to x_0}\left[f(x) \cdot g(x)\right] = \lim_{x \to x_0}f(x) \cdot \lim_{x \to x_0}g(x) \end{equation}$$ 特别的，当\(\underset{x \to x_0}{\lim}g(x) \neq 0\)时，有：
$$ \begin{equation}\label{f3} \lim_{x \to x_0}\left[\cfrac{f(x)}{g(x)}\right] = \cfrac{\underset{x \to x_0}{\lim}f(x)}{\underset{x \to x_0}{\lim}g(x)} \end{equation}$$我们称式(\(\ref{f1}, \ref{f2}, \ref{f3}\))所描述的关系为函数极限的四则运算。 根据Haine定理和数列极限的四则运算我们可以很容易证明之。
对于复合函数\(f \circ g = f(g(x))\)，若函数\(f(t)\)在\(t_0\)的去心邻域中有极限\(l\)，函数\(g(x)\)在\(x_0\)的去心邻域中有极限\(t_0\)，\(g(t) \neq t_0\)， 则有\(\underset{x \to x_0}{\lim} f \circ g(x) = l\)。我称这一定理为函数极限的复合法则。
因为函数\(f(t)\)在\(t_0\)的去心邻域中有极限\(l\)，所以对于任意的\(\varepsilon > 0\)，我们总可以找到一个\(\delta_0 > 0\)， 使得\(|t - t_0| < \delta_0\)时都有\(|f(t) - l| < \varepsilon\)。 又因为函数\(g(x)\)在\(x_0\)的去心邻域中有极限\(t_0\)，所以对于\(\delta_0 > 0\)我们总可以找到一个\(\delta > 0\)使得\(|x - x_0| < \delta\)时\(|g(x) - t_0| < \delta_0\)。 以至于对于任意的\(\varepsilon > 0\)，总存在\(\delta > 0\)，使得\(|x - x_0| < \delta\)时有\(|f(g(t)) - l| < \varepsilon\)成立。
需要强调的是，复合法则中的条件\(g(t)\)在\(t_0\)的去心邻域内\(g(t) \neq t_0\)是不能少的。习题【607】就是一个这样的反例。
【411】(1)\(\underset{x \to 0}{\lim}\frac{x^2 - 1}{2x^2 - x -1}\); (2)\(\underset{x \to 1}{\lim}\frac{x^2 - 1}{2x^2 - x -1}\); (3)\(\underset{x \to \infty}{\lim}\frac{x^2 - 1}{2x^2 - x -1}\)
解：(1) 容易证明\(\underset{x \to 0}{\lim} \left(x^2 - 1\right) = -1,\underset{x \to 0}{\lim} \left(2x^2 - x -1\right) = -1\)，所以 \(\underset{x \to 0}{\lim}\cfrac{x^2 - 1}{2x^2 - x -1} = \cfrac{-1}{-1} = 1\)
(2) 根据【410】的讨论， 有\(\underset{x \to 1}{\lim}\cfrac{x^2 - 1}{2x^2 - x -1} = \underset{x \to 1}{\lim}\cfrac{(x-1)(x+1)}{(x-1)(2x+1)} = \cfrac{2}{3}\)
(3) 根据【409】的讨论，有 \(\underset{x \to \infty}{\lim}\cfrac{x^2 - 1}{2x^2 - x -1} = \cfrac{1}{2}\)
【412】\(\underset{x \to 0}{\lim}\cfrac{(1+x)(1+2x)(1+3x) - 1}{x}\)
解：显然\(\underset{x\to 0}{\lim} x = 0\)。但是\(\cfrac{(1+x)(1+2x)(1+3x) - 1}{x} = \cfrac{6x^3 + 11x^2 + 6x + 1 - 1}{x} = \cfrac{x(6x^2 + 11x + 6)}{x}\)
所以在\(x = 0\)的去心邻域上，\(\underset{x \to 0}{\lim}\cfrac{(1+x)(1+2x)(1+3x) - 1}{x} = \underset{x \to 0}{\lim} 6x^2 + 11x + 6 = 6\)。
【413】\(\underset{x \to 0}{\lim}\cfrac{(1+x)^5 - (1+5x)}{x^2 + x^5}\)
解：\(\underset{x \to 0}{\lim}\cfrac{(1+x)^5 - (1+5x)}{x^2 + x^5} = \underset{x \to 0}{\lim}\cfrac{x^5 + 5x^4 + 10x^3 + 10x^2}{x^2(x^3 + 1)} = 10\)
【414】\(\underset{x \to 0}{\lim}\cfrac{(1 + mx)^n - (1 + nx)^m}{x^2}\)，\(m, n\)是正整数。
解：\(\cfrac{(1 + mx)^n - (1 + nx)^m}{x^2} = \cfrac{\sum_{i= 0}^n \mathrm{C}_n^i (mx)^{(n - i)} - \sum_{j = 0}^m \mathrm{C}_m^j (nx)^{(m - j)}}{x^2}\)。关于该式分子的常数项、一次项和二次项有:
$$ \begin{cases} \mathrm{C}_n^n (mx)^0 - \mathrm{C}_m^m (nx)^0 & = 1 - 1 = 0 \\ \mathrm{C}_n^{n-1} (mx)^1 - \mathrm{C}_m^{m-1} (nx)^1 & = nmx - mnx = 0 \\ \mathrm{C}_n^{n-2} (mx)^2 - \mathrm{C}_m^{m-2} (nx)^2 & = \cfrac{n(n-1)}{2}m^2x^2 - \cfrac{m(m-1)}{2}n^2x^2 = \cfrac{mn^2 - nm^2}{2}x^2 \\ \end{cases} $$所以\(\underset{x \to 0}{\lim}\cfrac{(1 + mx)^n - (1 + nx)^m}{x^2} = \cfrac{mn^2 - nm^2}{2}\)。
【415】\(\underset{x \to \infty}{\lim}\cfrac{(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)}{(5x - 1)^5}\)
解：\((x - 1)(x - 2)(x - 3)(x - 4)(x - 5)\)的最高次项为\(x^5\)，\((5x - 1)^5\)的最高次项为\(5^5x^5\)， 所以\(\underset{x \to \infty}{\lim}\cfrac{(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)}{(5x - 1)^5} = \cfrac{1}{5^5}\)。
【416】\(\underset{x \to \infty}{\lim}\cfrac{(2x - 3)^{20}(3x + 2)^{30}}{(2x + 1)^{50}}\)
解：因为\((2x - 3)^{20}(3x + 2)^{30}\)的最高次项为\(2^{20}x^{20}\cdot 3^{30}x^{30} = 2^{20}\cdot3^{30}\cdot x^{50}\)， \((2x + 1)^{50}\)的最高次项为\(2^{50}x^{50}\)。
所以\(\underset{x \to \infty}{\lim}\cfrac{(2x - 3)^{20}(3x + 2)^{30}}{(2x + 1)^{50}} = \cfrac{2^{20}\cdot3^{30}}{2^{50}} = 1.5^{30}\)。
【417】\(\underset{x \to \infty}{\lim}\cfrac{(x+1)(x^2+1)\cdots(x^n+1)}{\left((nx)^n + 1\right)^{\frac{n+1}{2}}}\)
解：因为\((x+1)(x^2+1)\cdots(x^n+1)\)的最高次项为\(x^{\sum_{i=1}^n i} = x^{\frac{n(n+1)}{2}}\)， \(\left((nx)^n + 1\right)^{\frac{n+1}{2}}\)的最高次项为\((nx)^{n\frac{n+1}{2}} = n^{\frac{n(n+1)}{2}}x^{\frac{n(n+1)}{2}}\)。
