一些重要的函数极限

根据洛必达法则，可以很容易计算得到\(\underset{x \to 0}{\lim}\cfrac{\sin{x}}{x} = 1\)。 关于自然对数底\(e\)，我们有极限\(\underset{x \to 0}{\lim}\left(1 + x \right)^{\frac{1}{x}} = e\)， \(\underset{x \to \infty}{\lim}\left(1 + \cfrac{1}{x} \right)^{x} = e\)。通过一些变量替换的技巧，这两个函数极限可以建立起计算幂函数、指数函数、对数函数、 三角函数以及反三角函数极限的桥梁。

根据极限\(\underset{x \to 0}{\lim}\cfrac{\sin{x}}{x} = 1\)，我们可以很容易得出如下的推论：

$$ \underset{x \to 0}{\lim}\cfrac{\tan{x}}{x} = \underset{x \to 0}{\lim}\cfrac{\sin{x}}{x\cos x} = 1 $$ $$ \underset{x \to 0}{\lim}\cfrac{\arcsin{x}}{x} \overset{\underline{y = \arcsin(x)}}{} \underset{y \to 0}{\lim}\cfrac{y}{\sin y} = 1 $$ $$ \underset{x \to 0}{\lim}\cfrac{\arctan{x}}{x} \overset{\underline{y = \arctan(x)}}{} \underset{y \to 0}{\lim}\cfrac{y}{\tan y} = 1 $$

根据极限\(\underset{x \to 0}{\lim}\left(1 + x \right)^{\frac{1}{x}} = e\)，\(\underset{x \to \infty}{\lim}\left(1 + \cfrac{1}{x} \right)^{x} = e\)， 可以得到\(\underset{x \to 0}{\lim} \cfrac{\ln \left(1 + x \right)}{x} = 1, \underset{x \to \infty}{\lim} x\ln \left(1 + \cfrac{1}{x} \right) = 1\)。经过变量替换， 我们还可以得到习题【541】的极限\(\underset{x \to 0}{\lim} \cfrac{a^x - 1}{x} = \ln a, a > 0\)。

【471】\(\underset{x \to 0}{\lim}\cfrac{\sin(5x)}{x}\)

解：\(\underset{x \to 0}{\lim}\cfrac{\sin(5x)}{x} = \underset{x \to 0}{\lim}\cfrac{5\sin(5x)}{5x} = 5\)

【473】\(\underset{x \to \pi}{\lim}\cfrac{\sin(mx)}{\sin(nx)}\)，\(m,n\)均为整数。

解：令\(y = x - \pi\)，有:\(\underset{x \to \pi}{\lim}\cfrac{\sin(mx)}{\sin(nx)} = \underset{y \to 0}{\lim}\cfrac{\sin(my - m\pi)}{\sin(ny - n\pi)} = \underset{y \to 0}{\lim}\cfrac{\sin(my)\cos(m\pi) - \sin(m\pi)\cos(my)}{\sin(ny)\cos(n\pi) - \sin(n\pi)\cos(ny)} = \underset{y \to 0}{\lim}\cfrac{-1^m\sin(my)}{-1^n\sin(ny)} = \underset{y \to 0}{\lim}\left[-1^{m - n} · \cfrac{m}{n} · \cfrac{\sin(my)}{my} · \cfrac{ny}{\sin(ny)} \right]\)

所以，\(\underset{x \to \pi}{\lim}\cfrac{\sin(mx)}{\sin(nx)} = -1^{m - n}·\cfrac{m}{n}\)。

【474】\(\underset{x \to 0}{\lim}\cfrac{1 - \cos x}{x^2}\)

解：\(\underset{x \to 0}{\lim}\cfrac{1 - \cos x}{x^2} = \underset{x \to 0}{\lim}\cfrac{2\sin^2{\frac{x}{2}}}{x^2} = \underset{x \to 0}{\lim}\cfrac{\sin^2{\frac{x}{2}}}{2\frac{x^2}{4}} = \cfrac{1}{2}\)

【475】\(\underset{x \to 0}{\lim}\cfrac{\tan x - \sin x}{\sin^3 x}\)

解：\(\underset{x \to 0}{\lim}\cfrac{\tan x - \sin x}{\sin^3 x} = \underset{x \to 0}{\lim}\cfrac{\frac{\sin x}{\cos x} - \sin x}{\sin^3 x} = \underset{x \to 0}{\lim}\cfrac{1 - \cos x}{\sin^2 x \cos x} = \underset{x \to 0}{\lim}\cfrac{1 - \cos x}{(1 - \cos^2 x) \cos x} = \underset{x \to 0}{\lim}\cfrac{1}{(1 + \cos x) \cos x} = \cfrac{1}{2} \)

【476】\(\underset{x \to 0}{\lim}\cfrac{\sin(5x) - \sin(3x)}{\sin x}\)

解：\(\underset{x \to 0}{\lim}\cfrac{\sin(5x) - \sin(3x)}{\sin x} = \underset{x \to 0}{\lim}\cfrac{\sin(3x)\cos(2x) + \cos(3x)\sin(2x) - \sin(3x)}{\sin x} = \underset{x \to 0}{\lim}\cfrac{\sin(3x)\left[\cos(2x) - 1\right] + 2\cos(3x)\sin x \cos x}{\sin x} = \underset{x \to 0}{\lim}\cfrac{-2\sin(3x)\sin^2(x) + 2\cos(3x)\sin x \cos x}{\sin x} = \underset{x \to 0}{\lim}\left[-2\sin(3x)\sin x + 2\cos(3x) \cos x\right] = 2\)

【477】\(\underset{x \to 0}{\lim}\cfrac{\cos x - \cos(3x)}{x^2}\)

解：因为\(\cos(3x) = \cos(x)\cos(2x) - \sin(x)\sin(2x) = \cos(x)\left[\cos^2(x) - \sin^2(x)\right] - 2\sin^2(x)\cos(x) = \cos(x)\left[\cos^2(x) - 3\sin^2(x)\right] \)

所以\(\underset{x \to 0}{\lim}\cfrac{\cos x - \cos(3x)}{x^2} = \underset{x \to 0}{\lim}\cfrac{\cos(x)\left[1 - \cos^2(x) + 3\sin^2(x)\right]}{x^2} = \underset{x \to 0}{\lim}4 \cos(x)\cfrac{\sin^2(x)}{x^2} = 4\)

【478】\(\underset{x \to 0}{\lim}\cfrac{1 + \sin x - \cos x}{1 + \sin(px) - \cos(px)}\)

解：\(\underset{x \to 0}{\lim}\cfrac{1 + \sin x - \cos x}{1 + \sin(px) - \cos(px)} = \underset{x \to 0}{\lim}\cfrac{2\sin^2{\frac{x}{2} + 2\sin \frac{x}{2} \cos \frac{x}{2}}}{2\sin^2\frac{px}{2} + 2\sin \frac{px}{2}\cos \frac{px}{2}} = \underset{x \to 0}{\lim}\cfrac{1}{p} · \cfrac{\sin \frac{x}{2}}{\frac{x}{2}} · \cfrac{\frac{px}{x}}{\sin \frac{px}{2}} · \cfrac{\sin \frac{x}{2} + \cos \frac{x}{2}}{\sin \frac{px}{2} + \cos \frac{px}{2}} = \cfrac{1}{p}\)

【479】\(\underset{x \to \frac{\pi}{4}}{\lim} \tan(2x)\tan\left(\frac{\pi}{4} - x\right)\)

解：\(\underset{x \to \frac{\pi}{4}}{\lim} \tan(2x)\tan\left(\frac{\pi}{4} - x\right) = \underset{y \to 0}{\lim} \tan(2y + \frac{\pi}{2})\tan(-y) = \underset{y \to 0}{\lim} \cfrac{\sin(2y)\cos \frac{\pi}{2} + \cos(2y)\sin\frac{\pi}{2}}{\cos(2y)\cos\frac{\pi}{2} - \sin(2y)\sin\frac{\pi}{2}} · \cfrac{-\sin y}{\cos y} = \underset{y \to 0}{\lim} \cfrac{\cos(2y)}{- 2\sin y\cos y} · \cfrac{-\sin y}{\cos y} = \cfrac{1}{2}\)

【480】\(\underset{x \to 1}{\lim} (1 - x)\tan\left(\cfrac{\pi x}{2}\right)\)

解：\(\underset{x \to 1}{\lim} (1 - x)\tan\left(\cfrac{\pi x}{2}\right) = \underset{y \to 0}{\lim} \left[-y \tan \left(\cfrac{\pi(y + 1)}{2} \right) \right] = \underset{y \to 0}{\lim} \left[-y \cfrac{\cos \frac{\pi y}{2}}{-\sin \frac{\pi y}{2}}\right] = \cfrac{2}{\pi} \)

【482】\(\underset{x \to a}{\lim} \cfrac{\sin x - \sin a}{x - a}\)

解：\(\underset{x \to a}{\lim} \cfrac{\sin x - \sin a}{x - a} = \underset{x \to a}{\lim} \cfrac{2\sin\left(\frac{x - a}{2}\right)\cos\left(\frac{x + a}{2}\right)}{x - a} = \cos a\)

【483】\(\underset{x \to a}{\lim} \cfrac{\cos x - \cos a}{x - a}\)

解：\(\underset{x \to a}{\lim} \cfrac{\cos x - \cos a}{x - a} = \underset{x \to a}{\lim} \cfrac{-2\sin\left(\frac{x + a}{2}\right)\sin\left(\frac{x - a}{2}\right)}{x - a} = -\sin a\)

【484】\(\underset{x \to a}{\lim} \cfrac{\tan x - \tan a}{x - a}, a \neq \left(k + \cfrac{1}{2}\right)\pi, k \in N\)

解：\(\underset{x \to a}{\lim} \cfrac{\tan x - \tan a}{x - a} = \underset{x \to a}{\lim} \cfrac{\cfrac{\sin x}{\cos x} - \cfrac{\sin a}{\cos a}}{x - a} = \underset{x \to a}{\lim} \cfrac{\sin x \cos a - \sin a \cos x}{\cos x \cos a (x - a)} = \underset{x \to a}{\lim} \cfrac{\sin (x - a)}{\cos x \cos a (x - a)} = \cfrac{1}{\cos^2 a}\)

【485】\(\underset{x \to a}{\lim} \cfrac{\cot x - \cot a}{x - a}, a \neq k\pi, k \in N\)

解：\(\underset{x \to a}{\lim} \cfrac{\cot x - \cot a}{x - a} = \underset{x \to a}{\lim} \cfrac{\sin (a - x)}{\sin x \sin a (x - a)} = -\cfrac{1}{\sin^2 a}\)

【486】\(\underset{x \to a}{\lim} \cfrac{\sec x - \sec a}{x - a}\)

解：\(\underset{x \to a}{\lim} \cfrac{\sec x - \sec a}{x - a} = \underset{x \to a}{\lim} \cfrac{\cfrac{1}{\cos x} - \cfrac{1}{\cos a}}{x - a} = \underset{x \to a}{\lim} \cfrac{\cos a - \cos x}{\cos x \cos a (x - a)} = \underset{x \to a}{\lim} \cfrac{-2\sin\left(\frac{a + x}{2}\right)\sin\left(\frac{a - x}{2}\right)}{\cos x \cos a (x - a)} = \cfrac{\sin a}{\cos^2 a}\)

【487】\(\underset{x \to a}{\lim} \cfrac{\csc x - \csc a}{x - a}\)

