2022数学分析【南昌大学】
2022 数学分析
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利用极限定义证明： lim n → ∞ 4 n 3 + n − 2 2 n 3 − 10 = 2 \mathop {\lim }\limits_{n \to \infty } \frac{ {4{n^3} + n - 2}}{ {2{n^3} - 10}} = 2 n→∞lim2n3−104n3+n−2=2
∀ ε > 0 \forall \varepsilon>0 ∀ε>0 要使不等式成立，限制 n > 18 n>18 n>18
∣ 4 n 3 + n − 2 2 n 3 − 10 − 2 ∣ = ∣ 4 n 3 + n − 2 2 n 3 − 10 − 2 ∣ = ∣ n + 18 2 n 3 − 10 ∣ ≤ ∣ 2 n 2 n 3 − 2 n ∣ = ∣ 1 n 2 − 1 ∣ < ε \begin{align*} \left| {\frac{ {4{n^3} + n - 2}}{ {2{n^3} - 10}} - 2} \right| &= \left| {\frac{ {4{n^3} + n - 2}}{ {2{n^3} - 10}} - 2} \right|\\ &= \left| {\frac{ {n + 18}}{ {2{n^3} - 10}}} \right|\\ &\le \left| {\frac{ {2n}}{ {2{n^3} - 2n}}} \right|\\ &= \left| {\frac{1}{ { {n^2} - 1}}} \right|\\ &< \varepsilon \end{align*} 2n3−104n3+n−2−2 = 2n3−104n3+n−2−2 = 2n3−10n+18 ≤ 2n3−2n2n = n2−11 <ε
解得 n > 1 ε + 1 n>\sqrt{\frac{1}{\varepsilon}+1} n>ε1+1 ，取 N = max { 18 , 1 ε + 1 } N=\max\{18,\sqrt{\frac{1}{\varepsilon}+1}\} N=max{ 18,ε1+1}
于是有， ∀ ε > 0 \forall \varepsilon>0 ∀ε>0, ∃ N = max { 18 , 1 ε + 1 } \exist N=\max\{18,\sqrt{\frac{1}{\varepsilon}+1}\} ∃N=max{ 18,ε1+1} , 当 n > N n>N n>N, 有 ∣ 4 n 3 + n − 2 2 n 3 − 10 − 2 ∣ < ε \left| {\frac{ {4{n^3} + n - 2}}{ {2{n^3} - 10}} - 2} \right| < \varepsilon 2n3−104n3+n−2−2 <ε。即 lim n → ∞ 4 n 3 + n − 2 2 n 3 − 10 = 2 \mathop {\lim }\limits_{n \to \infty } \frac{ {4{n^3} + n - 2}}{ {2{n^3} - 10}} = 2 n→∞lim2n3−104n3+n−2=2
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求下列极限。
- lim n → ∞ ( 1 n + 1 + 1 n + 2 + ⋯ + 1 n + n ) ; \lim_{n \to \infty} \left( \frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{n + n} \right); n→∞lim(n+11+n+21+⋯+n+n1);
- lim x → 0 [ sin x − sin ( sin x ) ] sin x x 4 ; \lim_{x \to 0} \frac{[\sin x - \sin(\sin x)] \sin x}{x^4}; x→0limx4[sinx−sin(sinx)]sinx;
- lim x → 0 + ∫ 0 2 x ∣ t − x ∣ sin t d t x 3 . \lim_{x \to 0^+} \frac{\int_0^{2x} \left|t - x\right| \sin t \, dt}{x^3}. x→0+limx3∫02x∣t−x∣sintdt.
解答 1
lim n → ∞ ( 1 n + 1 + 1 n + 2 + ⋯ + 1 n + n ) = lim n → ∞ 1 n ∑ k = 1 n 1 1 + k n = ∫ 0 1 1 1 + x d x = ln ( 1 + x ) ∣ 0 1 = ln 2 \begin{align*} \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{ {n + 1}} + \frac{1}{ {n + 2}} + \cdots + \frac{1}{ {n + n}}} \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{ {1 + \frac{k}{n}}}} \\ &= \int_0^1 {\frac{1}{ {1 + x}}dx} \\ &= \ln \left( {1 + x} \right)|{}_0^1\\ &= \ln 2 \end{align*} n→∞lim(n+11+n+21+⋯+n+n1)=n→∞limn1k=1∑n1+nk1=∫011+x1dx=ln(1+x)∣01=ln2
解答 2
lim x → 0 [ sin x − sin ( sin x ) ] sin x x 4 = lim x → 0 1 6 sin 3 x sin x x 4 = 1 6 \begin{align*} \mathop {\lim }\limits_{x \to 0} \frac{ {[\sin x - \sin (\sin x)]\sin x}}{ { {x^4}}}& = \mathop {\lim }\limits_{x \to 0} \frac{ {\frac{1}{6}{ {\sin }^3}x\sin x}}{ { {x^4}}} \\ &= \frac{1}{6} \\ \end{align*} x→0limx4[sinx−sin(sinx)]sinx=x→0limx461sin3xsinx=61
解答 3
lim x → 0 + ∫ 0 2 x ∣ t − x ∣ sin t d t x 3 = lim x → 0 + ∫ 0 x ( x − t ) sin t d t + ∫ x 2 x ( t − x ) sin t d t x 3 = lim x → 0 + ∫ 0 x sin t d t + 2 x sin 2 x + ∫ x 2 x − sin t d t 3 x 2 = lim x → 0 + sin x + 2 sin 2 x + 4 x cos 2 x − 2 sin 2 x + sin x 6 x = lim x → 0 + 4 x cos 2 x 6 x + lim x → 0 + 2 sin x 6 x = 1 \begin{align*} \lim_{x \to 0^+} \frac{\int_0^{2x} |t - x| \sin t \, dt}{x^3} &= \lim_{x \to 0^+} \frac{\int_0^x (x - t) \sin t \, dt + \int_x^{2x} (t - x) \sin t \, dt}{x^3} \\ &= \lim_{x \to 0^+} \frac{\int_0^x \sin t \, dt + 2x \sin 2x + \int_x^{2x} -\sin t \, dt}{3x^2} \\ &= \lim_{x \to 0^+} \frac{\sin x + 2 \sin 2x + 4x \cos 2x - 2 \sin 2x + \sin x}{6x} \\ &= \lim_{x \to 0^+} \frac{4x \cos 2x}{6x} + \lim_{x \to 0^+} \frac{2 \sin x}{6x} \\ &= 1 \end{align*} x→0+limx3∫02x∣t−x∣sintdt=x→0+limx3∫0x(x−t)sintdt+∫x2x(t−x)sintdt=x→0+lim3x2∫0xsintdt+2xsin2x+∫x2x−sintdt=x→0+lim6xsinx+2sin2x+4xcos2x−2sin2x+sinx=x→0+lim6x4xcos2x+x→0+lim6x2sinx=1
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证明：函数 f ( x ) = x sin x f(x) = x \sin x f(x)=xsinx 在 ( − ∞ , + ∞ ) (-\infty, +\infty) (−∞,+∞) 上不一致连续。
证明：反证法
设 f ( x ) = x sin x f(x) = x \sin x f(x)=xsinx 在 ( − ∞ , + ∞ ) (-\infty, +\infty) (−∞,+∞) 上一致连续。则 ∀ ε > 0 \forall \varepsilon>0 ∀ε>0， ∃ δ > 0 \exist\delta>0 ∃δ>0 当， ∀ x 1 , x 2 : ∣ x 1 − x 2 ∣ < δ \forall x_1,x_2:\left|x_1-x_2\right|<\delta ∀x1,x2:∣x<