所以\(\underset{x \to \infty}{\lim}\cfrac{(x+1)(x^2+1)\cdots(x^n+1)}{\left((nx)^n + 1\right)^{\frac{n+1}{2}}} = n^{\frac{n(n+1)}{2}}\)。
【418】\(\underset{x \to 3}{\lim}\cfrac{x^2 - 5x + 6}{x^2 -8x + 15}\)
解：因为\(x^2 - 5x + 6 = (x - 3)(x - 2)\)，并且\(x^2 -8x + 15 = (x - 3)(x - 5)\)。 所以\(\underset{x \to 3}{\lim}\cfrac{x^2 - 5x + 6}{x^2 -8x + 15} = \cfrac{3 - 2}{3 - 5} = -0.5\)。
【419】\(\underset{x \to 1}{\lim}\cfrac{x^3 - 3x + 2}{x^4 -4x + 3}\)
解：因为\(x^3 - 3x + 2 = (x - 1)^2(x + 2)\)，并且\(x^4 -4x + 3 = (x - 1)^2(x^2 + 2x + 3)\)。 所以\(\underset{x \to 1}{\lim}\cfrac{x^3 - 3x + 2}{x^4 -4x + 3} = \cfrac{1 + 2}{1 + 2 + 3} = 0.5\)。
【420】\(\underset{x \to 1}{\lim}\cfrac{x^3 - 3x + 2}{x^4 - x^3 - x + 1}\)
解：因为\(x^3 - 3x + 2 = (x - 1)^2(x + 2)\)，并且\(x^4 - x^3 - x + 1 = (x - 1)^2(x^2 + x + 1)\)。 所以\(\underset{x \to 1}{\lim}\cfrac{x^3 - 3x + 2}{x^4 - x^3 - x + 1} = \cfrac{1 + 2}{1 + 1 + 1} = 1\)。
【421】\(\underset{x \to 2}{\lim}\cfrac{x^3 - 2x^2 - 4x + 8}{x^4 - 8x^2 + 16}\)
解：因为\(x^3 - 2x^2 - 4x + 8 = (x - 2)^2(x + 2)\)，并且\(x^4 - 8x^2 + 16 = (x - 2)^2(x + 2)^2\)。 所以\(\underset{x \to 2}{\lim}\cfrac{x^3 - 2x^2 - 4x + 8}{x^4 - 8x^2 + 16} = \cfrac{2 + 2}{(2 + 2)^2} = 0.25\)。
【422】\(\underset{x \to -1}{\lim}\cfrac{x^3 - 2x - 1}{x^5 - 2x - 1}\)
解：因为\(x^3 - 2x - 1 = (x + 1)(x^2 - x - 1)\)，并且\(x^5 - 2x - 1 = (x + 1)(x^4 - x^3 + x^2 - x - 1)\)。 所以\(\underset{x \to -1}{\lim}\cfrac{x^3 - 2x - 1}{x^5 - 2x - 1} = \cfrac{1 + 1 - 1}{1 + 1 + 1 + 1 - 1} = \cfrac{1}{3}\)。
【423】\(\underset{x \to 2}{\lim}\cfrac{(x^2 - x - 2)^{20}}{(x^3 - 12x + 16)^{10}}\)
解：因为\((x^2 - x - 2)^{20} = (x - 2)^{20}(x + 1)^{20}\)，并且\((x^3 - 12x + 16)^{10} = (x - 2)^{20}(x + 4)^{10}\)。 所以\(\underset{x \to 2}{\lim}\cfrac{(x^2 - x - 2)^{20}}{(x^3 - 12x + 16)^{10}} = \cfrac{(2 + 1)^{20}}{(2 + 4)^{10}} = 1.5^{10}\)。
【424】\(\underset{x \to 1}{\lim}\cfrac{x + x^2 + \cdots + x^n - n}{x - 1}\)
解：因为\(x + x^2 + \cdots + x^n - n = (x - 1)(x^{n-1} + 2x^{n-2} + \cdots + (n-1)x + n)\)。 所以\(\underset{x \to 1}{\lim}\cfrac{x + x^2 + \cdots + x^n - n}{x - 1} = 1 + 2 + \cdots + n = \cfrac{n(n+1)}{2}\)。
【425】\(\underset{x \to 1}{\lim}\cfrac{x^m - 1}{x^n - 1}\)，\(m,n\)均为正整数。
解：因为\(x^m - 1 = (x - 1)(x^{m-1} + x^{m-2} + \cdots + x + 1)\)。 所以\(\underset{x \to 1}{\lim}\cfrac{x^m - 1}{x^n - 1} = \cfrac{m}{n}\)。
【426】\(\underset{x \to a}{\lim}\cfrac{(x^n - a^n)-na^{n-1}(x - a)}{(x - a)^2}\)，\(n\)为正整数。
解：令\(y = x - a\)，有\(\underset{x \to a}{\lim}\cfrac{(x^n - a^n)-na^{n-1}(x - a)}{(x - a)^2} = \underset{y \to 0}{\lim}\cfrac{(y + a)^n - a^n -na^{n-1}y}{y^2}\)。
又因为\((y + a)^n - a^n -na^{n-1}y\)的常数项为\(a^n - a^n = 0\)，一次项为\(\mathrm{C}_n^{n-1}a^{n-1}y - na^{n-1}y = 0\)， 二次项为\(\mathrm{C}_n^{n-2}a^{n-2}y^2\)。
所以\(\underset{y \to 0}{\lim}\cfrac{(y + a)^n - a^n -na^{n-1}y}{y^2} = \mathrm{C}_n^{n-2}a^{n-2} = \cfrac{n(n-1)}{2}a^{n-2}\)。
【427】\(\underset{x \to 1}{\lim}\cfrac{x^{n+1} - (n + 1)x + n}{(x - 1)^2}\)，\(n\)为正整数。
解：令\(y = x - 1\)，有\(\underset{x \to 1}{\lim}\cfrac{x^{n+1} - (n + 1)x + n}{(x - 1)^2} = \underset{y \to 0}{\lim}\cfrac{(y + 1)^{n+1}-(n+1)y - 1}{y^2} = \cfrac{n(n+1)}{2}\)
【428】\(\underset{x \to 1}{\lim}\left(\cfrac{m}{1 - x^m} - \cfrac{n}{1 - x^n}\right)\)，\(m,n\)为正整数。
解：\(\cfrac{m}{1 - x^m} - \cfrac{n}{1 - x^n} = \cfrac{m(x^{n-1} + x^{n-2} + \cdots + 1) - n(x^{m -1} + x^{m-2} + \cdots + 1)}{(1 - x)(x^{n - 1} + \cdots + 1)(x^{m - 1} + \cdots + 1)}\)
当\(m > n\)时，有\(\cfrac{m(x^{n-1} + x^{n-2} + \cdots + 1) - n(x^{m -1} + x^{m-2} + \cdots + 1)}{(1 - x)(x^{n - 1} + \cdots + 1)(x^{m - 1} + \cdots + 1)} = \cfrac{(m - n)(x^{n-1} + x^{n-2} + \cdots + 1) - n(x^{m-1} + \cdots + x^{n})}{(1 - x)(x^{n - 1} + \cdots + 1)(x^{m - 1} + \cdots + 1)}\)
对上式的分子进行因式分解有：