解：\(\underset{x \to a}{\lim} \cfrac{\csc x - \csc a}{x - a} = \underset{x \to a}{\lim} \cfrac{\cfrac{1}{\sin x} - \cfrac{1}{\sin a}}{x - a} = \underset{x \to a}{\lim} \cfrac{\sin a - \sin x}{\sin x \sin a (x - a)} = \underset{x \to a}{\lim} \cfrac{2\sin\left(\frac{a - x}{2}\right)\cos\left(\frac{a + x}{2}\right)}{\sin x \sin a (x - a)} = -\cfrac{\cos a}{\sin^2 a}\)

【488】\(\underset{x \to 0}{\lim} \cfrac{\sin\left(a + 2x\right) - 2\sin\left(a + x\right) + \sin a}{x^2}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\sin\left(a + 2x\right) - 2\sin\left(a + x\right) + \sin a}{x^2} = \underset{x \to 0}{\lim} \cfrac{2\sin\left(\frac{a + 2x + a}{2}\right)\cos\left(\frac{a + 2x - a}{2}\right) - 2\sin\left(a + x\right)}{x^2} \underset{x \to 0}{\lim} \cfrac{2\sin\left(a + x\right)\left(\cos x - 1 \right)}{x^2} \underset{x \to 0}{\lim} \cfrac{2\sin\left(a + x\right)\left(-2\sin^2 \frac{x}{2} \right)}{x^2} = -\sin a \)

【489】\(\underset{x \to 0}{\lim} \cfrac{\cos\left(a + 2x\right) - 2\cos\left(a + x\right) + \cos a}{x^2}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\cos\left(a + 2x\right) - 2\cos\left(a + x\right) + \cos a}{x^2} = \underset{x \to 0}{\lim} \cfrac{2\cos\left(a + x\right)\cos x - 2\cos\left(a + x\right)}{x^2} \underset{x \to 0}{\lim} \cfrac{2\cos\left(a + x\right)\left(\cos x - 1 \right)}{x^2} \underset{x \to 0}{\lim} \cfrac{2\cos\left(a + x\right)\left(-2\sin^2 \frac{x}{2} \right)}{x^2} = -\cos a \)

【490】\(\underset{x \to 0}{\lim} \cfrac{\tan\left(a + 2x\right) - 2\tan\left(a + x\right) + \tan a}{x^2}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\tan\left(a + 2x\right) - 2\tan\left(a + x\right) + \tan a}{x^2} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\sin (a + 2x)}{\cos (a + 2x)} - \cfrac{2 \sin (a + x)}{\cos(a + x)} + \cfrac{\sin a}{\cos a}}{x^2} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\sin x}{\cos (a + 2x)\cos (a + x)} - \cfrac{\sin x}{\cos(a + x)\cos a}}{x^2} = \underset{x \to 0}{\lim} \cfrac{\sin x · \cfrac{\cos a - \cos (a + 2x)}{\cos (a + 2x)\cos (a + x)\cos a}}{x^2} = \underset{x \to 0}{\lim} \cfrac{-2\sin x · \sin (a + x) · \sin(-x)}{x^2\cos (a + 2x) \cos(a+x)\cos a} = \cfrac{2\sin a}{\cos^3 a} \)

【491】\(\underset{x \to 0}{\lim} \cfrac{\cot\left(a + 2x\right) - 2\cot\left(a + x\right) + \cot a}{x^2}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\cot\left(a + 2x\right) - 2\cot\left(a + x\right) + \cot a}{x^2} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\cos (a + 2x)}{\sin (a + 2x)} - \cfrac{2 \cos (a + x)}{\sin(a + x)} + \cfrac{\cos a}{\sin a}}{x^2} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\sin (-x)}{\sin (a + 2x)\sin (a + x)} - \cfrac{\sin (-x)}{\sin(a + x)\sin a}}{x^2} = \underset{x \to 0}{\lim} \cfrac{\sin (-x) · \cfrac{\sin a - \sin (a + 2x)}{\sin (a + 2x)\sin (a + x)\sin a}}{x^2} = \underset{x \to 0}{\lim} \cfrac{2\sin(-x) · \sin (-x) · \cos(a + x)}{x^2\sin (a + 2x) \sin(a+x)\sin a} = \cfrac{2\cos a}{\sin^3 a} \)

【492】\(\underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin(a + 2x) - \sin^2 a}{x}\)

解：

$$\begin{array}{rcl} \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin(a + 2x) - \sin^2 a}{x} & = & \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\left[\sin a \cos(2x) + \sin(2x) \cos a\right] - \sin^2 a}{x} \\ & = & \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin a \cos(2x) - \sin^2 a}{x} + \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin(2x) \cos a}{x} \\ & = & \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin a (1 - 2\sin^2 x) - \sin^2 a}{x} + 2\sin a \cos a \\ & = & \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin a - \sin^2 a}{x} - \underset{x \to 0}{\lim} \cfrac{\sin(a + x)\sin a · 2\sin^2 x}{x} + 2\sin a \cos a \\ & = & \underset{x \to 0}{\lim} \cfrac{\sin^2 a \cos x + \sin x \cos a \sin a - \sin^2 a}{x} - 0 + 2\sin a \cos a \\ & = & \underset{x \to 0}{\lim} \cfrac{\sin^2 a \left[\cos x - 1\right]}{x} + \underset{x \to 0}{\lim} \cfrac{\sin x \cos a \sin a}{x} - 0 + 2\sin a \cos a \\ & = & \underset{x \to 0}{\lim} \cfrac{-2\sin^2 a \sin^2{\frac{x}{2}}}{x} + \sin a \cos a - 0 + 2\sin a \cos a \\ & = & \cfrac{3}{2}\sin(2a) \end{array} $$

【493】\(\underset{x \to \frac{\pi}{6}}{\lim} \cfrac{2\sin^2 x + \sin x - 1}{2\sin^2x - 3\sin x + 1}\)

解：\( \underset{x \to \frac{\pi}{6}}{\lim} \cfrac{2\sin^2 x + \sin x - 1}{2\sin^2x - 3\sin x + 1} = \underset{x \to \frac{\pi}{6}}{\lim} \cfrac{(2\sin x - 1)(\sin x + 1)}{(2\sin x - 1)(\sin x - 1)} = -3 \)

【494】\(\underset{x \to 0}{\lim} \cfrac{1 - \cos (x) \cos (2x) \cos (3x)}{1 - \cos (x)}\)

解：因为\(\cos(3x) = \cos x\cos(2x) - \sin x\sin(2x) = \cos^3 x - 3\sin^2 x\cos x = 4\cos^3 x - 3\cos x\)，所以有

\(\underset{x \to 0}{\lim} \cfrac{1 - \cos (x) \cos (2x) \cos (3x)}{1 - \cos (x)} = \underset{x \to 0}{\lim} \cfrac{1 - \cos x \left(2\cos x - 1 \right)\left(4\cos^3 x - 3\cos x \right)}{1 - \cos x} = \underset{x \to 0}{\lim} \cfrac{-8\cos^6 x + 10\cos^4 x - 3\cos^2 x + 1}{1 - \cos x}\)

对上式的分子进行因式分解有：\(-8\cos^6 x + 10\cos^4 x - 3\cos^2 x + 1 = \left(1 - \cos x\right)\left(8\cos^5 x + 8\cos^4 x - 2\cos^3 x -2\cos^2 x + \cos x + 1\right)\)

所以\(\underset{x \to 0}{\lim} \cfrac{1 - \cos (x) \cos (2x) \cos (3x)}{1 - \cos (x)} = 8 + 8 - 2 - 2 + 1 + 1 = 14\)

【495】\(\underset{x \to \frac{\pi}{3}}{\lim} \cfrac{\sin \left(x - \frac{\pi}{3}\right)}{1 - 2\cos (x)}\)

解：令\(y = x - \cfrac{\pi}{3}\)，有 \(\underset{x \to \frac{\pi}{3}}{\lim} \cfrac{\sin \left(x - \frac{\pi}{3}\right)}{1 - 2\cos (x)} = \underset{y \to 0}{\lim} \cfrac{\sin y}{1 - 2\cos \left(y + \frac{\pi}{3}\right)} = \underset{y \to 0}{\lim} \cfrac{\sin y}{1 - \cos y + \sqrt{3}\sin y} = \underset{y \to 0}{\lim} \cfrac{2\sin \frac{y}{2}\cos \frac{y}{2}}{2\sin^2{\frac{y}{2}} + 2\sqrt{3}\sin \frac{y}{2}\cos \frac{y}{2}} = \cfrac{\sqrt{3}}{3}\)

【496】\(\underset{x \to \frac{\pi}{3}}{\lim} \cfrac{\tan^3 x - 3\tan x}{\cos \left(x + \frac{\pi}{6}\right)}\)

解：\(\underset{x \to \frac{\pi}{3}}{\lim} \cfrac{\tan^3 x - 3\tan x}{\cos \left(x + \frac{\pi}{6}\right)} = \underset{x \to \frac{\pi}{3}}{\lim} \cfrac{\tan x(\tan x - \sqrt{3})(\tan x + \sqrt{3})}{\sin \left(\frac{\pi}{3} - x\right)} = \underset{x \to \frac{\pi}{3}}{\lim} \cfrac{\tan x\tan (x - \frac{\pi}{3})(1 + \tan x \tan {\frac{\pi}{3}})(\tan x + \sqrt{3})}{\sin \left(\frac{\pi}{3} - x\right)} = -24 \)

【497】\(\underset{x \to 0}{\lim} \cfrac{\tan(a + x)\tan(a - x) - \tan^2 a}{x^2}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\tan(a + x)\tan(a - x) - \tan^2 a}{x^2} = \underset{x \to 0}{\lim} \cfrac{\frac{\tan a + \tan x}{1 - \tan a \tan x}\frac{\tan a - \tan x}{1 + \tan a \tan x} - \tan^2 a}{x^2} = \underset{x \to 0}{\lim} \cfrac{(\tan^4 a - 1)\tan^2 x}{x^2(1 - \tan^2 a \tan^2 x)} = \tan^4 a - 1 \)

【498】\(\underset{x \to \frac{\pi}{4}}{\lim} \cfrac{1 - \cot^3 x}{2 - \cot x - \cot^3 x}\)

解：令\(y = \cot x\)，有\(\underset{x \to \frac{\pi}{4}}{\lim} \cfrac{1 - \cot^3 x}{2 - \cot x - \cot^3 x} = \underset{y \to 1}{\lim} \cfrac{(1 - x)(x^2 + x + 1)}{(1 - x)(x^2 + x + 2)} = \cfrac{3}{4}\)

【499】\(\underset{x \to 0}{\lim} \cfrac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} = \underset{x \to 0}{\lim} \cfrac{\tan x - \sin x}{x^3\left(\sqrt{1 + \tan x} + \sqrt{1 + \sin x}\right)} = \underset{x \to 0}{\lim} \cfrac{2\sin x \sin^2{\frac{x}{2}}}{x^3\left(\sqrt{1 + \tan x} + \sqrt{1 + \sin x}\right)\cos x} = \cfrac{1}{4}\)

【500】\(\underset{x \to 0}{\lim} \cfrac{x^2}{\sqrt{1 + x\sin x} - \sqrt{\cos x}}\)

解：\(\underset{x \to 0}{\lim} \cfrac{x^2}{\sqrt{1 + x\sin x} - \sqrt{\cos x}} = \underset{x \to 0}{\lim} \cfrac{x^2\left(\sqrt{1 + x\sin x} + \sqrt{\cos x} \right)}{1 + x\sin x - \cos x} = \underset{x \to 0}{\lim} \cfrac{\left(\sqrt{1 + x\sin x} + \sqrt{\cos x} \right)}{\cfrac{2\sin^2{\frac{x}{2}}}{x^2} + \cfrac{x\sin x}{x^2}} = \cfrac{4}{3}\)

