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4.2
1.
证明：
\((1) \lim_{x \to +\infty}\frac{x^k}{a^x} = 0(a > 1, k > 0)\)
\[\begin{aligned}
\lim_{x \to +\infty}\frac{x^k}{a^x} = \lim_{x \to +\infty} \left( \frac{x}{a^{\frac{x}{k}}} \right)^k
\end{aligned}
\]
令 \(b = a^\frac{1}{k} > 1, f(x) = \frac{x}{b^x},f'(x) = \frac{1 - x\ln b}{b^{x}}\),故 \(f(x)\) 在 \((\frac{1}{\ln b}, +\infty)\) 上单减，由单调函数单侧极限存在可知，\(\lim_{x \to +\infty}f(x)\) 存在，由 \(\mathrm{Heine}\) 归结原理有
\[\begin{aligned}
\lim_{x \to +\infty}\frac{x^k}{a^x} = \lim_{x \to +\infty} \left( \frac{x}{b^{x}} \right)^k = \lim_{n \to +\infty} \left( \frac{n}{b^n} \right)^k = 0
\end{aligned}
\]
\((2)\lim_{x \to +\infty} \frac{\ln x}{x^k} = 0\)
\[\begin{aligned}
\lim_{x \to +\infty} \frac{\ln x}{x^k} = \lim_{x \to +\infty} \frac{1}{k} \frac{\ln x^k}{x^k}
\end{aligned}
\]
令 \(t = \ln x^k, x^k = \exp(t)\).
\[\begin{aligned}
\lim_{x \to +\infty} \frac{\ln x}{x^k} = \lim_{x \to +\infty} \frac{1}{k} \frac{\ln x^k}{x^k} = \frac{1}{k} \lim_{x \to +\infty} \frac{t}{e^t} = 0
\end{aligned}
\]
\((3) \lim_{x \to \infty} \sqrt[x]{a} = 1, (a > 0)\)
\[\begin{aligned}
\lim_{x \to \infty} e^{\frac{\ln a}{x}} = e^{\lim_{x \to \infty} \frac{\ln a}{x}} = 1
\end{aligned}
\]
\((4) \lim_{x \to +\infty} \sqrt[x]{x} = 1\)
\[\begin{aligned}
\lim_{x \to \infty} e^{\frac{\ln x}{x}} = e^{\lim_{x \to \infty} \frac{\ln x}{x}} = 1
\end{aligned}
\]
2
求 \(\lim_{y \to +\infty}\frac{\sqrt{1 + y^3}}{\sqrt{y^2 + y^3} + y}\)
\[\begin{aligned}
\lim_{y \to +\infty}\frac{\sqrt{1 + y^3}}{\sqrt{y^2 + y^3} + y} &= \lim_{y \to +\infty}\frac{\sqrt{\frac{1}{y^3} + 1}}{\sqrt{\frac{1}{y} + 1} + \sqrt{\frac{1}{y}}}\\ &= \frac{\lim_{y \to +\infty}\sqrt{\frac{1}{y^3} + 1}}{\lim_{y \to +\infty}\sqrt{\frac{1}{y} + 1} + \lim_{y \to +\infty}\sqrt{\frac{1}{y}}}\\ &＝ \frac{1}{1 + 0} = 1
\end{aligned}
\]
3
求 \(\lim_{x \to +\infty} \left( \frac{x^2 - 1}{x^2 + 1} \right)^{\frac{x - 1}{x + 2}}\)
\[\begin{aligned}
\lim_{x \to +\infty} \left( \frac{x^2 - 1}{x^2 + 1} \right)^{\frac{x - 1}{x + 2}} &＝ \lim_{x \to +\infty} e^{\frac{x - 1}{x + 2} \ln \left( \frac{x^2 - 1}{x^2 + 1} \right)} \\
&= e^{\lim_{x \to +
\infty}\frac{x - 1}{x + 2} \lim_{x \to +
\infty} \ln \left( \frac{x^2 - 1}{x^2 + 1} \right)} \\
&= e^{1 \cdot 0} \\
&= 1
\end{aligned}
\]
4
求 \(\lim_{x \to 0} \frac{\sqrt[n]{x + 1} - 1}{x}\),其中 \(n\) 为正整数
令 \(t = \sqrt[n]{x + 1}\),原式为 \(\lim_{t \to 1} \frac{t - 1}{t^n - 1} = \lim_{t \to 1} \frac{1}{t^{n - 1} + t^{n - 2} + \cdots + t + 1} = \frac{1}{n}\).