$$ (m - n)(x^{n-1} + x^{n-2} + \cdots + 1) - n(x^{m-1} + \cdots + x^{n}) = (1 - x)\left(nx^{m-2} + 2nx^{m-3} + \cdots + (m-n)nx^{n-1} + (m - n)(n-1)x^{n-2} + \cdots + (m-n)\right) $$所以，当\(m > n\)时，\(\underset{x \to 1}{\lim}\left(\cfrac{m}{1 - x^m} - \cfrac{n}{1 - x^n}\right) = \cfrac{n\sum_1^{m-n} i + (m-n)\sum_1^{n-1} i}{nm} = \cfrac{m - n}{2}\)
当\(m < n\)时，有\(\cfrac{m(x^{n-1} + x^{n-2} + \cdots + 1) - n(x^{m -1} + x^{m-2} + \cdots + 1)}{(1 - x)(x^{n - 1} + \cdots + 1)(x^{m - 1} + \cdots + 1)} = -\cfrac{(n - m)(x^{m-1} + x^{m-2} + \cdots + 1) - m(x^{n-1} + x^{n-2} + \cdots + x^m)}{(1 - x)(x^{n - 1} + \cdots + 1)(x^{m - 1} + \cdots + 1)}\)
所以，当\(m < n\)时，\(\underset{x \to 1}{\lim}\left(\cfrac{m}{1 - x^m} - \cfrac{n}{1 - x^n}\right) = -\cfrac{m\sum_1^{n-m} i + (n-m)\sum_1^{m-1} i}{nm} = \cfrac{m - n}{2}\)
当\(m = n\)时，\(\underset{x \to 1}{\lim}\left(\cfrac{m}{1 - x^m} - \cfrac{n}{1 - x^n}\right) = \cfrac{(n-m)\sum_1^{m-1} i}{nm} = 0\)
综上，\(\underset{x \to 1}{\lim}\left(\cfrac{m}{1 - x^m} - \cfrac{n}{1 - x^n}\right) = \cfrac{m - n}{2}\)
【435】\(\underset{x \to +\infty}{\lim}\cfrac{\sqrt{x + \sqrt{x + \sqrt{x}}}}{\sqrt{x + 1}}\)。
解：\(\underset{x \to +\infty}{\lim}\cfrac{\sqrt{x + \sqrt{x + \sqrt{x}}}}{\sqrt{x + 1}} = \underset{x \to +\infty}{\lim}\cfrac{\sqrt{1 + \sqrt{\cfrac{1}{x} + \cfrac{\sqrt{x}}{x}}}}{\sqrt{1 + \cfrac{1}{x}}} = 1\)
【436】\(\underset{x \to +\infty}{\lim}\cfrac{\sqrt{x} + \sqrt[3]{x} + \sqrt[4]{x}}{\sqrt{2x + 1}}\)。
解：\(\underset{x \to +\infty}{\lim}\cfrac{\sqrt{x} + \sqrt[3]{x} + \sqrt[4]{x}}{\sqrt{2x + 1}} = \cfrac{\sqrt{2}}{2}\)
【437】\(\underset{x \to 4}{\lim}\cfrac{\sqrt{1 + 2x} - 3}{\sqrt{x} - 2}\)。
解：\(\underset{x \to 4}{\lim}\cfrac{\sqrt{1 + 2x} - 3}{\sqrt{x} - 2} = \underset{x \to 4}{\lim}\cfrac{(1 + 2x - 9)(\sqrt{x} + 2)}{(x - 4)(\sqrt{1 + 2x} + 3)} = \cfrac{2(2 + 2)}{3 + 3} = \cfrac{4}{3}\)
【438】\(\underset{x \to -8}{\lim}\cfrac{\sqrt{1 - x} - 3}{2 + \sqrt[3]{x}}\)。
解：\(\underset{x \to -8}{\lim}\cfrac{\sqrt{1 - x} - 3}{2 + \sqrt[3]{x}} = \underset{x \to -2}{\lim}\cfrac{\sqrt{1 - x^3} - 3}{2 + x} = \underset{x \to -2}{\lim}\cfrac{1 - x^3 - 9}{(2 + x)(\sqrt{1 - x^3} + 3)} = -\underset{x \to -2}{\lim}\cfrac{(x + 2)(x^2 - 2x + 4)}{(2 + x)(\sqrt{1 - x^3} + 3)} = 2\)
【439】\(\underset{x \to a}{\lim}\cfrac{\sqrt{x} - \sqrt{a} + \sqrt{x - a}}{\sqrt{x^2 - a^2}}\)。
解：\(\underset{x \to a}{\lim}\cfrac{\sqrt{x} - \sqrt{a} + \sqrt{x - a}}{\sqrt{x^2 - a^2}} = \underset{x \to a}{\lim}\cfrac{(x - a) + \sqrt{x - a}(\sqrt{x} + \sqrt{a})}{\sqrt{x^2 - a^2}(\sqrt{x} + \sqrt{a})} = \underset{x \to a}{\lim}\cfrac{(x - a)\sqrt{x^2 - a^2}+(x - a)\sqrt{x + a}(\sqrt{x} + \sqrt{a})}{(x-a)(x+a)(\sqrt{x} + \sqrt{a})} = \cfrac{1}{\sqrt{2a}} \)
【440】\(\underset{x \to 3}{\lim}\cfrac{\sqrt{x + 13} - 2\sqrt{x + 1}}{x^2 - 9}\)。
解：\(\underset{x \to 3}{\lim}\cfrac{\sqrt{x + 13} - 2\sqrt{x + 1}}{x^2 - 9} = \underset{x \to 3}{\lim}\cfrac{(x + 13) - 4(x + 1)}{(x^2 - 9)(\sqrt{x + 13} + 2\sqrt{x + 1})} = \underset{x \to 3}{\lim}\cfrac{9 - 3x}{(x^2 - 9)(\sqrt{x + 13} + 2\sqrt{x + 1})} = \cfrac{1}{16}\)
【441】\(\underset{x \to -2}{\lim}\cfrac{\sqrt[3]{x - 6} + 2}{x^3 + 8}\)。
解：\(\underset{x \to -2}{\lim}\cfrac{\sqrt[3]{x - 6} + 2}{x^3 + 8} = \underset{y \to -2}{\lim}\cfrac{y + 2}{(y^3 + 6)^3 + 8}\) \(\underset{y \to -2}{\lim}\cfrac{y + 2}{(y^3 + 6)^3 + 8} = \underset{y \to -2}{\lim}\cfrac{y + 2}{y^9 + 18y^6 + 108 y^3 + 224} = \underset{y \to -2}{\lim}\cfrac{y + 2}{(y + 2)(y^8 -2y^7 + 4y^6 + 10y^5 - 20y^4 + 40y^3 + 28y^2 - 56y + 112)} = \cfrac{1}{144}\)
【442】\(\underset{x \to 16}{\lim}\cfrac{\sqrt[4]{x} - 2}{\sqrt{x} - 4}\)。
解: \(\underset{x \to 16}{\lim}\cfrac{\sqrt[4]{x} - 2}{\sqrt{x} - 4} = \underset{y \to 2}{\lim}\cfrac{y - 2}{y^2 - 4} = \underset{y \to 2}{\lim}\cfrac{1}{y + 2} = \cfrac{1}{4}\)
【443】\(\underset{x \to 8}{\lim}\cfrac{\sqrt{9 + 2x} - 5}{\sqrt[3]{x} - 2}\)。