【501】\(\underset{x \to 0}{\lim} \cfrac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}\)

解：令\(y = \sqrt[6]{\cos x}\)，有： \(\underset{x \to 0}{\lim} \cfrac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} = \underset{y \to 1}{\lim} \cfrac{y^3 - y^2}{1 - y^{12}} = \underset{y \to 1}{\lim} \cfrac{y^2(y - 1)}{(1 - y)\underset{i = 0}{\overset{11}{\Sigma}} y^i} = -\cfrac{1}{12}\)

【502】\(\underset{x \to 0}{\lim} \cfrac{\sqrt{1 - \cos{x^2}}}{1 - \cos x}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\sqrt{1 - \cos{x^2}}}{1 - \cos x} = \underset{x \to 0}{\lim} \cfrac{\sqrt{2\sin^2{\frac{x^2}{2}}}}{2\sin^2 {\frac{x}{2}}} = \underset{x \to 0}{\lim} \cfrac{\sqrt{2}}{2} \cfrac{\sin{\frac{x^2}{2}}}{\frac{x^2}{2}} \cfrac{\left(\frac{x}{2}\right)^2}{\sin^2 {\frac{x}{2}}} \cfrac{\frac{x^2}{2}}{\frac{x^2}{4}} = \sqrt{2}\)

【503】\(\underset{x \to 0}{\lim} \cfrac{1 - \sqrt{\cos x}}{1 - \cos (\sqrt{x})}\)

解：\(\underset{x \to 0}{\lim} \cfrac{1 - \sqrt{\cos x}}{1 - \cos (\sqrt{x})} = \underset{x \to 0}{\lim} \cfrac{1 - \cos x}{\left(1 - \cos (\sqrt{x})\right)\left(1 + \cos x\right)} = \underset{x \to 0}{\lim} \cfrac{2\sin^2 {\frac{x}{2}}}{\left(\frac{x}{2}\right)^2} \cfrac{\left(\frac{\sqrt {x}}{2}\right)^2}{2\sin^2 {\frac{\sqrt{x}}{2}}} \cfrac{\left(\frac{x}{2}\right)^2}{\left(\frac{\sqrt {x}}{2}\right)^2\left(1 + \cos x\right)} = 0\)

【504】\(\underset{x \to 0}{\lim} \cfrac{1 - \cos x\sqrt{\cos {2x}}\sqrt[3]{\cos {3x}}}{x^2}\)

解： $$ \begin{array}{rcl} \underset{x \to 0}{\lim} \cfrac{1 - \cos x\sqrt{\cos {2x}}\sqrt[3]{\cos {3x}}}{x^2} & = & \underset{x \to 0}{\lim} \cfrac{1 - \cos x}{x^2} + \underset{x \to 0}{\lim} \cfrac{\cos x - \cos x\sqrt{\cos {2x}}\sqrt[3]{\cos {3x}}}{x^2} \\ & = & \cfrac{1}{2} + \underset{x \to 0}{\lim} \cos x\cfrac{1 - \sqrt{\cos {2x}}}{x^2} + \underset{x \to 0}{\lim} \cos x\sqrt{\cos {2x}}\cfrac{1 - \sqrt[3]{\cos {3x}}}{x^2} \\ & = & \cfrac{1}{2} + \underset{x \to 0}{\lim} \cfrac{2\sin^2 x\cos x}{x^2(1 + \cos {2x})} + \underset{x \to 0}{\lim} \cos x\sqrt{\cos {2x}}\cfrac{1 - \sqrt[3]{\cos {3x}}}{x^2} \\ & = & \cfrac{1}{2} + 1 + \underset{x \to 0}{\lim} \cos x\sqrt{\cos {2x}}\cfrac{1 - \cos {3x}}{x^2(\cos^{\frac{2}{3}}x + \cos^{\frac{1}{3}}x + 1)} \\ & = & \cfrac{1}{2} + 1 + \underset{x \to 0}{\lim} \cos x\sqrt{\cos {2x}}\cfrac{1 - \cos^3 x + 3\cos x\sin^2 x}{x^2(\cos^{\frac{2}{3}}x + \cos^{\frac{1}{3}}x + 1)} \\ & = & \cfrac{1}{2} + 1 + \underset{x \to 0}{\lim} \cos x\sqrt{\cos {2x}}\cfrac{(1 - \cos x)(\cos^2 x + \cos x + 1)}{x^2(\cos^{\frac{2}{3}}x + \cos^{\frac{1}{3}}x + 1)} + 1 \\ & = & \cfrac{1}{2} + 1 + \cfrac{1}{2} + 1 = 3 \\ \end{array} $$

【510】\(\underset{x \to \frac{\pi}{4} + 0}{\lim} \left[\tan \left(\frac{\pi}{8} + x\right)\right]^{\tan {2x}}\)

解：因为\(\underset{x \to \frac{\pi}{4} + 0}{\lim} \tan \left(\frac{\pi}{8} + x\right) = \tan \cfrac{5\pi}{8} > 1\)， 并且\(\underset{x \to \frac{\pi}{4} + 0}{\lim} \tan {2x} = -\infty\)。 所以\(\underset{x \to \frac{\pi}{4} + 0}{\lim} \left[\tan \left(\frac{\pi}{8} + x\right)\right]^{\tan {2x}}=\left(\tan \cfrac{5\pi}{8}\right)^{-\infty} = 0\)。

【511】\(\underset{x \to \infty}{\lim} \left(\cfrac{x^2 - 1}{x^2 + 1}\right)^{\frac{x - 1}{x + 1}}\)

解：因为\(\underset{x \to \infty}{\lim} \cfrac{x^2 - 1}{x^2 + 1} = 1\)，并且\(\underset{x \to \infty}{\lim} \cfrac{x - 1}{x + 1} = 1\)， 所以\(\underset{x \to \infty}{\lim} \left(\cfrac{x^2 - 1}{x^2 + 1}\right)^{\frac{x - 1}{x + 1}} = 1^1 = 1\)。

【512】\(\underset{x \to \infty}{\lim} \left(\cfrac{x^2 + 1}{x^2 - 1}\right)^{x^2}\)

解：\(\underset{x \to \infty}{\lim} \left(\cfrac{x^2 + 1}{x^2 - 1}\right)^{x^2} = \underset{x \to \infty}{\lim} \left(1 + \cfrac{2}{x^2 - 1}\right)^{x^2} = \underset{x \to \infty}{\lim} \left(1 + \cfrac{1}{\frac{x^2 - 1}{2}}\right)^{2\left(\frac{x^2 - 1}{2}\right)} · \left(1 + \cfrac{1}{\frac{x^2 - 1}{2}}\right) = e^2·1 = e^2 \)

【513】\(\underset{x \to \infty}{\lim} \left(\cfrac{x^2 + 2x - 1}{2x^2 - 3x - 2}\right)^{\frac{1}{x}}\)

解：因为\(\underset{x \to \infty}{\lim} \cfrac{x^2 + 2x - 1}{2x^2 - 3x - 2} = \cfrac{1}{2}\)，并且\(\underset{x \to \infty}{\lim} \cfrac{1}{x} = 0\)， 所以\(\underset{x \to \infty}{\lim} \left(\cfrac{x^2 + 2x - 1}{2x^2 - 3x - 2}\right)^{\frac{1}{x}} = \left(\cfrac{1}{2}\right)^0 = 1\)

【514】\(\underset{x \to 0}{\lim} \sqrt[x]{1 - 2x}\)

解：\(\underset{x \to 0}{\lim} \sqrt[x]{1 - 2x} = \underset{x \to 0}{\lim} \left(1 - 2x\right)^{\frac{1}{x}} = \underset{x \to 0}{\lim} \left[1 + (-2x)\right]^{\frac{1}{-2x} ·(-2)} = e^{-2}\)

【515】\(\underset{x \to \infty}{\lim} \left(\cfrac{x + a}{x - a}\right)^x\)

解：\(\underset{x \to \infty}{\lim} \left(\cfrac{x + a}{x - a}\right)^x = \underset{x \to \infty}{\lim} \left(1 + \cfrac{1}{\frac{x - a}{2a}}\right)^{\frac{x - a}{2a} · 2a}\left(1 + \cfrac{1}{\frac{x - a}{2a}}\right)^{a} = e^{2a}\)

【516】\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1 x + b_1}{a_2 x + b_2}\right)^x, a_1 > 0, a_2 > 0\)

解： $$ \begin{array}{rcl} \underset{x \to +\infty}{\lim} \left(\cfrac{a_1 x + b_1}{a_2 x + b_2}\right)^x & = & \underset{x \to +\infty}{\lim} \left(\cfrac{a_1}{a_2}\right)^x\left(\cfrac{x + \frac{b_1}{a_1}}{x + \frac{b_2}{a_2}}\right)^x \\ & = & \underset{x \to +\infty}{\lim} \left(\cfrac{a_1}{a_2}\right)^x \left(1 + \cfrac{1}{\cfrac{x + \frac{b_2}{a_2}}{\frac{b_1}{a_1} - \frac{b_2}{a_2}}}\right)^{\left[ \cfrac{x + \frac{b_2}{a_2}}{\frac{b_1}{a_1} - \frac{b_2}{a_2}} \right]·\left[\cfrac{b_1}{a_1} - \cfrac{b_2}{a_2} \right] - \cfrac{b_2}{a_2}} \\ & = & \underset{x \to +\infty}{\lim} \left(\cfrac{a_1}{a_2}\right)^x e^{\left[\frac{b_1}{a_1} - \frac{b_2}{a_2} \right]} \end{array} $$

当\(a_1 > a_2\)时，有\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1}{a_2}\right)^x = +\infty\)， 则\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1 x + b_1}{a_2 x + b_2}\right)^x = +\infty\)。

当\(a_1 = a_2\)时，有\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1}{a_2}\right)^x = 1\)， 则\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1 x + b_1}{a_2 x + b_2}\right)^x = e^{\left[\frac{b_1}{a_1} - \frac{b_2}{a_2} \right]}\)。

当\(a_1 < a_2\)时，有\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1}{a_2}\right)^x = 0\)， 则\(\underset{x \to +\infty}{\lim} \left(\cfrac{a_1 x + b_1}{a_2 x + b_2}\right)^x = 0\)。

【517】\(\underset{x \to 0}{\lim} \left(1 + x^2\right)^{\cot^2 x}\)

解：\(\underset{x \to 0}{\lim} \left(1 + x^2\right)^{\cot^2 x} = \underset{x \to 0}{\lim} \left(1 + x^2\right)^{\frac{1}{x^2}\frac{x^2}{\sin^2 x}\cos^2 x} = e\)

【518】\(\underset{x \to 1}{\lim} \left(1 + \sin {\pi x}\right)^{\cot {\pi x}}\)

解：\(\underset{x \to 1}{\lim} \left(1 + \sin {\pi x}\right)^{\cot {\pi x}} = \underset{x \to 1}{\lim} \left(1 + \sin {\pi x}\right)^{\frac{1}{\sin {\pi x}}\cos {\pi x}} = e^{-1}\)

【519】\(\underset{x \to 0}{\lim} \left(\cfrac{1 + \tan x}{1 + \sin x}\right)^{\frac{1}{\sin x}}\)

解：\(\underset{x \to 0}{\lim} \left(\cfrac{1 + \tan x}{1 + \sin x}\right)^{\frac{1}{\sin x}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{1}{\frac{1 + \sin x}{\tan x - \sin x}}\right)^{\cfrac{1 + \sin x}{\tan x - \sin x}\cfrac{\tan x - \sin x}{1 + \sin x}\cfrac{1}{\sin x}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{1}{\frac{1 + \sin x}{\tan x - \sin x}}\right)^{\cfrac{1 + \sin x}{\tan x - \sin x}\cfrac{1 - \cos x}{(1 + \sin x)\cos x}} = e^0 = 1\)