5
设已知 \(\lim_{x \to 0} \frac{f(x)}{x} = l, b \not = 0\), 求 \(\lim_{x \to 0}\frac{f(bx)}{x}\)
\[\begin{aligned}
\lim_{x \to 0} \frac{f(bx)}{x} &= \lim_{x \to 0} \frac{f(bx)}{bx} \times b \\
&= b \lim_{t \to 0} \frac{f(t)}{t} \\
&= bl
\end{aligned}
\]
6
证明: \(\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sin x} = 1\)
\[\begin{aligned}
\lim_{x \to 0} \frac{2 \sin x}{(\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) \sin x} &= \lim_{x \to 0} \frac{2}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \\
&= \frac{2}{\lim_{x \to 0} \sqrt{1 + \sin x} + \lim_{x \to 0} \sqrt{1 - \sin x}} \\
&= \frac{2}{1 + 1} \\
&= 1
\end{aligned}
\]
7
证明:在区间 \((a, +\infty)\) 上单调有界函数 \(f\)，一定存在极限 \(\lim_{x \to +\infty} f(x)\)
不妨设 \(f\) 单调增,值域为 \(S\), \(\beta = \sup S\),则 \(\forall \varepsilon > 0, \exists x_0 \in(a, +\infty)\)，使得 \(f(x_0) > \beta - \varepsilon\)，则 \(\forall \varepsilon > 0\)，取 \(M = x_0, \forall x > M\)
\[\beta - \varepsilon < f(x_0) < f(x) < \beta
\]
故 \(\lim_{ x \to +\infty } f(x) = \beta\)
8
设 \(f(x)\) 在区间 \((a, b)\) 上为单调增加函数，且存在一个数列 \(\{ x_n \} \in (a, b)\),使得 \(\lim_{ n \to \infty }x_n = b, \lim_{ n \to \infty }f(x_n) = A\). 证明:
\((1) f\) 在区间 \((a, b)\) 上以 \(A\) 为上界
\((2)\lim_{ x \to b^- }f(x) = A\)
\(f\) 是 \((a, b)\) 上单调递增函数，则 \(\lim_{ x \to b }f(x)\) 存在，由 \(\text{Heine}\) 归结原理有
\[\lim_{ x \to b^- } f(x) = \lim_{ n \to \infty } f(x_n) = A
\]
设 \(S\) 为 \(f(x)\) 在 \((a, b)\) 上的值域，\(B = \sup S\)，有 \(\forall \varepsilon > 0, \exists x_0 \ s.t. \ B - \varepsilon < f(x_0) < B\)，由 \(\varepsilon\) 的任意性可知，\(\lim_{ x \to b }f(x) = B\)，再由极限的唯一性得到 \(A = B = \sup S\)。
9
设 \(\displaystyle\lim_{x\to+\infty} f(x)=A>0\)。证明：对每个 \(c\in(0,A)\)，存在 \(M>0\)，当 \(x>M\) 时，成立 \(f(x)>c\)。
（这是关于极限类型为 \(\displaystyle\lim_{x\to+\infty} f(x)\) 的保号性定理。）
取 \(\varepsilon = A - C\)，由极限的定义有 \(\exists M > 0, \ s.t. \ x > M, \left| A - f(x) \right| < \varepsilon \Rightarrow A - \varepsilon = C < f(x) < A + \varepsilon\).
10
设 \(f(a^{-}) < f(a^{+})\) 证明：存在 \(\delta >0\)，当 \(x\in(a-\delta,a)\) 和 \(y\in(a,a+\delta)\) 时,\(f(y)>f(x)\).
\(\lim_{ x \to a^{-}} f(x) = S, \lim_{ x \to a^{+}} = T\),取 \(\varepsilon = \frac{T - S}{2} > 0\).
\(\exists \delta_{1} > 0, x_{1} \in(a - \delta_{1}, a), \left| f(x_{1}) - S \right| < \varepsilon, i.e. \ f(x_{1}) < \frac{T + S}{2}\).