解：\(\underset{x \to 8}{\lim}\cfrac{\sqrt{9 + 2x} - 5}{\sqrt[3]{x} - 2} = \underset{y \to 2}{\lim}\cfrac{9 + 2y^3 - 25}{(y - 2)(\sqrt{9 + 2y^3} + 5)} = \underset{y \to 2}{\lim}\cfrac{2(y^2 + 2y + 4)}{\sqrt{9 + 2y^3} + 5} = 2.4\)
【444】\(\underset{x \to 0}{\lim}\cfrac{\sqrt[n]{1 + x} - 1}{x}\)。
解：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[n]{1 + x} - 1}{x} = \underset{y \to 1}{\lim}\cfrac{y - 1}{y^n - 1} = \underset{y \to 1}{\lim}\cfrac{1}{y^{n-1} + y^{n+2} + \cdots + 1} = \cfrac{1}{n}\)
【445】\(\underset{x \to 0}{\lim}\cfrac{\sqrt{1 - 2x - x^2} - (1 + x)}{x}\)。
解：\(\underset{x \to 0}{\lim}\cfrac{\sqrt{1 - 2x - x^2} - (1 + x)}{x} = \underset{x \to 0}{\lim}\cfrac{1 - 2x - x^2 - (1 + 2x + x^2)}{x\left(\sqrt{1 - 2x - x^2} + (1 + x) \right)} = \underset{x \to 0}{\lim}\cfrac{-2x(2 + x)}{x\left(\sqrt{1 - 2x - x^2} + (1 + x) \right)} = -2\)
【446】\(\underset{x \to 0}{\lim}\cfrac{\sqrt[3]{8 + 3x - x^2} - 2}{x + x^2}\)。
解：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[3]{8 + 3x - x^2} - 2}{x + x^2} = \underset{x \to 0}{\lim}\cfrac{8 + 3x - x^2 - 8}{x(1 + x)(\sqrt[3]{(8 + 3x - x^2)^2} + 2\sqrt[3]{8 + 3x - x^2} + 4)} = \underset{x \to 0}{\lim}\cfrac{3}{4 + 4 + 4} = \cfrac{1}{4} \)
【447】\(\underset{x \to 0}{\lim}\cfrac{\sqrt[3]{27 + x} - \sqrt[3]{27 - x}}{x + 2 \sqrt[3]{x^4}}\)。
解：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[3]{27 + x} - \sqrt[3]{27 - x}}{x + 2 \sqrt[3]{x^4}} = \underset{x \to 0}{\lim}\cfrac{(27 + x) - (27 - x)}{x\left(1 + 2 \sqrt[3]{x}\right)\left(\sqrt[3]{(27 + x)^2} + \sqrt[3]{(27 + x)(27 - x) + \sqrt[3]{(27 - x)^2}}\right)} = \cfrac{2}{27}\)
【448】\(\underset{x \to 0}{\lim}\cfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt[3]{1 + x} - \sqrt[3]{1 - x}}\)。
解：\(\underset{x \to 0}{\lim}\cfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt[3]{1 + x} - \sqrt[3]{1 - x}} = \underset{x \to 0}{\lim}\cfrac{\left((1 + x) - (1 - x)\right)\left(\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 - x^2} + \sqrt[3]{(1 - x)^2}\right)} {\left(\left(1 + x\right) - \left(1 - x\right)\right)\left(\sqrt{1 + x} + \sqrt{1 - x}\right)} = \cfrac{3}{2}\)
【449】\(\underset{x \to 7}{\lim}\cfrac{\sqrt{x + 2} - \sqrt[3]{x + 20}}{\sqrt[4]{x + 9} - 2}\)。
解：\(\underset{x \to 7}{\lim}\cfrac{\sqrt{x + 2} - \sqrt[3]{x + 20}}{\sqrt[4]{x + 9} - 2} = \underset{x \to 7}{\lim}\cfrac{\left((x + 2) - \sqrt[3]{(x + 20)^2}\right)\left(\sqrt[4]{x + 9} + 2\right)\left(\sqrt{x + 9} + 4\right)} {(x + 9 - 16)(\sqrt{x + 2} + \sqrt[3]{x + 20})} = \underset{x \to 7}{\lim}\cfrac{(x - 7)(x^2 + 12x + 56)\left(\sqrt[4]{x + 9} + 2\right)\left(\sqrt{x + 9} + 4\right)} {(x - 7)(\sqrt{x + 2} + \sqrt[3]{x + 20})((x + 2)^2 + (x+2)\sqrt[3]{(x+20)^2}+\sqrt[3]{(x+20)^4})} = \underset{x \to 7}{\lim}\cfrac{(x^2 + 12x + 56)\left(\sqrt[4]{x + 9} + 2\right)\left(\sqrt{x + 9} + 4\right)} {(\sqrt{x + 2} + \sqrt[3]{x + 20})((x + 2)^2 + (x+2)\sqrt[3]{(x+20)^2}+\sqrt[3]{(x+20)^4})} = \cfrac{6048}{1458} \)
【450】\(\underset{x \to 0}{\lim}\cfrac{\sqrt[3]{1 + \cfrac{x}{3}} - \sqrt[4]{1 + \cfrac{x}{4}}}{1 - \sqrt{1 - \cfrac{x}{2}}}\)。
解：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[3]{1 + \cfrac{x}{3}} - \sqrt[4]{1 + \cfrac{x}{4}}}{1 - \sqrt{1 - \cfrac{x}{2}}} = \underset{x \to 0}{\lim}\cfrac{\left[\left(1 + \cfrac{x}{3}\right)^4 - \left(1 + \cfrac{x}{4}\right)^3\right]\left(1 + \sqrt{1 - \cfrac{x}{2}}\right)} {\left(1 - 1 + \cfrac{x}{2}\right)\left[\left(1 + \cfrac{x}{3}\right)^{\frac{1}{3}} + \left(1 + \cfrac{x}{4}\right)^{\frac{1}{4}}\right] \left[\left(1 + \cfrac{x}{3} \right)^{\frac{2}{3}} + \left(1 + \cfrac{x}{4}\right)^{\frac{1}{2}}\right] \left[\left(1 + \cfrac{x}{3} \right)^{\frac{8}{3}} + \left(1 + \cfrac{x}{4}\right)\left(1 + \cfrac{x}{3}\right)^{\frac{4}{3}} + \left(1 + \cfrac{x}{3} \right)^{\frac{8}{3}} \right]} = \underset{x \to 0}{\lim}\cfrac{2\left[\left(\cfrac{x}{3}\right)^4 + 4\left(\cfrac{x}{3}\right)^3 + 6\left(\cfrac{x}{3}\right)^2 + 4\left(\cfrac{x}{3}\right) + 1 - \left(\cfrac{x}{4} \right)^3 - 3\left(\cfrac{x}{4}\right)^2 - 3\left(\cfrac{x}{4}\right) - 1\right]\left(1 + \sqrt{1 - \cfrac{x}{2}}\right)} {x\left[\left(1 + \cfrac{x}{3}\right)^{\frac{1}{3}} + \left(1 + \cfrac{x}{4}\right)^{\frac{1}{4}}\right] \left[\left(1 + \cfrac{x}{3} \right)^{\frac{2}{3}} + \left(1 + \cfrac{x}{4}\right)^{\frac{1}{2}}\right] \left[\left(1 + \cfrac{x}{3} \right)^{\frac{8}{3}} + \left(1 + \cfrac{x}{4}\right)\left(1 + \cfrac{x}{3}\right)^{\frac{4}{3}} + \left(1 + \cfrac{x}{3} \right)^{\frac{8}{3}} \right]} = \cfrac{7}{36} \)