【520】\(\underset{x \to a}{\lim} \left(\cfrac{\sin x}{\sin a}\right)^{\frac{1}{x - a}}\)

解：\(\underset{x \to a}{\lim} \left(\cfrac{\sin x}{\sin a}\right)^{\frac{1}{x - a}} = \underset{x \to a}{\lim} \left(1 + \cfrac{1}{\frac{\sin a}{\sin x - \sin a}}\right)^{\cfrac{\sin a}{\sin x - \sin a}\cfrac{\sin x - \sin a}{(x - a)\sin a}} = \underset{x \to a}{\lim} \left(1 + \cfrac{1}{\frac{\sin a}{\sin x - \sin a}}\right)^{\cfrac{\sin a}{\sin x - \sin a}\cfrac{2\sin \frac{x-a}{2} \cos \frac{x + a}{2}}{(x - a)\sin a}} = e^{\frac{\cos a}{\sin a}} \)

【521】\(\underset{x \to 0}{\lim} \left(\cfrac{\cos x}{\cos {2x}}\right)^{\frac{1}{x^2}}\)

解：\(\underset{x \to 0}{\lim} \left(\cfrac{\cos x}{\cos {2x}}\right)^{\frac{1}{x^2}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{1}{\frac{\cos {2x}}{\cos x - \cos {2x}}}\right)^{\cfrac{\cos {2x}}{\cos x - \cos {2x}}\cfrac{\cos - \cos{2x}}{x^2 \cos {2x}}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{1}{\frac{\cos {2x}}{\cos x - \cos {2x}}}\right)^{\cfrac{\cos {2x}}{\cos x - \cos {2x}}\cfrac{2\sin{\frac{3x}{2}}\sin{\frac{x}{2}}}{x^2 \cos {2x}}} = e^{1.5} \)

【522】\(\underset{x \to \frac{\pi}{4}}{\lim} \left(\tan x\right)^{\tan{2x}}\)

解：\(\underset{x \to \frac{\pi}{4}}{\lim} \left(\tan x\right)^{\tan{2x}} = \underset{x \to \frac{\pi}{4}}{\lim} \left(1 + \cfrac{\sin x - \cos x}{\cos x}\right)^{\cfrac{\cos x}{\sin x - \cos x}\cfrac{\sin x - \cos x}{\cos x}\cfrac{2\sin x \cos x}{(\cos x + \sin x)(\cos x - \sin x)}} = \underset{x \to \frac{\pi}{4}}{\lim} \left(1 + \cfrac{\sin x - \cos x}{\cos x}\right)^{\cfrac{\cos x}{\sin x - \cos x}\cfrac{-2\sin x}{\cos x + \sin x}} = e^{-1}\)

【523】\(\underset{x \to \frac{\pi}{2}}{\lim} \left(\sin x\right)^{\tan{x}}\)

解：\(\underset{x \to \frac{\pi}{2}}{\lim} \left(\sin x\right)^{\tan{x}} = \underset{x \to \frac{\pi}{2}}{\lim} \left(1 + \cfrac{\cos^2 x}{\sin^2 x}\right)^{\tan^2{x}·(-\frac{1}{2\tan x})} = e^0 = 1 \)

【524】\(\underset{x \to 0}{\lim} \left[\tan \left(\cfrac{\pi}{4} - x\right)\right]^{\cot{x}}\)

解：\(\underset{x \to 0}{\lim} \left[\tan \left(\cfrac{\pi}{4} - x\right)\right]^{\cot{x}} = \underset{x \to 0}{\lim} \left(\cfrac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4}\tan x}\right)^{\cot{x}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{-2\tan x}{1 + \tan x}\right)^{\cfrac{1 + \tan x}{-2\tan x}\cfrac{-2\tan x}{1 + \tan x}\cfrac{1}{\tan x}} = e^{-2}\)

【525】\(\underset{x \to \infty}{\lim} \left(\sin \cfrac{1}{x} + \cos \cfrac{1}{x}\right)^{x}\)

解：\(\underset{x \to \infty}{\lim} \left(\sin \cfrac{1}{x} + \cos \cfrac{1}{x}\right)^{x} = \underset{y \to 0}{\lim} \left(\sin y + \cos y\right)^{\frac{1}{y}} = \underset{y \to 0}{\lim} \left(\sin y + \cos y\right)^{2 · \frac{1}{2y}} = \underset{y \to 0}{\lim} \left(1 + \sin {2y}\right)^{\frac{1}{\sin {2y}}·\frac{\sin {2y}}{2y}} = e^1 = 1\)

【526】\(\underset{x \to 0}{\lim} \sqrt[x]{\cos \sqrt{x}}\)

解：\(\underset{x \to 0}{\lim} \sqrt[x]{\cos \sqrt{x}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{\sin^2 \sqrt{x}}{\cos^2 \sqrt{x}}\right)^{-\frac{1}{2}\frac{1}{x}\frac{\cos^2 \sqrt{x}}{\sin^2 \sqrt{x}}\frac{\sin^2 \sqrt{x}}{\cos^2 \sqrt{x}}} = e^{-\frac{1}{2}} \)

【527】\(\underset{n \to \infty}{\lim} \left(\cfrac{n + x}{n - 1}\right)^n\)

解：\(\underset{n \to \infty}{\lim} \left(\cfrac{n + x}{n - 1}\right)^n = \underset{n \to \infty}{\lim} \left(1 + \cfrac{x + 1}{n - 1}\right)^{\frac{n - 1}{x + 1}\frac{n(x + 1)}{n - 1}} = e^{x + 1}\)

【528】\(\underset{n \to \infty}{\lim} \cos^n \cfrac{x}{\sqrt{n}}\)

解：\(\underset{n \to \infty}{\lim} \cos^n \cfrac{x}{\sqrt{n}} = \underset{n \to \infty}{\lim} \left(1 + \cfrac{\sin^2 \frac{x}{\sqrt{n}}}{\cos^2 \frac{x}{\sqrt{n}}}\right)^{\cfrac{\cos^2 \frac{x}{\sqrt{n}}}{\sin^2 \frac{x}{\sqrt{n}}} \cfrac{\sin^2 \frac{x}{\sqrt{n}}}{\cos^2 \frac{x}{\sqrt{n}}}\cfrac{-n}{2}} = e^{-\frac{x^2}{2}}\)

【529】\(\underset{x \to 0}{\lim} \cfrac{\ln (1 + x)}{x}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\ln (1 + x)}{x} = \underset{x \to 0}{\lim} \ln (1 + x)^{\frac{1}{x}} = \ln e = 1\)

【530】\(\underset{x \to +\infty}{\lim} x\left[\ln (x + 1) - \ln x\right]\)

解：\(\underset{x \to +\infty}{\lim} x\left[\ln (x + 1) - \ln x\right] = \underset{x \to +\infty}{\lim} \ln (1 + \frac{1}{x})^{x} = \ln e = 1 \)

【531】\(\underset{x \to a}{\lim} \cfrac{\ln x - \ln a}{x - a}\)

解：\(\underset{x \to a}{\lim} \cfrac{\ln x - \ln a}{x - a} = \underset{x \to a}{\lim} \ln \left(1 + \cfrac{x - a}{a}\right)^{\frac{1}{x - a}} = \ln e^{\frac{1}{a}} = \cfrac{1}{a} \)

【532】\(\underset{x \to +\infty}{\lim} \left[\sin \ln (x + 1) - \sin \ln x \right]\)

解：\(\underset{x \to +\infty}{\lim} \left[\sin \ln (x + 1) - \sin \ln x \right] = \underset{x \to +\infty}{\lim} 2\sin \left(\cfrac{\ln (x + 1) - \ln x}{2}\right) \cos \left(\cfrac{\ln (x + 1) + \ln x}{2}\right)\)

又因为\(\underset{x \to +\infty}{\lim} 2\sin \left(\cfrac{\ln (x + 1) - \ln x}{2}\right) = 0\)，并且\(\left|\cos \left(\cfrac{\ln (x + 1) + \ln x}{2}\right)\right| \leq 1\)， 所以\(\underset{x \to +\infty}{\lim} \left[\sin \ln (x + 1) - \sin \ln x \right] = 0\)

【533】\(\underset{x \to +\infty}{\lim} \cfrac{\ln (x^2 - x + 1)}{\ln (x^{10} + x + 1)}\)

解：\(\underset{x \to +\infty}{\lim} \cfrac{\ln (x^2 - x + 1)}{\ln (x^{10} + x + 1)} = \underset{x \to +\infty}{\lim} \cfrac{2\ln x + \ln (1 - \frac{1}{x} + \frac{1}{x^2})}{10\ln x + \ln (1 + \frac{1}{x^9} + \frac{1}{x^{10}})} = \cfrac{1}{5}\)

【534】\(\underset{x \to \infty}{\lim} \left(\lg \cfrac{100 + x^2}{1 + 100x^2} \right)\)

解：\(\underset{x \to \infty}{\lim} \left(\lg \cfrac{100 + x^2}{1 + 100x^2} \right) = \underset{x \to \infty}{\lim} \left(\lg \cfrac{\frac{100}{x^2} + 1}{\frac{1}{x^2} + 100} \right) = \lg \cfrac{1}{100} = -2\)

【535】\(\underset{x \to +\infty}{\lim} \cfrac{\ln (2 + e^{3x})}{\ln (3 + e^{2x})}\)

解：\(\underset{x \to +\infty}{\lim} \cfrac{\ln (2 + e^{3x})}{\ln (3 + e^{2x})} = \underset{x \to +\infty}{\lim} \cfrac{\ln {e^{3x}} + \ln (\frac{2}{e^{3x}} + 1)}{\ln {e^{2x}} + \ln (\frac{3}{e^{2x}})} = \underset{x \to +\infty}{\lim} \cfrac{3x + \ln (\frac{2}{e^{3x}} + 1)}{2x + \ln (\frac{3}{e^{2x}})} = \cfrac{3}{2}\)

【536】\(\underset{x \to +\infty}{\lim} \cfrac{\ln (1 + \sqrt{x} + \sqrt[3]{x})}{\ln (1 + \sqrt[3]{x} + \sqrt[4]{x})}\)

解：\(\underset{x \to +\infty}{\lim} \cfrac{\ln (1 + \sqrt{x} + \sqrt[3]{x})}{\ln (1 + \sqrt[3]{x} + \sqrt[4]{x})} = \underset{x \to +\infty}{\lim} \cfrac{\ln \sqrt{x} + \ln (x^{-\frac{1}{2}} + 1 + x^{-\frac{1}{6}})}{\ln \sqrt[3]{x} + \ln (x^{-\frac{1}{3}} + 1 + x^{-\frac{1}{12}})} = \cfrac{3}{2} \)

【537】\(\underset{h \to 0}{\lim} \cfrac{\lg (x + h) + \lg (x - h) - 2\lg x}{h^2}\)

解：\(\underset{h \to 0}{\lim} \cfrac{\lg (x + h) + \lg (x - h) - 2\lg x}{h^2} = \underset{h \to 0}{\lim} \lg (1 + \cfrac{-h^2}{x^2})^{\frac{1}{h^2}} = \underset{h \to 0}{\lim} \lg (1 + \cfrac{-h^2}{x^2})^{-\frac{x^2}{h^2} · -\frac{h^2}{x^2} · \frac{1}{h^2}} = -\cfrac{\lg e}{x^2} \)