\(\exists \delta_{2} > 0, x_{2} \in(a, a + \delta_{2}), \left| f(x_{2}) - T \right| < \varepsilon, i.e. \ f(x_{2}) > \frac{T + S}{2}\).
取 \(\delta = \min(\delta_{1}, \delta_{2})\)，满足条件。
11
试用 \(\text{Heine}\) 归结原理证明单调函数的单侧极限存在定理。
\(f(x)\) 在 \(x_0\) 左侧极限存在 \(\Leftrightarrow\) 存在 \(A\)，对 \(\forall \{ x_{n}\}\) 单增趋于 \(x_{0}\)，均有 \(\lim_{ n \to \infty }f(x_{n}) = A\).
必要性显然。下证充分性：
反证法，若 \(\lim_{ x \to x_{0}^- }f(x)\) 不存在，则 \(\exists \varepsilon > 0\)，使得 \(\forall \delta > 0, \exists x \in(x_{0} - \delta, x_{0})\) 有\(\left| f(x) - A \right| \ge \varepsilon_{0}\)
取 \(\delta_{1} = 1\), \(\exists x_{1}\) 使 \(\left| f(x_{1}) - A \right| \ge \varepsilon_{0}\)，取 \(\delta_{2} = \min\left\{ \frac{1}{2}, |x_{1} - x_{0}| \right\}\)，则 \(\exists x_{2} \in (x_{0} - \delta_{2}, x_{0})\)，使 \(\left| f(x_{2}) - A \right| \ge \varepsilon_{0}\)，依次有 \(\delta_{n} = \min\left\{ \frac{1}{n}, |x_{n - 1} - x_{0}| \right\}\)，则 \(\exists x_{n} \in (x_{0} - \delta_{2}, x_{0})\)，使 \(|f(x_{2}) - A| \ge x_{0}\) 矛盾。
4.3
两个重要极限
\[\begin{aligned}
\lim_{ x \to 0 } \frac{\sin x}{x} = 1 \ \ \ \ \ \lim_{ x \to 0 } (1 + x)^{\frac{1}{x}} = e
\end{aligned}
\]
一些极限关系
\[\begin{aligned}
e^x - 1 \sim x (x \to 0) &\ \ \ \ \ \ \ \ \ &\sin x \sim x(x \to 0) \\
\ln(x + 1) \sim x (x \to 0) & & 1 - \cos x \sim \frac{1}{2} x^2(x \to 0) \\
\ln x = o(x^{-\alpha}) (x\to 0^{+}) & & x^k = o(a^x)(x \to +\infty) (a > 1) \\
(1 + x)^{\alpha} - 1 \sim \alpha x(x \to 0) & & \arctan x \sim x (x \to 0)
\end{aligned}
\]
1
(1)
\[\begin{aligned}
\lim_{x \to +\infty} \left( \frac{2}{\pi} \arctan x \right)^x &= \lim_{ x \to +\infty } \left( \frac{2}{\pi} \left( \frac{\pi}{2} - \arctan \frac{1}{x} \right) \right)^x \\
&= \lim_{ x \to +\infty } \left(1 - \frac{2}{\pi} \arctan{\frac{1}{x}} \right)^{-\frac{\pi}{2\arctan \frac{1}{x}} (-\frac{2 \arctan\frac{1}{x}}{\pi})x} \\
&= \lim_{ x \to +\infty } e^{ -\frac{2 x\arctan\frac{1}{x}}{\pi} } \\
&= e^{ -\frac{2}{\pi}}
\end{aligned}
\]
(2)
\[\begin{aligned}
\lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x}& \ \ 令 \ t = \frac{\pi}{2} - x \\
\lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x} &= \lim_{ t \to 0 } (\cos t)^{\frac{1}{\tan t}} \\
&= \lim_{ t \to 0 } \left( 1 - \frac{1}{2}t^2 \right)^{\frac{\cos t}{\sin t}} \\
&= \lim_{ t \to 0 }\left( 1 - \frac{1}{2}t^2 \right)^{-\frac{2}{t^2} \left( -\frac{t^2}{2}\frac{\cos t}{\sin t} \right)} \\