【451】\(\underset{x \to 0}{\lim}\cfrac{x^2}{\sqrt[5]{1+5x} - (1 + x)}\)
解：\(\underset{x \to 0}{\lim}\cfrac{x^2}{\sqrt[5]{1+5x} - (1 + x)} = \underset{x \to 0}{\lim}\cfrac{x^2\left[\left(1 + 5x\right)^{\frac{4}{5}} + (1 + x)\left(1 + 5x\right)^{\frac{3}{5}} + (1 + x)^2\left(1 + 5x\right)^{\frac{2}{5}} + (1 + x)^3\left(1 + 5x\right)^{\frac{1}{5}} + (1 + x)^4\right]} {(1 + 5x) - (1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5)} = -\cfrac{1}{2}\)
【452】\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} - \sqrt[n]{1 + \beta x}}{x}\), (\(m\)及\(n\)为整数)。
解：当\(m > 0, n > 0\)时，有：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} - \sqrt[n]{1 + \beta x}}{x} = \underset{x \to 0}{\lim}\cfrac{\left[\left(1 + \alpha x\right)^n - \left(1 + \beta x\right)^m \right]} {x \sum_{i=0}^{mn - 1}\left(1 + \beta x\right)^{\frac{i}{n}} \left(1 + \alpha x\right)^{\frac{mn - 1 - i}{mn}}} = \underset{x \to 0}{\lim}\cfrac{\sum_{i = 0}^nC_n^i\left(\alpha x\right)^i - \sum_{i = 0}^mC_m^i\left(\beta x\right)^i} {x \sum_{i=0}^{mn - 1}\left(1 + \beta x\right)^{\frac{i}{n}} \left(1 + \alpha x\right)^{\frac{mn - 1 - i}{mn}}} = \cfrac{\alpha n - \beta m}{mn}\)
当\(m < 0, n < 0\)时，令\(k = -m, l = -n\)，有:
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} - \sqrt[n]{1 + \beta x}}{x} = \underset{x \to 0}{\lim}\cfrac{\left(1 + \alpha x\right)^{-\frac{1}{k}} - \left(1 + \beta x\right)^{-\frac{1}{l}}}{x} = \underset{x \to 0}{\lim}\cfrac{\left(1 + \beta x\right)^{\frac{1}{l}} - \left(1 + \alpha x\right)^{\frac{1}{k}}} {x \left(1 + \beta x\right)^{\frac{1}{l}} \left(1 + \alpha x\right)^{\frac{1}{k}}} = \cfrac{\beta k - \alpha l}{lk} = \cfrac{\alpha n - \beta m}{mn} $$当\(m < 0, n > 0\)时，令\(k = -m\)有：
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} - \sqrt[n]{1 + \beta x}}{x} = \underset{x \to 0}{\lim}\cfrac{1 - \left(1 + \beta x\right)^{\frac{1}{n}}\left(1 + \alpha x\right)^{\frac{1}{k}}}{x\left(1 + \alpha x\right)^{\frac{1}{k}}} = \underset{x \to 0}{\lim}\cfrac{1 - \left(1 + \beta x\right)^k \left(1 + \alpha x\right)^n} {x\left(1 + \alpha x\right)^{\frac{1}{k}} \sum_{i = 0}^{nk - 1}\left[\left(1 + \beta x\right)^{\frac{1}{n}}\left(1 + \alpha x\right)^{\frac{1}{k}}\right]^i} = \cfrac{-(\alpha n + \beta k)}{nk} = \cfrac{\alpha n - \beta m}{mn} $$当\(m > 0, n < 0\)时，令\(l = -n\)有：
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} - \sqrt[n]{1 + \beta x}}{x} = \underset{x \to 0}{\lim}\cfrac{\left(1 + \alpha x\right)^{\frac{1}{m}}\left(1 + \beta x\right)^{\frac{1}{l}} - 1}{x\left(1 + \beta x\right)^{\frac{1}{l}}} = \underset{x \to 0}{\lim}\cfrac{\left(1 + \alpha x\right)^l\left(1 + \beta x\right)^m - 1} {x\left(1 + \beta x\right)^{\frac{1}{l}} \sum_{i = 0}^{lm - 1}\left[\left(1 + \alpha x\right)^{\frac{1}{m}}\left(1 + \beta x\right)^{\frac{1}{l}}\right]^i} = \cfrac{\alpha l + \beta m}{lm} = \cfrac{\alpha n - \beta m}{mn} $$当\(m = 0\)或者\(n = 0\)时没有意义。所以综上所述，当\(m \neq 0 且 n \neq 0\)时，\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} - \sqrt[n]{1 + \beta x}}{x} = \cfrac{\alpha n - \beta m}{mn}\)。
【453】\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} · \sqrt[n]{1 + \beta x} - 1}{x}\), (\(m\)及\(n\)为整数)。