【538】\(\underset{x \to 0}{\lim} \cfrac{\ln \tan \left(\frac{\pi}{4} + ax\right)}{\sin {bx}}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\ln \tan \left(\frac{\pi}{4} + ax\right)}{\sin {bx}} = \underset{x \to 0}{\lim} \cfrac{\ln \cfrac{1 + \tan {ax}}{1 - \tan {ax}}}{\sin {bx}} = \underset{x \to 0}{\lim} \ln \left[\left(1 + \cfrac{2\tan {ax}}{1 - \tan {ax}}\right)^{\cfrac{1 - \tan {ax}}{2\tan {ax}}\cfrac{2\sin {ax}}{\sin {bx}\cos {ax}(1 - \tan {ax}) }}\right] = \cfrac{2a}{b}\)

【539】\(\underset{x \to 0}{\lim} \cfrac{\ln \cos {ax}}{\ln \cos {bx}}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\ln \cos {ax}}{\ln \cos {bx}} = \underset{x \to 0}{\lim} \cfrac{\ln \cos {ax}}{\cos {ax} - 1} · \cfrac{\cos {bx} - 1}{\ln \cos {bx}} · \cfrac{\cos {ax} - 1}{\cos {bx} - 1}\)

因为\(\underset{x \to 0}{\lim} \cfrac{\ln \cos {ax}}{\cos {ax} - 1} = \underset{x \to 0}{\lim} \left[\ln \left(1 + \cfrac{\sin^2 {ax}}{\cos^2 {ax}} \right)^{\cfrac{\cos^2 {ax}}{\sin^2 {ax}} · \cfrac{-1}{2} · \cfrac{\sin^2 {ax}}{\cos^2 {ax}} · \cfrac{1}{\cos {ax} - 1}} \right] = \underset{x \to 0}{\lim} \left[\cfrac{-1}{2} · \cfrac{\sin^2 {ax}}{\cos^2 {ax}} · \cfrac{1}{-2\sin^2 {\frac{ax}{2}}}\right] · \underset{x \to 0}{\lim} \left[ \ln \left(1 + \cfrac{\sin^2 {ax}}{\cos^2 {ax}} \right)^{\cfrac{\cos^2 {ax}}{\sin^2 {ax}}} \right] = 1 \)

又\(\underset{x \to 0}{\lim} \cfrac{\cos {ax} - 1}{\cos {bx} - 1} = \underset{x \to 0}{\lim} \cfrac{2\sin^2 \cfrac{ax}{2}}{2 \sin^2 \cfrac{bx}{2}} = \cfrac{a^2}{b^2} \)， 所以\(\underset{x \to 0}{\lim} \cfrac{\ln \cos {ax}}{\ln \cos {bx}} = \cfrac{a^2}{b^2} \)。

【540】\(\underset{x \to 0}{\lim} \left(\ln \cfrac{nx + \sqrt{1 - n^2x^2}}{x + \sqrt{1 - x^2}} \right)\)

解：\(\underset{x \to 0}{\lim} \left(\ln \cfrac{nx + \sqrt{1 - n^2x^2}}{x + \sqrt{1 - x^2}} \right) = \ln 1 = 0\)

【541】\(\underset{x \to 0}{\lim} \cfrac{a^x - 1}{x}, a > 0\)

解：令\(y = a^x - 1\)，有\(x = \log_a{y + 1}\)，则\(\underset{x \to 0}{\lim} \cfrac{a^x - 1}{x} = \underset{y \to 0}{\lim} \cfrac{y\ln a}{\ln (1 + y)} = \underset{y \to 0}{\lim} \cfrac{\ln a}{\ln \left[(1 + y)^{\frac{1}{y}}\right]} = \ln a\)

【542】\(\underset{x \to a}{\lim} \cfrac{a^x - x^a}{x - a}, a > 0\)

解：因为\(\cfrac{a^x - x^a}{x - a} = a^a\cfrac{a^{x - a} - \left(\frac{x}{a}\right)^a}{x - a} = a^a\cfrac{a^{x - a} - 1}{x - a} - a^a\cfrac{\left(\frac{x}{a}\right)^a - 1}{x - a} = a^a\cfrac{a^{x - a} - 1}{x - a} - a^a\cfrac{e^{a\ln {\frac{x}{a}}} - 1}{a\ln {\frac{x}{a}}}\cfrac{a\ln {\frac{x}{a}}}{x - a} = a^a\cfrac{a^{x - a} - 1}{x - a} - a^a\cfrac{e^{a\ln {\frac{x}{a}}} - 1}{a\ln {\frac{x}{a}}}\ln \left[\left(1 + \cfrac{x - a}{a}\right)^{\frac{a}{x - a}} \right]\)

而\(\underset{x \to a}{\lim} \cfrac{a^{x - a} - 1}{x - a} = \ln a\)，并且\(\underset{x \to a}{\lim} \cfrac{e^{a\ln {\frac{x}{a}}} - 1}{a\ln {\frac{x}{a}}} = \ln e = 1\) 且\(\underset{x \to a}{\lim} \ln \left[\left(1 + \cfrac{x - a}{a}\right)^{\frac{a}{x - a}} \right] = \ln e = 1\)

所以\(\underset{x \to a}{\lim} \cfrac{a^x - x^a}{x - a} = a^a (\ln a - 1)\)

【543】\(\underset{x \to a}{\lim} \cfrac{x^x - a^a}{x - a}, a > 0\)

解：因为\(\cfrac{x^x - a^a}{x - a} = a^a\cfrac{\left(\frac{x}{a}\right)^a · x^{x - a} - 1}{x - a} = a^a\cfrac{e^{x\ln x - a\ln a} - 1}{x - a} = a^a\cfrac{e^{x\ln x - a\ln a} - 1}{x\ln x - a\ln a}\cfrac{x\ln x - a\ln a}{x - a} \)

又因为\(\underset{x \to a}{\lim} \cfrac{x^x - a^a}{x - a} = \cfrac{e^{x\ln x - a\ln a} - 1}{x\ln x - a\ln a} = 1\)， 并且\(\underset{x \to a}{\lim} \cfrac{x\ln x - a\ln a}{x - a} = \underset{x \to a}{\lim} \left(\cfrac{x\ln x - x\ln a}{x - a} + \ln a\right) = \underset{x \to a}{\lim} \left(\cfrac{x}{a}\ln \left[\left(1 + \cfrac{x - a}{a} \right)^{\frac{a}{x - a}} \right] + \ln a\right) = 1 + \ln a\)

所以\(\underset{x \to a}{\lim} \cfrac{x^x - a^a}{x - a} = a^a(\ln a + 1)\)

【544】\(\underset{x \to 0}{\lim} \left(x + e^x\right)^{\frac{1}{x}}\)

解：\(\underset{x \to 0}{\lim} \left(x + e^x\right)^{\frac{1}{x}} = \underset{x \to 0}{\lim} e\left(1 + \cfrac{x}{e^x}\right)^{\frac{e^x}{x}e^x} = e · e = e^2\)

【545】\(\underset{x \to 0}{\lim} \left(\cfrac{1 + x · 2^x}{1 + x · 3^x}\right)^{\frac{1}{x^2}}\)

解：\(\underset{x \to 0}{\lim} \left(\cfrac{1 + x · 2^x}{1 + x · 3^x}\right)^{\frac{1}{x^2}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{x · 2^x - x · 3^x}{1 + x · 3^x}\right)^{\cfrac{1 + x · 3^x}{x · 2^x - x · 3^x}\cfrac{x · 2^x - x · 3^x}{x^2(1 + x · 3^x)}} = e^{\ln 2 - \ln 3} = \ln \cfrac{2}{3}\)

【546】\(\underset{n \to \infty}{\lim} \tan^n \left(\cfrac{\pi}{4} + \cfrac{1}{n}\right)\)

解：\(\underset{n \to \infty}{\lim} \tan^n \left(\cfrac{\pi}{4} + \cfrac{1}{n}\right) = \underset{n \to \infty}{\lim} \left(1 + \cfrac{2\tan {\frac{1}{n}}}{1 - \tan {\frac{1}{n}}} \right)^n = \underset{n \to \infty}{\lim} \left(1 + \cfrac{2\tan {\frac{1}{n}}}{1 - \tan {\frac{1}{n}}} \right)^{\cfrac{1 - \tan {\frac{1}{n}}}{2\tan {\frac{1}{n}}} \cfrac{2n\tan {\frac{1}{n}}}{1 - \tan {\frac{1}{n}}}} = \underset{n \to \infty}{\lim} \left(1 + \cfrac{2\tan {\frac{1}{n}}}{1 - \tan {\frac{1}{n}}} \right)^{\cfrac{1 - \tan {\frac{1}{n}}}{2\tan {\frac{1}{n}}} \cfrac{2\sin {\frac{1}{n}}}{(1 - \tan {\frac{1}{n}}) \frac{1}{n} \cos {\frac{1}{n}}}} = e \)

【547】\(\underset{x \to 0}{\lim} \cfrac{e^{\alpha x} - e^{\beta x}}{\sin {\alpha x} - \sin {\beta x}}\)

解：\(\underset{x \to 0}{\lim} \cfrac{e^{\alpha x} - e^{\beta x}}{\sin {\alpha x} - \sin {\beta x}} = \underset{x \to 0}{\lim} \cfrac{e^{\beta x} \left(e^{(\alpha - \beta) x} - 1\right)}{2\sin {\cfrac{\alpha x - \beta x}{2}}\cos {\cfrac{\alpha x + \beta x}{2}}} = \underset{x \to 0}{\lim} \cfrac{\left(e^{(\alpha - \beta) x} - 1\right)}{\alpha x - \beta x} \bullet \underset{x \to 0}{\lim} \cfrac{\alpha x - \beta x}{2\sin {\cfrac{\alpha x - \beta x}{2}}} \bullet \underset{x \to 0}{\lim} \cfrac{e^{\beta x}}{\cos {\cfrac{\alpha x + \beta x}{2}}} = \ln e \bullet 1 \bullet 1 = 1\)

【548】\(\underset{x \to 0}{\lim} \cfrac{x^{\alpha} - a^{\alpha}}{x^{\beta} - a^{\beta}}, a > 0\)

解：\(\underset{x \to 0}{\lim} \cfrac{x^{\alpha} - a^{\alpha}}{x^{\beta} - a^{\beta}} = \underset{x \to 0}{\lim} \cfrac{a^{\alpha}\left(\left(\cfrac{x}{a}\right)^{\alpha} - 1\right)}{a^{\beta}\left(\left(\cfrac{x}{a}\right)^{\beta} - 1\right)} = a^{\alpha - \beta} \bullet \underset{x \to 0}{\lim} \cfrac{e^{\alpha \ln \left(\frac{x}{a}\right)} - 1}{\alpha \ln \left(\frac{x}{a}\right)} \cfrac{\beta \ln \left(\frac{x}{a}\right)}{e^{\beta \ln \left(\frac{x}{a}\right)} - 1} \cfrac{\alpha}{\beta} = a^{\alpha - \beta}\cfrac{\alpha}{\beta} \)

【549】\(\underset{x \to 0}{\lim} \cfrac{a^x - a^b}{x - b}, a > 0\)

解：\(\underset{x \to 0}{\lim} \cfrac{a^x - a^b}{x - b} = a^b \bullet \underset{x \to 0}{\lim} \cfrac{a^{x - b} - 1}{x - b} = a^b \bullet \underset{x \to 0}{\lim} \cfrac{e^{(x - b)\ln a} - 1}{x - b} = a^b \ln a \)

【550】\(\underset{h \to 0}{\lim} \cfrac{a^{x + h} + a^{x - h} - 2a^x}{h^2}, a > 0\)