&= \lim_{ t \to 0 } e^{-\frac{t^2\left( 1 - \frac{1}{2}t^2 \right)}{2t}} \\
&= \lim_{ t \to 0 } e^{\left( \frac{t^3}{4} - \frac{t}{2} \right)} \\
&= 1
\end{aligned}
\]
(3)
\[\begin{aligned}
\lim_{x \to \infty} \left( \frac{x^2 - 1}{x^2 + 1} \right)^{x^2} &= \lim_{ x \to \infty } \left( 1 - \frac{2}{x^2 + 1} \right)^{-\frac{x^2+1}{2} \left(\frac{-2x^2}{x^2+1} \right)} \\
&= \lim_{ x \to \infty } e^{\frac{-2}{1 + \frac{1}{x^2}}} \\
&= e^{-2}
\end{aligned}
\]
(4)
\[\begin{aligned}
\lim_{x \to \frac{\pi}{2}} (\cos x)^{\frac{\pi}{2} - x}& \ \ \ \ 令 t = \frac{\pi}{2} - x \\
\lim_{x \to \frac{\pi}{2}} (\cos x)^{\frac{\pi}{2} - x} &= \lim_{ t \to 0^{+} } (\sin t)^{t} \\
&= \lim_{ t \to 0^{+} } e^{t \ln(\sin t)} \\
&= \lim_{ t \to 0^{+} } e^{(t/\sin t) \sin t\ln(\sin t)} = 1
\end{aligned}
\]
(5)
\[\begin{aligned}
\lim_{x \to 0} \frac{\sin 2x - 2 \sin x}{x^3} &= \lim_{ x \to 0 } \frac{2(\cos x - 1) \sin x}{x^3} \\
&= \lim_{ x \to 0 } \frac{2\left( -\frac{1}{2}x^2 \right)x}{x^3} \\
&= -1
\end{aligned}
\]
(6)
\[\begin{aligned}
\lim_{x \to 1} (1 - x) \tan \left( \frac{\pi}{2} x \right) &\ \ \ \ 令 t = 1 - x \\
\lim_{x \to 1} (1 - x) \tan \left( \frac{\pi}{2} x \right) &= \lim_{ t \to 0 } \frac{t}{\tan \frac{\pi}{2}t} = \lim_{ t \to 0 } \frac{t\left( \cos \frac{\pi}{2}t \right)}{\sin \frac{\pi}{2}t} \\&= \lim_{ t \to 0 } \frac{t\left( 1 - \left( \frac{\pi}{2}t \right)^2 \right)}{\frac{\pi}{2}t} = \frac{2}{\pi}
\end{aligned}
\]
2
注意下列两个“不等式”并求出正确值:
(1)
\[\begin{aligned}
\lim_{x \to +\infty} \frac{\sin x}{x} \neq 1 \\
0 < \left| \frac{\sin x}{x} \right| \le \left| \frac{1}{x} \right| \Rightarrow \lim_{ x \to +\infty } \frac{\sin x}{x} = 0
\end{aligned}
\]
(2)
\[\begin{aligned}
\lim_{x \to +\infty} (1 + x)^{\frac{1}{x}} \neq e \\
\lim_{x \to +\infty} (1 + x)^{\frac{1}{x}} = \lim_{ x \to \infty } e^{ \frac{1}{x}\ln(1+x) } = e^{\lim_{ x \to \infty }\frac{1}{x+1}\ln(1+x) \frac{x+1}{x}} = 1
\end{aligned}
\]
3
设 \(( a > 0, b > 0)\)，求极限
\[\begin{aligned}
\lim_{n \to \infty} \left( \frac{\sqrt[n]{a} + \sqrt[n]{b}}{2} \right)^n
\end{aligned}
\]
（本题是数列极限问题，但现在可以用函数极限知识来解决。）
见下题
4
设 \(( a_1, \cdots, a_n)\) 为正数，\(( n \geq 2)\)
\[f(x) = \left( \frac{a_1^x + a_2^x + \cdots + a_n^x}{n} \right)^\frac{1}{x}
\]
求
\[\lim_{x \to 0} f(x)
\]