解：当\(m > 0, n > 0\)时，有：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} · \sqrt[n]{1 + \beta x} - 1}{x} = \underset{x \to 0}{\lim}\cfrac{\left(1 + \alpha x\right)^n\left(1 + \beta x\right)^m - 1} {x\sum_{i = 0}^{nm - 1}\left[\left(1 + \alpha x\right)^{\frac{1}{m}}\left(1 + \beta x\right)^{\frac{1}{n}}\right]^i} = \cfrac{\alpha n + \beta m}{mn}\)
当\(m < 0, n < 0\)时，令\(k = -m, l = -n\)，有:
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} · \sqrt[n]{1 + \beta x} - 1}{x} = \underset{x \to 0}{\lim}\cfrac{1 - (1+\alpha x)^{\frac{1}{k}} · (1 + \beta x)^{\frac{1}{l}}}{x(1+\alpha x)^{\frac{1}{k}} · (1 + \beta x)^{\frac{1}{l}}} = \underset{x \to 0}{\lim}\cfrac{1 - (1+\alpha x)^l · (1 + \beta x)^k}{x(1+\alpha x)^{\frac{1}{k}} · (1 + \beta x)^{\frac{1}{l}} \sum_{i = 0}^{kl - 1}\left[\left(1 + \alpha x\right)^{\frac{1}{k}}\left(1 + \beta x\right)^{\frac{1}{l}}\right]^i} = \cfrac{-\alpha l - \beta k}{kl} = \cfrac{\alpha n + \beta m}{mn} $$当\(m < 0, n > 0\)时，令\(k = -m\)有：
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} · \sqrt[n]{1 + \beta x} - 1}{x} = \underset{x \to 0}{\lim}\cfrac{(1 + \beta x)^{\frac{1}{n}} - (1+\alpha x)^{\frac{1}{k}}}{x(1+\alpha x)^{\frac{1}{k}}} = \cfrac{\beta k - \alpha n}{nk} = \cfrac{\alpha n + \beta m}{mn} $$当\(m > 0, n < 0\)时，令\(l = -n\)有：
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} · \sqrt[n]{1 + \beta x} - 1}{x} = \underset{x \to 0}{\lim}\cfrac{(1+\alpha x)^{\frac{1}{m}} - (1 + \beta x)^{\frac{1}{l}}}{x(1+ \beta x)^{\frac{1}{l}}} = \cfrac{\alpha l - \beta m}{lm} = \cfrac{\alpha n + \beta m}{mn} $$当\(m = 0\)或者\(n = 0\)时没有意义。所以综上所述，当\(m \neq 0 且 n \neq 0\)时，\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1+\alpha x} · \sqrt[n]{1 + \beta x} - 1}{x} = \cfrac{\alpha n + \beta m}{mn}\)。
【454】设\(P(x) = a_1 x + a_2 x^2 + \cdots + a_n x^n\)，又\(m\)为整数，求证：\(\underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1 + P(x)} - 1}{x} = \cfrac{a_1}{m}\)
证：当\(m > 0\)时，有：
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1 + P(x)} - 1}{x} = \underset{x \to 0}{\lim}\cfrac{1 + P(x) - 1}{x \sum_{i = 0}^{m - 1}\left(1 + P(x)\right)^{\frac{i}{m}}} = \underset{x \to 0}{\lim}\cfrac{a_1 + a_2 x + \cdots + a_n x^{n - 1}}{\sum_{i = 0}^{m - 1}\left(1 + P(x)\right)^{\frac{i}{m}}} = \cfrac{a_1}{m} $$当\(m < 0\)时，令\(k = -m\)有：
$$ \underset{x \to 0}{\lim}\cfrac{\sqrt[m]{1 + P(x)} - 1}{x} = \underset{x \to 0}{\lim}\cfrac{1 - \left(1 + P(x)\right)^{\frac{1}{k}}}{x\left(1 + P(x)\right)^{\frac{1}{k}}} = \underset{x \to 0}{\lim}\cfrac{1 - 1 - P(x)}{x \left(1 + P(x)\right)^{\frac{1}{k}} \sum_{i = 0}^{k - 1}\left(1 + P(x)\right)^{\frac{i}{k}}} = \cfrac{-a_1}{k} = \cfrac{a_1}{m} $$【455】\(\underset{x \to 1}{\lim}\cfrac{\sqrt[m]{x} - 1}{\sqrt[n]{x} - 1}\), (\(m\)及\(n\)为整数)。
解：当\(m > 0, n > 0\)时，有：
$$ \underset{x \to 1}{\lim}\cfrac{\sqrt[m]{x} - 1}{\sqrt[n]{x} - 1} = \underset{x \to 1}{\lim}\cfrac{(x - 1)\sum_{i = 0}^{n - 1}x^{\frac{i}{n}}}{(x - 1)\sum_{i = 0}^{m - 1}x^{\frac{i}{m}}} = \cfrac{n}{m} $$当\(m < 0, n < 0\)时，令\(k = -m, l = -n\)有：
$$ \underset{x \to 1}{\lim}\cfrac{\sqrt[m]{x} - 1}{\sqrt[n]{x} - 1} = \underset{x \to 1}{\lim}\cfrac{(1 - x^{\frac{1}{k}})x^{\frac{1}{l}}}{(1 - x^{\frac{1}{l}})x^{\frac{1}{k}}} = \underset{x \to 1}{\lim}\cfrac{(1 - x)x^{\frac{1}{l}}\sum_{i = 0}^{l-1}x^{\frac{i}{l}}}{(1 - x)x^{\frac{1}{k}}\sum_{i = 0}^{k-1}x^{\frac{i}{k}}} = \cfrac{l}{k} = \cfrac{n}{m} $$当\(m > 0, n < 0\)时，令\(l = -n\)有：
$$ \underset{x \to 1}{\lim}\cfrac{\sqrt[m]{x} - 1}{\sqrt[n]{x} - 1} = \underset{x \to 1}{\lim}\cfrac{(x^{\frac{1}{m}} - 1)x^{\frac{1}{l}}}{(1 - x^{\frac{1}{l}})} = \underset{x \to 1}{\lim}\cfrac{(x - 1)x^{\frac{1}{l}}\sum_{i = 0}^{l-1}x^{\frac{i}{l}}}{(1 - x)\sum_{i = 0}^{m-1}x^{\frac{i}{m}}} = \cfrac{-l}{m} = \cfrac{n}{m} $$当\(m > 0, n < 0\)时，令\(k = -m\)有：