解：\(\underset{h \to 0}{\lim} \cfrac{a^{x + h} + a^{x - h} - 2a^x}{h^2} = a^x \bullet \underset{h \to 0}{\lim} \cfrac{a^{2h} - 2a^h + 1}{a^h h^2} = a^x \bullet \underset{h \to 0}{\lim} \cfrac{(a^h - 1)^2}{a^h h^2} = a^x \bullet \underset{h \to 0}{\lim} \cfrac{(e^{h \ln a} - 1)^2}{a^h h^2} = a^x \bullet \underset{h \to 0}{\lim} \cfrac{(e^{h \ln a} - 1)^2\ln^2 a}{a^h h^2 \ln^2 a} = a^x \ln^2 a\)

【551】\(\underset{x \to \infty}{\lim} \cfrac{(x + a)^{x + a}(x + b)^{x + b}}{(x + a + b)^{2x + a + b}}\)

解：\(\underset{x \to \infty}{\lim} \cfrac{(x + a)^{x + a}(x + b)^{x + b}}{(x + a + b)^{2x + a + b}} = \underset{x \to \infty}{\lim} \cfrac{e^{(x + a)\ln (x + a) + (x + b)\ln (x + b)}}{e^{(2x + a + b)\ln(x + a + b)}} = \underset{x \to \infty}{\lim} e^{(x + a)\ln \frac{x + a}{x + a + b} + (x + b)\ln \frac{x + b}{x + a + b}} = \underset{x \to \infty}{\lim} e^{-\ln \left[\left(1 + \frac{b}{x + a}\right)^{x+a}\right] - \ln \left[\left(1 + \frac{a}{x + b}\right)^{x + b} \right]} = e^{-b - a} \)

【552】\(\underset{n \to \infty}{\lim} n(\sqrt[n]{x} - 1), x > 0\)

解：\(\underset{n \to \infty}{\lim} n(\sqrt[n]{x} - 1) = \underset{n \to \infty}{\lim} n(e^{\frac{1}{n}\ln x} - 1) = \underset{n \to \infty}{\lim} \cfrac{\left(e^{\frac{1}{n}\ln x} - 1\right)\ln x}{\frac{1}{n}\ln x} = \ln x\)

【553】\(\underset{n \to \infty}{\lim} n^2(\sqrt[n]{x} - \sqrt[n+1]{x}), x > 0\)

解：\(\underset{n \to \infty}{\lim} n^2(\sqrt[n]{x} - \sqrt[n+1]{x}) = \underset{n \to \infty}{\lim} n^2\left[\left(e^{\frac{1}{n}\ln x} - 1\right) - \left(e^{\frac{1}{n + 1}\ln x} - 1\right)\right] = \underset{n \to \infty}{\lim} n^2\left[\cfrac{\frac{1}{n}\ln x\left(e^{\frac{1}{n}\ln x} - 1\right)}{\frac{1}{n}\ln x} - \cfrac{\frac{1}{n + 1}\ln x\left(e^{\frac{1}{n + 1}\ln x} - 1\right)}{\frac{1}{n + 1}\ln x} \right]\)

因为\(\underset{n \to \infty}{\lim} \cfrac{e^{\frac{1}{n}\ln x} - 1}{\frac{1}{n}\ln x} = 1\)， 并且\(\underset{n \to \infty}{\lim} \cfrac{e^{\frac{1}{n+1}\ln x} - 1}{\frac{1}{n+1}\ln x} = 1\)。

所以\(\underset{n \to \infty}{\lim} n^2(\sqrt[n]{x} - \sqrt[n+1]{x}) = \underset{n \to \infty}{\lim} n^2\left(\cfrac{\ln x}{n} - \cfrac{\ln x}{n + 1}\right) = \underset{n \to \infty}{\lim} \cfrac{n^2 \ln x}{n^2 + n} = \ln x\)

【554】\(\underset{n \to \infty}{\lim} \left(\cfrac{a - 1 + \sqrt[n]{b}}{a}\right)^n, a > 0, b > 0\)

解：\(\underset{n \to \infty}{\lim} \left(\cfrac{a - 1 + \sqrt[n]{b}}{a}\right)^n = \underset{n \to \infty}{\lim} \left(1 + \cfrac{\sqrt[n]{b} - 1}{a}\right)^{\cfrac{a}{\sqrt[n]{b} - 1} · \cfrac{n(\sqrt[n]{b} - 1)}{a}}\)。 因为\(\underset{n \to \infty}{\lim} \cfrac{n(\sqrt[n]{b} - 1)}{a} = \underset{n \to \infty}{\lim} \cfrac{n(e^{\frac{1}{n}\ln b} - 1)}{a} = \cfrac{\ln b}{a}\)

所以\(\underset{n \to \infty}{\lim} \left(\cfrac{a - 1 + \sqrt[n]{b}}{a}\right)^n = e^{\cfrac{\ln b}{a}} = \sqrt[a]{b}\)。

【555】\(\underset{n \to \infty}{\lim} \left(\cfrac{\sqrt[n]{a} + \sqrt[n]{b}}{2}\right)^n, a > 0, b > 0\)

解：\(\underset{n \to \infty}{\lim} \left(\cfrac{\sqrt[n]{a} + \sqrt[n]{b}}{2}\right)^n = \underset{n \to \infty}{\lim} \left(1 + \cfrac{e^{\frac{1}{n}\ln a} - 1 + e^{\frac{1}{n}\ln b} - 1}{2}\right)^n = \underset{n \to \infty}{\lim} \left(1 + \cfrac{e^{\frac{1}{n}\ln a} - 1 + e^{\frac{1}{n}\ln b} - 1}{2}\right)^ {\frac{2}{e^{\frac{1}{n}\ln a} - 1 + e^{\frac{1}{n}\ln b} - 1}\frac{e^{\frac{1}{n}\ln a} - 1 + e^{\frac{1}{n}\ln b} - 1}{\frac{2}{n}}} = e^{\frac{\ln a + \ln b}{2}} = \sqrt{ab}\)

【556】\(\underset{x \to 0}{\lim} \left(\cfrac{a^x + b^x + c^x}{3}\right)^{\frac{1}{x}}, a > 0, b > 0, c > 0\)

解：\(\underset{x \to 0}{\lim} \left(\cfrac{a^x + b^x + c^x}{3}\right)^{\frac{1}{x}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{a^x - 1 + b^x - 1 + c^x - 1}{3}\right)^{\frac{3}{a^x - 1 + b^x - 1 + c^x - 1} \frac{a^x - 1 + b^x - 1 + c^x - 1}{3x}} = e^{\frac{\ln a + \ln b + \ln c}{3}} = \sqrt[3]{abc}\)

【557】\(\underset{x \to 0}{\lim} \left(\cfrac{a^{x+1} + b^{x+1} + c^{x+1}}{a + b + c}\right)^{\frac{1}{x}}, a > 0, b > 0, c > 0\)

解：\(\underset{x \to 0}{\lim} \left(\cfrac{a^{x+1} + b^{x+1} + c^{x+1}}{a + b + c}\right)^{\frac{1}{x}} = \underset{x \to 0}{\lim} \left(1 + \cfrac{a^{x+1} - a + b^{x+1} - b + c^{x+1} - c}{a + b + c}\right)^ {\frac{a + b + c}{a^{x+1} - a + b^{x+1} - b + c^{x+1} - c}\frac{a (a^x - 1) + b(b^x - 1) + c(c^x - 1)}{x(a + b + c)}} = \left(a^a b^b c^c\right)^{\frac{1}{a + b + c}}\)

【558】\(\underset{x \to 0}{\lim} \left(\cfrac{a^{x^2} + b^{x^2}}{a^x + b^x}\right)^{\frac{1}{x}}, a > 0, b > 0\)

解：

$$ \begin{array}{rcl} \underset{x \to 0}{\lim} \left(\cfrac{a^{x^2} + b^{x^2}}{a^x + b^x}\right)^{\frac{1}{x}} & = & \underset{x \to 0}{\lim} \left(1 + \cfrac{a^{x^2} - a^x + b^{x^2} - b^x}{a^x + b^x}\right)^ {\frac{a^x + b^x}{a^{x^2} - a^x + b^{x^2} - b^x}\frac{a^{x^2} - a^x + b^{x^2} - b^x}{x(a^x + b^x)}} \\ & = & \underset{x \to 0}{\lim} \left(1 + \cfrac{a^{x^2} - a^x + b^{x^2} - b^x}{a^x + b^x}\right)^ {\frac{a^x + b^x}{a^{x^2} - a^x + b^{x^2} - b^x} \left[ \frac{x^2(a^{x^2} - 1)}{x^2} - \frac{x(a^x - 1)}{x} + \frac{x^2(b^{x^2} - 1)}{x^2} - \frac{x(b^x - 1)}{x} \right] \frac{1}{x(a^x + b^x)}} \\ & = & e^{\frac{-\ln a - \ln b}{2}} = (ab)^{-\frac{1}{2}} \end{array} $$

【559】\(\underset{x \to 0}{\lim} \cfrac{a^{x^2} - b^{x^2}}{\left(a^x - b^x\right)^2}, a > 0, b > 0\)

解：\(\underset{x \to 0}{\lim} \cfrac{a^{x^2} - b^{x^2}}{\left(a^x - b^x\right)^2} = \underset{x \to 0}{\lim} \left(\cfrac{a^{x^2} - 1}{x^2} - \cfrac{b^{x^2} - 1}{x^2}\right)\cfrac{x^2}{\left(a^x - b^x\right)^2} = \underset{x \to 0}{\lim} \left(\cfrac{a^{x^2} - 1}{x^2} - \cfrac{b^{x^2} - 1}{x^2}\right) \left(\cfrac{a^x - 1}{x} - \cfrac{b^x - 1}{x}\right)^{-2} = \left(\ln a - \ln b\right)^{-1} \)

【560】\(\underset{x \to a}{\lim} \cfrac{a^{a^x} - a^{x^a}}{a^x - x^a}, a > 0\)

解：\(\underset{x \to a}{\lim} \cfrac{a^{a^x} - a^{x^a}}{a^x - x^a} = \underset{x \to a}{\lim} a^{x^a}\cfrac{a^{a^x - x^a} - 1}{a^x - x^a} = a^{a^a}\ln a\)

【561】(1)\(\underset{x \to -\infty}{\lim} \cfrac{\ln \left(1 + 3^x\right)}{\ln \left(1 + 2^x\right)}\)， (2)\(\underset{x \to +\infty}{\lim} \cfrac{\ln \left(1 + 3^x\right)}{\ln \left(1 + 2^x\right)}\)。

解：(1) \(\underset{x \to -\infty}{\lim} \cfrac{\ln \left(1 + 3^x\right)}{\ln \left(1 + 2^x\right)} = \underset{x \to -\infty}{\lim} \cfrac{\ln \left(1 + 3^x\right)}{3^x} \cfrac{2^x}{\ln \left(1 + 2^x\right)} \cfrac{3^x}{2^x} = 0\)

(2) \(\underset{x \to +\infty}{\lim} \cfrac{\ln \left(1 + 3^x\right)}{\ln \left(1 + 2^x\right)} = \underset{x \to +\infty}{\lim} \cfrac{x\ln 3 + \ln \left(1 + 3^{-x}\right)}{x\ln2 + \ln \left(1 + 2^{-x}\right)} = \underset{x \to +\infty}{\lim} \cfrac{\ln 3 + \cfrac{\ln \left(1 + 3^{-x}\right)}{3^{-x}}\cfrac{3^{-x}}{x}} {\ln 2 + \cfrac{\ln \left(1 + 2^{-x}\right)}{2^{-x}}\cfrac{2^{-x}}{x}} = \cfrac{\ln 3}{\ln 2} \)