\[\begin{aligned}
&\lim_{ x \to 0 } \left( 1 + \left( \frac{a_{1}^x+a_{2}^x+\cdots+a_{n}^x - n}{n} \right) \right)^{\frac{1}{x}} \\=& \lim_{ x \to 0 } \left( 1 + \frac{a_{1}^x+a_{2}^x+\cdots+a_{n}^x - n}{n} \right)^{\frac{n}{a_{1}^x+a_{2}^x+\cdots+a_{n}^x - n}\frac{a_{1}^x+a_{2}^x+\cdots+a_{n}^x - n}{nx}} \\
=&\lim_{ x \to 0 } e^{\frac{1}{n}\sum_{i = 1}^{n}\frac{a_{i}^x - 1}{x}} \\
=&\lim_{ x \to 0 } e^{\frac{1}{n}\sum_{i = 1}^{n} \ln a_{i}} \\
=&\sqrt[n]{ \sum_{i = 1}^{n} a_{i}}
\end{aligned}
\]
5
求极限
\[\begin{aligned}
\lim_{n \to \infty} \prod_{k=1}^n \cos \frac{x}{2^k} &= \lim_{ n \to \infty } \frac{\cos \frac{x}{2} \cos \frac{x}{2^2} \cdots \cos \frac{x}{2^n} \sin \frac{x}{2^n}}{\sin \frac{x}{2^n}} \\
&= \lim_{ n \to \infty } \frac{1}{2} \frac{\cos \frac{x}{2} \cos \frac{x}{2^2} \cdots \cos \frac{x}{2^{n-1}} \sin \frac{x}{2^{n - 1}}}{\sin \frac{x}{2^n}} \\
&= \lim_{ n \to \infty } \frac{\sin x}{2^n \sin \frac{x}{2^n}} \\
&= \lim_{ n \to \infty } \frac{\sin x}{2^n \frac{x}{2^n}} \\
&= \frac{\sin x}{x}
\end{aligned}
\]
证明 \(\text{Viete}\) 公式
\[\begin{aligned}
\frac{\pi}{2} = \frac{1}{\sqrt{ \frac{1}{2} } \cdot \sqrt{ \frac{1}{2} + \frac{1}{2}\sqrt{ \frac{1}{2} } } \cdot \sqrt{ \frac{1}{2} + \frac{1}{2}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} } } } \cdot \
\cdots }
\end{aligned}
\]
\[\begin{aligned}
\cos 2\theta = 2 \cos^2\theta - 1, \cos \theta = \sqrt{ \frac{1}{2} + \frac{1}{2}\cos 2\theta },\\ 取 x = \frac{\pi}{2}, \cos \frac{x}{2} = \sqrt{ \frac{1}{2} }, \cos \frac{x}{4} = \sqrt{ \frac{1}{2} + \frac{1}{2}\sqrt{ \frac{1}{2} } } \\
上式 = \frac{\frac{\pi}{2}}{\sin \frac{\pi}{2}} = \frac{\pi}{2}
\end{aligned}
\]
4.4
1
确定下列无穷小量的阶：
(1)
\[\begin{aligned}
\sqrt{1+\tan x} - \sqrt{1-\tan x} (x \to 0) \\
\sqrt{1+\tan x} - \sqrt{1-\tan x} = \frac{2\tan x}{\sqrt{ 1 + \tan x } + \sqrt{ 1 - \tan x }} \sim \tan x \sim x \\
1 阶
\end{aligned}
\]
(2)
\[\begin{aligned}
\ln x - \ln a (x \to a), a > 0 \\
\ln x - \ln a = \ln \frac{x}{a} = \ln \left( \frac{x}{a} - 1 + 1\right) \sim \frac{x}{a} - 1 \\
1 阶
\end{aligned}
\]
(3)
\[\begin{aligned}
a^x - 1 (x \to 0),a > 0 \\
a^x - 1 = e^{x \ln a} - 1 \sim x \ln a \\
1 阶
\end{aligned}
\]
(4)
\[\begin{aligned}
a^{x^2} - b^{x^2} (x \to 0), a, b > 0 \\
a^{x^2} - b^{x^2} = e^{x^2 \ln a} - e^{x^2 \ln b} = (e^{x^2 \ln a} - 1) - (e^{x^2 \ln b} - 1) \sim x^2 \ln \frac{a}{b} \\