$$ \underset{x \to 1}{\lim}\cfrac{\sqrt[m]{x} - 1}{\sqrt[n]{x} - 1} = \underset{x \to 1}{\lim}\cfrac{(1 - x^{\frac{1}{k}})}{(x^{\frac{1}{n}} - 1)x^{\frac{1}{k}}} = \underset{x \to 1}{\lim}\cfrac{(1 - x)\sum_{i = 0}^{n-1}x^{\frac{i}{n}}}{(x - 1)x^{\frac{1}{k}}\sum_{i = 0}^{k-1}x^{\frac{i}{k}}} = \cfrac{n}{-k} = \cfrac{n}{m} $$当\(m = 0\)或者\(n = 0\)时没有意义。所以综上所述，当\(m \neq 0 且 n \neq 0\)时，\(\underset{x \to 1}{\lim}\cfrac{\sqrt[m]{x} - 1}{\sqrt[n]{x} - 1} = \cfrac{n}{m}\)。
【456】\(\underset{x \to 1}{\lim}\cfrac{(1 - \sqrt{x})(1 - \sqrt[3]{x})\cdots(1-\sqrt[n]{x})}{(1 - x)^{n - 1}}\)
解：\(\underset{x \to 1}{\lim}\cfrac{(1 - \sqrt{x})(1 - \sqrt[3]{x})\cdots(1-\sqrt[n]{x})}{(1 - x)^{n - 1}} = \underset{x \to 1}{\lim}\cfrac{(1 - x)^{n-1}}{(1 - x)^{n-1} · (1 + x^{\frac{1}{2}}) · (1 + x^{\frac{1}{3}} + x^{\frac{2}{3}}) \cdots \sum_{i = 0}^{n-1}x^{\frac{i}{n}}} = \cfrac{1}{n!}\)
【457】\(\underset{x \to +\infty}{\lim}\left[\sqrt{(x + a)(x + b)} - x\right]\)
解：\(\underset{x \to +\infty}{\lim}\left[\sqrt{(x + a)(x + b)} - x\right] = \underset{x \to +\infty}{\lim}\cfrac{(a + b)x + ab}{\sqrt{(x + a)(x + b)} + x} = \underset{x \to +\infty}{\lim}\cfrac{(a + b) + \frac{ab}{x}}{\sqrt{1 + \frac{a + b}{x} + \frac{ab}{x^2}} + 1} = \cfrac{a + b}{2} \)
【458】\(\underset{x \to +\infty}{\lim}\left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}\right)\)
解：\(\underset{x \to +\infty}{\lim}\left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}\right) = \underset{x \to +\infty}{\lim}\cfrac{x + \sqrt{x + \sqrt{x}} - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} = \underset{x \to +\infty}{\lim}\cfrac{\sqrt{1 + \sqrt{\frac{1}{x}}}}{\sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}} + 1} = \cfrac{1}{2}\)
【459】\(\underset{x \to +\infty}{\lim}x\left(\sqrt{x^2 + 2x} - 2\sqrt{x^2 + x} + x\right)\)
解：\(\underset{x \to +\infty}{\lim}x\left(\sqrt{x^2 + 2x} - 2\sqrt{x^2 + x} + x\right) = \underset{y \to +0}{\lim}\cfrac{\sqrt{\frac{1}{y^2} + \frac{2}{y}} - 2 \sqrt{\frac{1}{y^2} + \frac{1}{y}} + \frac{1}{y}}{y} = \underset{y \to +0}{\lim}\cfrac{\sqrt{1 + 2y} - 2 \sqrt{1 + y} + 1}{y^2} = \underset{y \to +0}{\lim}\cfrac{2\sqrt{1 + 2y} - 2 (1 + y)}{y^2(\sqrt{1 + 2y} + 1 + 2\sqrt{1 + y})} = \underset{y \to +0}{\lim}\cfrac{4(1 + 2y) - 4 (1 + y)^2}{y^2(\sqrt{1 + 2y} + 1 + 2\sqrt{1 + y})(2\sqrt{1 + 2y} + 2(1 + y))} = -\cfrac{1}{4}\)
【460】\(\underset{x \to +0}{\lim}\left(\sqrt{\cfrac{1}{x} + \sqrt{\cfrac{1}{x} + \sqrt{\cfrac{1}{x}}}} - \sqrt{\cfrac{1}{x} - \sqrt{\cfrac{1}{x} + \sqrt{\cfrac{1}{x}}}}\right)\)
解：\(\underset{x \to +0}{\lim}\left(\sqrt{\cfrac{1}{x} + \sqrt{\cfrac{1}{x} + \sqrt{\cfrac{1}{x}}}} - \sqrt{\cfrac{1}{x} - \sqrt{\cfrac{1}{x} + \sqrt{\cfrac{1}{x}}}}\right) = \underset{y \to +\infty}{\lim}\cfrac{y + \sqrt{y + \sqrt{y}} - y + \sqrt{y + \sqrt{y}}}{\sqrt{y + \sqrt{y + \sqrt{y}}} + \sqrt{y - \sqrt{y + \sqrt{y}}}} = \underset{y \to +\infty}{\lim}\cfrac{2\sqrt{y + \sqrt{y}}}{\sqrt{y + \sqrt{y + \sqrt{y}}} + \sqrt{y - \sqrt{y + \sqrt{y}}}} = 1\)
【461】\(\underset{x \to \infty}{\lim}\left(\sqrt[3]{x^3 + x^2 + 1} - \sqrt[3]{x^3 - x^2 + 1}\right)\)
解：\(\underset{x \to \infty}{\lim}\left(\sqrt[3]{x^3 + x^2 + 1} - \sqrt[3]{x^3 - x^2 + 1}\right) = \underset{x \to \infty}{\lim}\cfrac{2x^2}{\left(x^3 + x^2 + 1\right)^{\frac{2}{3}} + \left(x^3 + x^2 + 1\right)^{\frac{1}{3}}\left(x^3 - x^2 + 1\right)^{\frac{1}{3}} + \left(x^3 - x^2 + 1\right)^{\frac{2}{3}}} = \cfrac{2}{3}\)
【462】\(\underset{x \to +\infty}{\lim}\left(\sqrt[3]{x^3 + 3x^2} - \sqrt{x^2 - 2x}\right)\)
解：\(\underset{x \to +\infty}{\lim}\left(\sqrt[3]{x^3 + 3x^2} - \sqrt{x^2 - 2x}\right) = \underset{x \to +\infty}{\lim}\cfrac{(x^3 + 3x^2)^2 - (x^2 - 2x)^3}{(x^3 + 3x^2)^{\frac{5}{3}} + (x^3 + 3x^2)^{\frac{4}{3}}(x^2 - 2x)^{\frac{1}{2}} + (x^3 + 3x^2)(x^2 - 2x) + (x^3 + 3x^2)^{\frac{2}{3}}(x^2 - 2x)^{\frac{3}{2}} + (x^3 + 3x^2)^{\frac{1}{3}}(x^2 - 2x)^2 + (x^2 - 2x)^{\frac{5}{2}}} = \underset{x \to +\infty}{\lim}\cfrac{12x^5 - 3x^4 + 8x^3}{6 x^5 + \cdots} = 2\)
【463】\(\underset{x \to \infty}{\lim}x^{\frac{1}{3}}\left[(x + 1)^{\frac{2}{3}} - (x - 1)^{\frac{2}{3}} \right]\)