【562】\(\underset{x \to +\infty}{\lim} \ln \left(1 + 2^x\right)\ln \left(1 + \cfrac{3}{x}\right)\)

解：\(\underset{x \to +\infty}{\lim} \ln \left(1 + 2^x\right)\ln \left(1 + \cfrac{3}{x}\right) = \underset{x \to +\infty}{\lim} \left[\ln 2^x + \ln \left(1 + 2^{-x}\right)\right]\cfrac{3}{x}\ln \left[\left(1 + \cfrac{3}{x}\right)^{\frac{x}{3}}\right] = 3\ln 2\)

【563】\(\underset{x \to 1}{\lim} \left(1 - x\right)\lg_x 2\)

解：\(\underset{x \to 1}{\lim} \left(1 - x\right)\lg_x 2 = \underset{x \to 1}{\lim} \left(1 - x\right)\cfrac{\ln 2}{\ln x} = \underset{x \to 1}{\lim} \ln 2 · \left(\cfrac{\ln x}{1 - x} \right)^{-1} = \underset{x \to 1}{\lim} \ln 2 · \left(\ln \left(1 + (x - 1)\right)^{\frac{1}{1 - x}} \right)^{-1} = -\ln 2\)

【566】(1)\(\underset{x \to 0}{\lim} \cfrac{\ln \left(x^2 + e^x\right)}{\ln \left(x^4 + e^{2x}\right)}\) (2)\(\underset{x \to +\infty}{\lim} \cfrac{\ln \left(x^2 + e^x\right)}{\ln \left(x^4 + e^{2x}\right)}\)

解：(1)\(\underset{x \to 0}{\lim} \cfrac{\ln \left(x^2 + e^x\right)}{\ln \left(x^4 + e^{2x}\right)} = \underset{x \to 0}{\lim} \cfrac{x + \ln \left(1 + e^{-x} x^2\right)}{2x + \ln \left(1 + e^{-2x}x^4\right)} = \underset{x \to 0}{\lim} \cfrac{x + \frac{\ln \left(1 + e^{-x} x^2\right)}{e^{-x}x^2} e^{-x}x^2}{2x + \frac{\ln \left(1 + e^{-2x}x^4\right)}{e^{-2x}x^4}e^{-2x}x^4} = \cfrac{1}{2}\)

(2)\(\underset{x \to +\infty}{\lim} \cfrac{\ln \left(x^2 + e^x\right)}{\ln \left(x^4 + e^{2x}\right)} = \underset{x \to +\infty}{\lim} \cfrac{x + \ln \left(1 + e^{-x} x^2\right)}{2x + \ln \left(1 + e^{-2x}x^4\right)} = \cfrac{1}{2}\)

【567】\(\underset{x \to 0}{\lim} \cfrac{\ln \left(1 + xe^x\right)}{\ln \left(x + \sqrt{1 + x^2}\right)}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\ln \left(1 + xe^x\right)}{\ln \left(x + \sqrt{1 + x^2}\right)} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\ln \left(1 + xe^x\right)}{xe^x}xe^x}{\ln \sqrt{1 + x^2} + \ln \left(\cfrac{x}{\sqrt{1 + x^2}} + 1\right)} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\ln \left(1 + xe^x\right)}{xe^x}xe^x}{\ln \sqrt{1 + x^2} + \cfrac{\ln \left(\cfrac{x}{\sqrt{1 + x^2}} + 1\right)} {\cfrac{x}{\sqrt{1 + x^2}}}\cfrac{x}{\sqrt{1 + x^2}}} = \underset{x \to 0}{\lim} \cfrac{xe^x}{\cfrac{x}{\sqrt{1 + x^2}}} = 1\)

【568】\(\underset{x \to \infty}{\lim} \left[(x + 2)\ln (x + 2) - 2(x + 1)\ln(x + 1) + x\ln x \right]\)

解：\(\underset{x \to \infty}{\lim} \left[(x + 2)\ln (x + 2) - 2(x + 1)\ln(x + 1) + x\ln x \right] = \underset{x \to \infty}{\lim} \left[\ln \left(1 + \cfrac{1}{x+1}\right)^{x + 2} - \ln \left(1 + \cfrac{1}{x}\right)^{x} \right] = 1 - 1 = 0\)

【569】\(\underset{x \to +0}{\lim} \left[\ln \left(x \ln a\right) · \ln \left(\cfrac{\ln {ax}}{\ln \frac{x}{a}} \right) \right]\)

解：\(\underset{x \to +0}{\lim} \left[\ln \left(x \ln a\right) · \ln \left(\cfrac{\ln {ax}}{\ln \frac{x}{a}} \right) \right] = \underset{x \to +0}{\lim} \left[\ln \left(x \ln a\right) · \ln \left(1 + \cfrac{2\ln a}{\ln x - \ln a} \right) \right] = \underset{x \to +0}{\lim} \ln \left[\left(1 + \cfrac{2\ln a}{\ln x - \ln a} \right)^{\frac{\ln x - \ln a}{2\ln a}\frac{2\ln a}{\ln x - \ln a}\ln \left(x \ln a\right)} \right] = \ln e^{2\ln a} = 2 \ln a\)

【570】\(\underset{x \to +\infty}{\lim} \left(\ln {\cfrac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 - 1}}} · \ln^{-2} {\cfrac{x + 1}{x - 1}} \right)\)

解： $$ \begin{array}{rcl} \underset{x \to +\infty}{\lim} \left(\ln {\cfrac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 - 1}}} · \ln^{-2} {\cfrac{x + 1}{x - 1}} \right) & = & \underset{x \to +\infty}{\lim} \left[\ln \left(1 + \cfrac{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}}\right) · \ln^{-2} \left(1 + \cfrac{2}{x - 1}\right) \right] \\ & = & \underset{x \to +\infty}{\lim} \left[\ln \left(1 + \cfrac{2}{(x + \sqrt{x^2 - 1})(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})}\right) · \ln^{-2} \left(1 + \cfrac{2}{x - 1}\right) \right] \\ & = & \underset{x \to +\infty}{\lim} \left[\cfrac{2}{(x + \sqrt{x^2 - 1})(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})} · \left(\cfrac{2}{x - 1} \right)^{-2} \right] \\ & = & \cfrac{1}{8} \end{array} $$

【571】\(\underset{x \to 0}{\lim} \cfrac{\sqrt{1 + x\sin x} - 1}{e^{x^2} - 1}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\sqrt{1 + x\sin x} - 1}{e^{x^2} - 1} = \underset{x \to 0}{\lim} \cfrac{x\sin x}{(e^{x^2} - 1)(\sqrt{1 + x\sin x} + 1)} = \underset{x \to 0}{\lim} \cfrac{\cfrac{\sin x}{x}}{\cfrac{(e^{x^2} - 1)}{x^2}(\sqrt{1 + x\sin x} + 1)} = \cfrac{1}{2}\)

【572】\(\underset{x \to 0}{\lim} \cfrac{\cos \left(xe^x\right) - \cos \left(xe^{-x}\right)}{x^3}\)

解： $$ \begin{array}{rcl} \underset{x \to 0}{\lim} \cfrac{\cos \left(xe^x\right) - \cos \left(xe^{-x}\right)}{x^3} & = & \underset{x \to 0}{\lim} \cfrac{-2\sin \left(\cfrac{xe^x + xe^{-x}}{2}\right)\sin \left(\cfrac{xe^x - xe^{-x}}{2}\right)}{x^3} \\ & = & \underset{x \to 0}{\lim} \cfrac{\sin \left(\cfrac{xe^x + xe^{-x}}{2}\right)}{\cfrac{xe^x + xe^{-x}}{2}} · \cfrac{\sin \left(\cfrac{xe^x - xe^{-x}}{2}\right)}{\cfrac{xe^x - xe^{-x}}{2}} · \cfrac{-2(x^2e^{2x} - x^2 e^{-2x})}{4x^3} \\ & = & -2 \end{array} $$

【573】\(\underset{x \to 0}{\lim} \left(2e^{\frac{x}{x + 1}} - 1\right)^{\frac{x^2 + 1}{x}}\)

解：

$$ \begin{array}{rcl} \underset{x \to 0}{\lim} \left(2e^{\frac{x}{x + 1}} - 1\right)^{\frac{x^2 + 1}{x}} & = & \underset{x \to 0}{\lim} \left(1 + 2\left(e^{\frac{x}{x + 1}} - 1\right)\right)^{\frac{1}{2\left(e^{\frac{x}{x + 1}} - 1\right)}\frac{(x^2 + 1)2\left(e^{\frac{x}{x + 1}} - 1\right)}{x}} \\ & = & \underset{x \to 0}{\lim} e^{2\frac{x^2 + 1}{x+1}\frac{\left(e^{\frac{x}{x + 1}} - 1\right)}{\frac{x}{x + 1}}} \\ & = & e^2 \end{array} $$

【574】\(\underset{x \to 1}{\lim} \left(2 - x\right)^{\sec{\frac{\pi x}{2}}}\)

解：\(\underset{x \to 1}{\lim} \left(2 - x\right)^{\sec{\frac{\pi x}{2}}} = \underset{x \to 1}{\lim} \left(1 + (1 - x)\right)^{\frac{1}{1 - x}\frac{1 - x}{\cos {\frac{\pi x}{2}}}} = \underset{x \to 1}{\lim} e^{\frac{1 - x}{\sin \left(\frac{\pi(1 - x)}{2}\right)}} = e^{\frac{2}{\pi}}\)

【575】\(\underset{x \to \frac{\pi}{2}}{\lim} \cfrac{1 - \sin^{\alpha + \beta} x}{\sqrt{(1 - \sin^{\alpha} x)(1 - \sin^{\beta} x)}}, \alpha > 0, \beta > 0\)

解：\(\underset{x \to \frac{\pi}{2}}{\lim} \cfrac{1 - \sin^{\alpha + \beta} x}{\sqrt{(1 - \sin^{\alpha} x)(1 - \sin^{\beta} x)}} = \underset{x \to \frac{\pi}{2}}{\lim} \cfrac{1 - (\alpha + \beta)e^{\ln {\sin x}}}{\sqrt{(1 - \alpha e^{\ln {\sin x}})(1 - \beta e^{\ln {\sin x}})}} = \underset{x \to \frac{\pi}{2}}{\lim} \left[\cfrac{1 - (\alpha + \beta)e^{\ln {\sin x}}}{(\alpha + \beta)e^{\ln {\sin x}}}\right] \left[\cfrac{\alpha e^{\ln {\sin x}}}{1 - \alpha e^{\ln {\sin x}}} \right]^{\frac{1}{2}} \left[\cfrac{\beta e^{\ln {\sin x}}}{1 - \beta e^{\ln {\sin x}}} \right]^{\frac{1}{2}} \left[\cfrac{(\alpha + \beta)e^{\ln {\sin x}}}{\sqrt{\alpha \beta} e^{\ln {\sin x}}} \right] = \cfrac{\alpha + \beta}{\sqrt{\alpha \beta}} \)

【576】\(\underset{x \to 0}{\lim} \cfrac{\sinh^2 x}{\ln (\cosh {3x})}\)