2 阶
\end{aligned}
\]
(5)
\[\begin{aligned}
\ln\left( x + \cos \frac{\pi}{2}x \right) (x \to 1) &\ \ \ \ 令 t = 1 - x, t \to 0 \\
\ln\left( x + \cos \frac{\pi}{2}x \right) &= \ln\left( 1 - t + \cos \left( \frac{\pi}{2} - \frac{\pi}{2}t\right) \right) \\
&= \ln\left( 1 - t + \sin\left( \frac{\pi}{2} t\right) \right) \\
&\sim \sin\left( \frac{\pi}{2}t \right) - t \sim \left( \frac{\pi}{2} - 1 \right) t \\
1 阶
\end{aligned}
\]
(6)
\[\begin{aligned}
\ln x \ln(x-1) (x \to 1^{+}) &\ \ \ \ \ 令 t = x - 1, t \to 0^{+} \\
\ln x \ln(x-1) &= \ln (t + 1) \ln t = \ln(t + 1)^{\ln t} = \ln(t + 1)^{\frac{1}{t + 1} ((t + 1) \ln t)} \\
&\sim \ln e^{(t + 1)\ln t} = (t + 1) \ln t \\
\ln t = o(t^{-\alpha}) \ (t \to 0^{+}), &(\alpha > 0)，阶数为 1 - \varepsilon, \forall \varepsilon > 0
\end{aligned}
\]
2
设存在极限 \(\lim_{x \to 0} \frac{f(x)}{x}\), 又有 \(f(x) - f\left(\frac{x}{2}\right) = o(x) (x \to 0)\), 证明：\(f(x) = o(x) (x \to 0)\)。
\[\begin{aligned}
\lim_{ x \to 0 } \frac{f(x) - f\left( \frac{x}{2} \right)}{x} = 0 \\
\lim_{ x \to 0 } \frac{f(x)}{x} = \lim_{ x \to 0 } \frac{f\left( \frac{x}{2} \right)}{x}=\lim_{ x \to 0 } \frac{1}{2}\frac{f\left( \frac{x}{2} \right)}{\frac{x}{2}} = \lim_{ x \to 0 } \frac{1}{2} \frac{f(x)}{x} \Rightarrow \lim_{ x \to 0 } \frac{f(x)}{x} = 0 \\
f(x) = o(x) (x \to 0)
\end{aligned}
\]
3
与数列中的几个常见的无穷大量之间的关系 \(\ln n \ll n^\varepsilon \ll a^n \ll n! \ll n^n (a > 1, \varepsilon > 0)\) 相类似，证明当 \(x \to +\infty\) 时，有
\[\ln x \ll x^\varepsilon \ll a^x \ll x^x (a > 1, \varepsilon > 0),
\]
其中 \(u \ll v\) 的定义是 \(\lim \frac{u}{v} = 0\)。
即证 \(\lim_{ x \to +\infty }\frac{\ln x}{x^{\epsilon}} = \lim_{ x \to +\infty } \frac{x^\epsilon}{a^x} = \lim_{ x \to +\infty } \frac{a^x}{x^x} = 0\)，前两个等式见 4.2 第一题，下证 \(\lim_{ x \to +\infty } \frac{a^x}{x^x} = 0\).
在 \(x\) 足够大的时候， \(\frac{a}{x} < 1, x > 1\) 此时 \(0 < \left| \left( \frac{a}{x} \right)^x \right| < \frac{a}{x}\)，当 \(x \to \infty\) 时由迫敛性得到 \(\lim_{ x \to +\infty } \left( \frac{a}{x} \right)^x = 0\)
4
(1)
\[\begin{aligned}
\lim_{x \to 0} \frac{\ln (\sin^2 x + e^x) - x}{\ln (x^2 + e^{2x}) - 2x} = \lim_{ x \to 0 } \frac{\ln\left( \frac{\sin^2x + e^x}{e^x} \right)}{\ln\left( \frac{x^2+e^{2x}}{e^{2x}} \right)} \lim_{ x \to 0 } \frac{\ln\left( 1 + \frac{\sin^2x}{e^x} \right)}{\ln\left( 1 + \frac{x^2}{e^{2x}} \right)} = \lim_{ x \to 0 } \frac{\sin^2x}{x^2} e^x = 1