解：\(\underset{x \to \infty}{\lim}x^{\frac{1}{3}}\left[(x + 1)^{\frac{2}{3}} - (x - 1)^{\frac{2}{3}} \right] = \underset{x \to \infty}{\lim}\cfrac{x^{\frac{1}{3}}\left[(x+1)^2 - (x - 1)^2\right]}{(x + 1)^{\frac{4}{3}} + (x + 1)^{\frac{2}{3}}(x - 1)^{\frac{2}{3}} + (x - 1)^{\frac{4}{3}}} = \underset{x \to \infty}{\lim}\cfrac{4x^{\frac{4}{3}}}{3x^{\frac{4}{3}} + \cdots } = \cfrac{4}{3}\)
【464】\(\underset{x \to +\infty}{\lim}x^{\frac{3}{2}}\left(\sqrt{x+2} - 2\sqrt{x + 1} + \sqrt{x}\right)\)
解：\(\underset{x \to +\infty}{\lim}x^{\frac{3}{2}}\left(\sqrt{x+2} - 2\sqrt{x + 1} + \sqrt{x}\right) = \underset{x \to +\infty}{\lim}\cfrac{x^{\frac{3}{2}}\left[2\sqrt{x(x+2)} - 2(x+1)\right]}{\sqrt{x+2} + \sqrt{x} + 2\sqrt{x+1}} = \underset{x \to +\infty}{\lim}\cfrac{x^{\frac{3}{2}}\left[4x(x + 2) - 4(x+1)^2\right]}{\left(\sqrt{x+2} + \sqrt{x} + 2\sqrt{x+1}\right)\left(2\sqrt{x(x+2)} + 2(x+1)\right)} = \underset{x \to +\infty}{\lim}\cfrac{-4x^{\frac{3}{2}}}{16x^{\frac{3}{2}} + \cdots} = \cfrac{1}{4}\)
【465】\(\underset{x \to +\infty}{\lim}\left[\sqrt[n]{(x + a_1) \cdots (x + a_n)} - x\right]\)
解：\(\underset{x \to +\infty}{\lim}\left[\sqrt[n]{(x + a_1) \cdots (x + a_n)} - x\right] = \underset{x \to +\infty}{\lim}\cfrac{(x + a_1)\cdots (x + a_n) - x^n}{\sum_{i = 0}^{n-1}a^{\frac{n-1-i}{n}}x^i} = \sum a_i\)
【466】\(\underset{x \to \infty}{\lim}\cfrac{(x - \sqrt{x^2 - 1})^n + (x + \sqrt{x^2 - 1})^n}{x^n}\)
解：\(\underset{x \to \infty}{\lim}\cfrac{(x - \sqrt{x^2 - 1})^n + (x + \sqrt{x^2 - 1})^n}{x^n} = \underset{x \to \infty}{\lim}\left[\left(1 - \sqrt{1 - \cfrac{1}{x^2}}\right)^n + \left(1 + \sqrt{1 - \cfrac{1}{x^2}}\right)^n\right] = 2^n\)
【467】\(\underset{x \to 0}{\lim}\cfrac{(\sqrt{1 + x^2} + x)^n - (\sqrt{1 + x^2} - x)^n}{x}\)
解：\(\underset{x \to 0}{\lim}\cfrac{(\sqrt{1 + x^2} + x)^n - (\sqrt{1 + x^2} - x)^n}{x} = \underset{x \to 0}{\lim}\cfrac{(\sqrt{1 + x^2} + x - \sqrt{1 + x^2} + x)\sum_{i = 0}^{n-1}\left[\left(\sqrt{1 + x^2} + x\right)^{n-1-i}\left(\sqrt{1 + x^2} - x\right)^i\right]}{x} = 2n\)
【506】(1) \(\underset{x \to 0}{\lim} \left(\cfrac{1 + x}{2 + x}\right)^{\frac{1 - \sqrt{x}}{1 - x}}\) (2) \(\underset{x \to 1}{\lim} \left(\cfrac{1 + x}{2 + x}\right)^{\frac{1 - \sqrt{x}}{1 - x}}\) (3) \(\underset{x \to +\infty}{\lim} \left(\cfrac{1 + x}{2 + x}\right)^{\frac{1 - \sqrt{x}}{1 - x}}\)
解：(1) \(\underset{x \to 0}{\lim} \left(\cfrac{1 + x}{2 + x}\right)^{\frac{1 - \sqrt{x}}{1 - x}} = \cfrac{1}{2}\) (2) \(\underset{x \to 1}{\lim} \left(\cfrac{1 + x}{2 + x}\right)^{\frac{1 - \sqrt{x}}{1 - x}} = \sqrt{\cfrac{2}{3}}\) (3) \(\underset{x \to +\infty}{\lim} \left(\cfrac{1 + x}{2 + x}\right)^{\frac{1 - \sqrt{x}}{1 - x}} = 1\)
【508】\(\underset{x \to \infty}{\lim} \left(\cfrac{3x^2 - x + 1}{2x^2 + x + 1}\right)^{\frac{x^3}{1 - x}}\)
解：因为\(\underset{x \to \infty}{\lim} \cfrac{3x^2 - x + 1}{2x^2 + x + 1} = \cfrac{3}{2}\)，并且\(\underset{x \to \infty}{\lim} \cfrac{x^3}{1 - x} = -\infty\)， 所以\(\underset{x \to \infty}{\lim} \left(\cfrac{3x^2 - x + 1}{2x^2 + x + 1}\right)^{\frac{x^3}{1 - x}} = \left(\cfrac{3}{2}\right)^{-\infty} = 0\)
【607】设\(\underset{x \to a}{\lim} \varphi (x) = A, \underset{x \to A}{\lim} \psi (x) = B\)，是否可由此推出\(\underset{x \to a}{\lim} \psi (\varphi (x)) = B\)？
研究例子\(x \to 0\)时\(\psi (\varphi (x))\)的极限，\(\psi(x) = \begin{cases} 1, & x \neq 0 \\ 0, & x = 0 \end{cases}\),
\(\varphi (x) = \begin{cases} \cfrac{1}{q}, & x = \cfrac{p}{q} (p,q是互素的整数) \\
0，& x为无理数 \end{cases}\)。
解：不一定能推出，还需要\(\psi(x)\)在\(A\)的去心邻域内有\(\psi(x) \neq a\)。对于函数
$$ \varphi (x) = \begin{cases} \cfrac{1}{q}, & x = \cfrac{p}{q} (p,q是互素的整数) \\ 0，& x为无理数 \end{cases} $$ $$ \psi(x) = \begin{cases} 1, & x \neq 0 \\ 0, & x = 0 \end{cases} $$有\(\underset{x \to 0}{\lim} \varphi (x) = 0, \underset{x \to 0}{\lim} \psi (x) = 1\)。但是\(\psi (\varphi (x))\)是著名的Dirichlet函数，是处处不存在极限的。