解：

$$ \begin{array}{rcl} \underset{x \to 0}{\lim} \cfrac{\sinh^2 x}{\ln (\cosh {3x})} & = & \underset{x \to 0}{\lim} \cfrac{\left(e^x - e^{-x}\right)^2}{4\ln \left(\cfrac{1}{2}(e^{3x} + e^{-3x})\right)} \\ & = & \underset{x \to 0}{\lim} \cfrac{\left(e^{2x} - 1\right)^2}{4e^{2x} \ln \left(1 + \cfrac{1}{2}\left(e^{\frac{3}{2}x} - e^{-\frac{3}{2}x}\right)^2\right)} \\ & = & \underset{x \to 0}{\lim} \left[\cfrac{\left(e^{2x} - 1\right)^2}{2x}\right]^2 \left[\cfrac{\cfrac{1}{2}\left(e^{\frac{3}{2}x} - e^{-\frac{3}{2}x}\right)^2}{\ln \left(1 + \cfrac{1}{2}\left(e^{\frac{3}{2}x} - e^{-\frac{3}{2}x}\right)^2\right)} \right] \left[\cfrac{x^2e^{3x}}{\cfrac{1}{2}\left(e^{3x} - 1\right)^2 e^{2x}} \right] \\ & = & \cfrac{2}{9} \end{array} $$

【577】\(\underset{x \to +\infty}{\lim} \cfrac{\sinh \sqrt{x^2 + x} - \sinh \sqrt{x^2 - x}}{\cosh x}\)

解：因为\(\sinh \sqrt{x^2 + x} - \sinh \sqrt{x^2 - x} = 2\cosh \cfrac{\sqrt{x^2 + x} + \sqrt{x^2 - x}}{2}\sinh \cfrac{\sqrt{x^2 + x} - \sqrt{x^2 - x}}{2}\)

并且\(\underset{x \to +\infty}{\lim} \sinh \cfrac{\sqrt{x^2 + x} - \sqrt{x^2 - x}}{2} = \underset{x \to +\infty}{\lim} \sinh \cfrac{x}{\sqrt{x^2 + x} + \sqrt{x^2 - x}} = \underset{x \to +\infty}{\lim} \sinh \cfrac{1}{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}} = \sinh \cfrac{1}{2}\)

而\(\underset{x \to +\infty}{\lim} \cfrac{2\cosh \cfrac{\sqrt{x^2 + x} + \sqrt{x^2 - x}}{2}}{\cosh x} = 2\underset{x \to +\infty}{\lim} \cfrac{e^{\frac{\sqrt{x^2 + x} + \sqrt{x^2 - x}}{2}} + e^{\frac{-\sqrt{x^2 + x} - \sqrt{x^2 - x}}{2}}}{e^x + e^{-x}}\)

搞不懂了。

【578】\(\underset{x \to +\infty}{\lim} \left(x - \ln \cosh x\right)\)

解：\(\underset{x \to +\infty}{\lim} \left(x - \ln \cosh x\right) = \underset{x \to +\infty}{\lim} \left(\ln e^x - \ln \cfrac{e^x + e^{-x}}{2}\right) = \underset{x \to +\infty}{\lim} \ln \cfrac{2e^x}{e^x + e^{-x}} = \underset{x \to +\infty}{\lim} \ln \cfrac{2}{1 + e^{-2x}} = \ln 2\)

【579】\(\underset{x \to 0}{\lim} \cfrac{e^{\sin {2x}} - e^{\sin x}}{\tanh x}\)

解：\(\underset{x \to 0}{\lim} \cfrac{e^{\sin {2x}} - e^{\sin x}}{\tanh x} = \underset{x \to 0}{\lim} \cfrac{e^{\sin {2x} - \sin x} - 1}{\sin {2x} - \sin x} \cfrac{2x}{e^{2x} - 1}\cfrac{\sin {2x} - \sin x}{2x}(e^{2x} + 1)e^{\sin x} = 1 · 1 · (1 - \cfrac{1}{2}) · 2 · 1 = 1\)

【580】\(\underset{n \to \infty}{\lim} \left(\cfrac{\cosh {\cfrac{\pi}{n}}}{\cos {\cfrac{\pi}{n}}}\right)^{n^2}\)

解：因为\(\underset{n \to \infty}{\lim} \left(\cfrac{\cosh {\cfrac{\pi}{n}}}{\cos {\cfrac{\pi}{n}}}\right)^{n^2} = \underset{n \to \infty}{\lim} \left(1 + \cfrac{\cosh {\cfrac{\pi}{n}} - \cos {\cfrac{\pi}{n}}}{\cos {\cfrac{\pi}{n}}}\right)^{n^2} = e^{\underset{n \to \infty}{\lim} \cfrac{n^2(\cosh {\cfrac{\pi}{n}} - \cos {\cfrac{\pi}{n}})}{\cos {\cfrac{\pi}{n}}}}\)

又，\(\underset{n \to \infty}{\lim} \cfrac{n^2(\cosh {\cfrac{\pi}{n}} - \cos {\cfrac{\pi}{n}})}{\cos {\cfrac{\pi}{n}}} = \underset{n \to \infty}{\lim} \cfrac{(e^{\frac{\pi}{2n}} - e^{-\frac{\pi}{2n}})^2 + 2 - 2(1 - 2\sin^2 {\cfrac{\pi}{2n}})}{2\cfrac{1}{n^2}} = \underset{n \to \infty}{\lim} \left(\cfrac{e^{\frac{\pi}{2n}} + 1}{\cfrac{1}{n}} - \cfrac{e^{-\frac{\pi}{2n}} + 1}{-\cfrac{1}{n}}\right)^2\cfrac{1}{2} + 4\cfrac{\sin^2 {\cfrac{\pi}{2n}}}{2\cfrac{1}{n^2}} = \pi^2\)

所以\(\underset{n \to \infty}{\lim} \left(\cfrac{\cosh {\cfrac{\pi}{n}}}{\cos {\cfrac{\pi}{n}}}\right)^{n^2} = e^{\pi^2}\)

【581】\(\underset{x \to \infty}{\lim} \arcsin \cfrac{1 - x}{1 + x}\)

解：\(\underset{x \to \infty}{\lim} \arcsin \cfrac{1 - x}{1 + x} = \arcsin {-1} = -\cfrac{\pi}{2}\)

【582】\(\underset{x \to +\infty}{\lim} \arccos \left(\sqrt{x^2 + x} - x\right)\)

解：\(\underset{x \to +\infty}{\lim} \arccos \left(\sqrt{x^2 + x} - x\right) = \arccos \cfrac{1}{2} = \cfrac{\pi}{3}\)

【583】\(\underset{x \to 2}{\lim} \arctan \cfrac{x-4}{(x - 2)^2}\)

解：\(\underset{x \to 2}{\lim} \arctan \cfrac{x-4}{(x - 2)^2} = \arctan -\infty = -\cfrac{\pi}{2}\)

【584】\(\underset{x \to -\infty}{\lim} \text{arccot} \cfrac{x}{\sqrt{1 + x^2}}\)

解：\(\underset{x \to -\infty}{\lim} \text{arccot} \cfrac{x}{\sqrt{1 + x^2}} = \text{arccot} (-1) = \cfrac{3}{4}\pi\)

【585】\(\underset{h \to 0}{\lim} \cfrac{\arctan (x + h) - \arctan x}{h}\)

解：因为\(\arctan (x + h) - \arctan x = \arctan \left[\tan \left(\arctan (x + h) - \arctan x\right)\right] = \arctan \left(\cfrac{x+h -x}{1 + x(x + h)}\right)\)，

所以\(\underset{h \to 0}{\lim} \cfrac{\arctan (x + h) - \arctan x}{h} = \underset{h \to 0}{\lim} \cfrac{\arctan \left(\cfrac{h}{1 + x(x + h)}\right)}{h} = \cfrac{1}{1 + x^2}\)

【586】\(\underset{x \to 0}{\lim} \cfrac{\ln \cfrac{1 + x}{1 - x}}{\arctan (1 + x) - \arctan (1 - x)}\)

解：\(\underset{x \to 0}{\lim} \cfrac{\ln \cfrac{1 + x}{1 - x}}{\arctan (1 + x) - \arctan (1 - x)} = \underset{x \to 0}{\lim} \cfrac{\ln (1 + \cfrac{2x}{1-x})}{\arctan \cfrac{2x}{1 + (1+x)(1-x)}} = \underset{x \to 0}{\lim} \left[\cfrac{\ln (1 + \cfrac{2x}{1-x})}{\cfrac{2x}{1 - x}}\right] \left[\cfrac{\cfrac{2x}{2 - x^2}}{\arctan \cfrac{2x}{2 - x^2}} \right] \left[\cfrac{\cfrac{2x}{1 - x}}{\cfrac{2x}{2 - x^2}} \right] = 2\)

【587】\(\underset{n \to \infty}{\lim} n\arctan \cfrac{1}{n(x^2 + 1) + x} · \tan^n \left(\cfrac{\pi}{4} + \cfrac{x}{2n}\right)\)

解：因为\(\underset{n \to \infty}{\lim} n\arctan \cfrac{1}{n(x^2 + 1) + x} = \underset{n \to \infty}{\lim} \left[\cfrac{\arctan \cfrac{1}{n(x^2 + 1) + x}}{\cfrac{1}{n(x^2 + 1) + x}}\right]\left[\cfrac{n}{n(x^2 + 1) + x}\right] = \cfrac{1}{x^2 + 1}\)

又因为\(\underset{n \to \infty}{\lim} \tan^n \left(\cfrac{\pi}{4} + \cfrac{x}{2n}\right) = \underset{n \to \infty}{\lim} e^{n \ln \left(1 + \cfrac{2 \tan \frac{x}{2n}}{1 - \tan \frac{x}{2n}} \right)} = \underset{n \to \infty}{\lim} e^{\ln \left(1 + \cfrac{2 \tan \frac{x}{2n}}{1 - \tan \frac{x}{2n}} \right)^{\frac{1 - \tan \frac{x}{2n}}{2 \tan \frac{x}{2n}} \frac{2n\tan \frac{x}{2n}}{1 - \tan \frac{x}{2n}}}} = e^{\ln e^x} = e^x\)

所以\(\underset{n \to \infty}{\lim} n\arctan \cfrac{1}{n(x^2 + 1) + x} · \tan^n \left(\cfrac{\pi}{4} + \cfrac{x}{2n}\right) = \cfrac{e^x}{x^2 + 1}\)

【588】\(\underset{x \to \infty}{\lim} x\left(\cfrac{\pi}{4} - \arctan \cfrac{x}{x + 1}\right)\)

解：\(\underset{x \to \infty}{\lim} x\left(\cfrac{\pi}{4} - \arctan \cfrac{x}{x + 1}\right) = \underset{x \to \infty}{\lim} x\arctan \cfrac{1}{2x + 1} = \underset{x \to \infty}{\lim} \cfrac{\arctan \cfrac{1}{2x + 1}}{\cfrac{1}{2x + 1}}\cfrac{x}{2x + 1} = \cfrac{1}{2}\)

【589】\(\underset{x \to +\infty}{\lim} x\left(\cfrac{\pi}{2} - \arcsin \cfrac{x}{\sqrt{x^2 + 1}}\right)\)

解：因为\(\sin \left(\cfrac{\pi}{2} - \arcsin \cfrac{x}{\sqrt{x^2 + 1}}\right) = \cos \left(\arcsin \cfrac{x}{\sqrt{x^2 + 1}}\right) = \sqrt{1 - \cfrac{x^2}{x^2 + 1}} = \cfrac{1}{\sqrt{x^2 + 1}}\)

所以\(\underset{x \to +\infty}{\lim} x\left(\cfrac{\pi}{2} - \arcsin \cfrac{x}{\sqrt{x^2 + 1}}\right) = \underset{x \to +\infty}{\lim} x\arcsin \left(\cfrac{1}{\sqrt{x^2 + 1}}\right) = 1\)

【590】\(\underset{n \to \infty}{\lim} \left[1 + \cfrac{(-1)^n}{n}\right]^{\csc (\pi \sqrt{1 + n^2})}\)

不会，严重怀疑书上的解法有问题。