\end{aligned}
\]
(2)
\[\begin{aligned}
\lim_{x \to +\infty} (x + 1)[\ln (x^2 + x) - 2 \ln (x + 1)] &= \lim_{ x \to +\infty } (x + 1)\ln\left( \frac{x(x + 1)}{(x + 1)(x + 1)} \right) \\
&= \lim_{ x \to +\infty } (x + 1)\ln\left( 1 - \frac{1}{x + 1} \right) \\
&= \lim_{ x \to +\infty } -(x + 1)\left( \frac{1}{x + 1} \right) \\
&= -1
\end{aligned}
\]
(3)
\[\begin{aligned}
\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[6]{1+x}}{\sqrt[3]{1+x} - 1} = \lim_{ x \to 0 } \frac{\frac{x}{2} - \frac{x}{6}}{\frac{x}{3}} = 1
\end{aligned}
\]
(4)
\[\begin{aligned}
\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} &= \lim_{ x \to 0 } \frac{\sqrt{ 1 + (\cos x - 1) } - \sqrt[3]{ 1 + (\cos x - 1) }}{\sin^2 x} \\
&= \lim_{ x \to 0 } \frac{\frac{\cos x - 1}{2} - \frac{\cos x - 1}{3}}{x^2} \\
&= \lim_{ x \to 0 } \frac{\cos x - 1}{6x^2} \\
&= \lim_{ x \to 0 } \frac{-\frac{1}{2}x^2}{6x^2} \\
&= -\frac{1}{12}
\end{aligned}
\]
(5)
\[\begin{aligned}
\lim_{x \to 0} \frac{(3 + 2 \sin x)^x - 3^x}{\tan^2 x} &= \lim_{ x \to 0 } \frac{3^x\left( 1 + \frac{2}{3} \sin x\right)^x - 3^x}{\tan^2x}\\
&= \lim_{ x \to 0 } \frac{3^x\left( 1 + \frac{2}{3}\sin x \right)^{\frac{3}{2\sin x} \left( x\frac{2}{3}\sin x \right)}- 3^x}{\tan^2 x} \\
&=\lim_{ x \to 0 } \frac{3^x \left( e^{\frac{2}{3}x\sin x} - 1 \right)}{\tan^2x} \\
&=\lim_{ x \to 0 } \frac{2 \cdot 3^{x - 1} x\sin x}{\tan^2x} \\
&=\frac{2}{3}
\end{aligned}
\]
(6)
\[\begin{aligned}
\lim_{t \to 0} \left( \frac{\arcsin t}{t} \right)^{\frac{1}{t^2}} &= e^{\lim_{ t \to 0 } \ln\left(\frac{\frac{\arcsin t}{t^2} - 1}{t^2}\right) } \\
&= e^{\lim_{ t \to 0 } \frac{\arcsin t - t}{\arcsin^3 t}} \\
令 x = \arcsin t, x \to 0 \\
&=e^{\lim_{ x \to 0 } \frac{x - \sin x}{x^3}} \\
&=e^{\lim_{ x \to 0 } \frac{1 - \cos x}{3x^2}} \\
&=e^{\lim_{ x \to 0 } \frac{2\sin^2 \frac{x}{2}}{3x^2}} \\
&=\sqrt[6]{ e }
\end{aligned}
\]
5.1
1
\((1)\) 将对偶法则用于连续性的第一定义和第二定义，写出函数 \(f\) 在点 \(a\) 出不连]续的两个正面叙述.
\[\begin{aligned}
(1) \lim_{ x \to a } f(x) \not= f(a) \\
(2) \exists \{ x_n \}, \lim_{ n \to \infty } x_n = a, \lim_{ n \to \infty } f(x_n) \not= f(a)
\end{aligned}
\]
\((2)\) 证明连续性的两个定义的等价性
即证 \(\lim_{ x \to a } f(x) = f(a) \Leftrightarrow\) 任何收敛于 \(a\) 的数列 \(\{x_n\}\)，有 \(\lim_{ n \to \infty }f(x_n) = f(a)\).
必要性：因为 \(\lim_{ x \to a }f(x) = f(a)\)，\(\forall \varepsilon > 0, \exists \delta > 0, s.t. \ 0 < \left| x - a \right| < \delta, \left| f(x) - f(a) \right| < \varepsilon\)，当 \(x_n \to a\)，存在 \(N\)，使得 \(n > N\) 时，\(0 < \left| x_n - a \right| < \delta\)，也就有 \(\left| f(x_n) - f(a) \right| < \varepsilon\)，故 \(\lim_{ n \to \infty } f(x_n) = f(a)\) 成立。
充分性：反证法，假设 \(\lim_{ x \to a }f(x) \not=f(a)\)，即是 \(\forall \varepsilon > 0,\forall \delta >0, \exists x_0\) 满足 \(\left| x_0 - a \right| < \delta,|f(x_0)-f(a)|>\varepsilon\)，依次取 \(\delta_i = \frac{1}{i},\exists x_i, |x_i - a|<\delta_i, \ s.t. |f(x_i) - f(a)|>\varepsilon\)，故 \(\lim_{ n \to \infty }x_n=a, \lim_{ n \to \infty }f(x_n) \not= f(a)\)，矛盾。
2
讨论下列函数与其间断点类型
\[\begin{aligned}
(1) f(x) = [x], x \in \mathbb{Z} 第一类(跳跃) \\
(2) f(x) = \begin{cases}\frac{e^{\sin x} - 1}{x}, x \in(-\infty, 0) \cup (0, +\infty) \\ 1, x = 0 \end{cases} 没有间断点 \\
(3)f(x)=[x] + [-x], x \in \mathbb{Z} 第一类(可去) \\
(4) f(x) = \begin{cases}
\frac{x^2-x}{|x|(x^2-1)},|x|\not=0,1 \\
1, |x|=0, 1 & x = -1, 第二类, x = 0, 1, 第一类
\end{cases}
\end{aligned}
\]
3
设函数 \(f\in C[a, b]\)，若有数列 \(\{ x_n \} \in [a, b]\)，使得 \(\lim_{ n \to \infty }f(x_n)=A\)，证明：存在 \(\xi\in [a, b]\)，使得 \(f(\xi)=A\).
反证法：若不存在 \(\xi\in [a, b], f(\xi)=A\)，记 \(f(x)\) 在 \(x = x_0\) 处取得最小值，在 \(x = x_1\) 处取得最大值，由函数的连续性可知，在 \([a, b]\) 上， \(f(x)\) 的值域为 \([f(x_0), f(x_1)]\)，故 \(A > f(x_1)\) 或 \(f(x_0)>A\)，不妨设 \(A>f(x_1)\)，令 \(t=A-f(x_1)\)，取 \(\varepsilon=\frac{t}{2},\forall x\in[a, b], |f(x)-A| \ge t = 2\varepsilon > \varepsilon\)，与题设不符，矛盾。
4
设函数 \(f\) 在 \((-\infty, +\infty)\) 上有定义，在 \(x = 0, 1\) 两点连续，且满足 \(f(x)=f(x^2), x\in \mathbf{R}\)，证明： \(f\) 是常值函数。
因为 \(f(-x)=f(x)=f(x^2)\)，故证明 \(f\) 在 \([0, +\infty)\) 为常值函数即可。
\(\forall x \ge 0, f(x)=f(x^{\frac{1}{2^n}}), \lim_{ n \to \infty }x^{\frac{1}{2^n}}=1,\lim_{ n \to \infty }f(x^\frac{1}{2^n})=f(1)\)。
5
设在区间 \(I\) 上函数 \(f, g\) 连续，证明：\(\max\{ f, g \}, \min\{ f,g \}\in C(I)\).
\[\begin{aligned}
\max\{ f, g \} = \frac{f + g}{2} + \left| \frac{f - g}{2} \right| \\
\min\{ f, g \} = \frac{f + g}{2} - \left| \frac{f - g}{2} \right|
\end{aligned}
\]
故结论成立
6
设有三个函数 \(f_1, f_2, f_3 \in C[a,b]\)，对于每个 \(x\in[a, b]\)，定义 \(f(x)\) 是是哪个函数值 \(f_1(x), f_2(x), f_3(x)\) 中处于中间的一个值，证明：\(f\in C[a, b]\).
\(f(x) = f_1(x)+f_2(x)+f_3(x) - \max\{ f_1(x),f_2(x),f_3(x) \} - \min\{ f_1(x), f_2(x), f_3(x) \}\),由上一题结论可得 \(f(x) \in C[a, b]\)
7
证明： \(f\) 为区间上连续函数的充分必要条件是：对每个正整数 \(n\)，函数
\[\begin{aligned}
f_n(x) =
\begin{cases}
-n, f(x) \le -n \\
f(x), -n < f(x) \le n, \\
n, f(x) > n
\end{cases}
\end{aligned}
\]
连续。
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