人邮高数 第1章 第1-5-2题

[AI解答]

[AI解答]

以下为各小题的详细解答，均利用重要极限 $\displaystyle{\lim_{u \to 0} (1+u)^{1/u} = e}$ 或其变形。

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(1) $\displaystyle{\lim_{x \to \infty} \left(1+\frac{2}{x}\right)^{x}}$

令 $\displaystyle u = \frac{2}{x}$，则当 $x \to \infty$ 时 $u \to 0$，且 $\displaystyle x = \frac{2}{u}$，于是
$$
\lim_{x \to \infty} \left(1+\frac{2}{x}\right)^{x} = \lim_{u \to 0} (1+u)^{\frac{2}{u}} = \left[\lim_{u \to 0} (1+u)^{1/u}\right]^2 = e^2.
$$

答案： $e^2$

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(2) $\displaystyle{\lim_{x \to 0} (1+3x)^{1/x}}$

令 $u = 3x$，则 $x \to 0$ 时 $u \to 0$，且 $\displaystyle \frac{1}{x} = \frac{3}{u}$，
$$
\lim_{x \to 0} (1+3x)^{1/x} = \lim_{u \to 0} (1+u)^{3/u} = e^3.
$$

答案： $e^3$

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(3) $\displaystyle{\lim_{x \to 0} (1-4x)^{1/x}}$

令 $u = -4x$，则 $x \to 0$ 时 $u \to 0$，且 $\displaystyle \frac{1}{x} = -\frac{4}{u}$，
$$
\lim_{x \to 0} (1-4x)^{1/x} = \lim_{u \to 0} (1+u)^{-4/u} = e^{-4}.
$$

答案： $e^{-4}$

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(4) $\displaystyle{\lim_{x \to \infty} \left(1-\frac{1}{2x}\right)^{x}}$

令 $\displaystyle u = -\frac{1}{2x}$，则 $x \to \infty$ 时 $u \to 0$，且 $\displaystyle x = -\frac{1}{2u}$，
$$
\lim_{x \to \infty} \left(1-\frac{1}{2x}\right)^{x} = \lim_{u \to 0} (1+u)^{-\frac{1}{2u}} = e^{-1/2}.
$$

答案： $e^{-1/2}$

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(5) $\displaystyle{\lim_{x \to \infty} \left(\frac{1+x}{x}\right)^{2x+1}}$

先化简：$\displaystyle \frac{1+x}{x} = 1 + \frac{1}{x}$，于是
$$
\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^{2x+1} = \lim_{x \to \infty} \left(1+\frac{1}{x}\right)^{2x} \cdot \left(1+\frac{1}{x}\right) = e^2 \cdot 1 = e^2.
$$

答案： $e^2$

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(6) $\displaystyle{\lim_{x \to \infty} \left(\frac{2x+1}{2x-3}\right)^{x}}$

改写：$\displaystyle \frac{2x+1}{2x-3} = 1 + \frac{4}{2x-3}$，令 $\displaystyle u = \frac{4}{2x-3}$，则 $\displaystyle x = \frac{3}{2} + \frac{2}{u}$，当 $x \to \infty$ 时 $u \to 0$，
$$
\lim_{x \to \infty} \left(1+\frac{4}{2x-3}\right)^{x} = \lim_{u \to 0} (1+u)^{\frac{3}{2} + \frac{2}{u}} = \lim_{u \to 0} (1+u)^{3/2} \cdot \lim_{u \to 0} (1+u)^{2/u} = 1 \cdot e^2 = e^2.
$$

答案： $e^2$

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(7) $\displaystyle{\lim_{x \to 0} \left(1+3\tan^2 x\right)^{\cot^2 x}}$

令 $u = 3\tan^2 x$，则 $\displaystyle \cot^2 x = \frac{1}{\tan^2 x} = \frac{3}{u}$，当 $x \to 0$ 时 $u \to 0$，
$$
\lim_{x \to 0} (1+3\tan^2 x)^{\cot^2 x} = \lim_{u \to 0} (1+u)^{3/u} = e^3.
$$

答案： $e^3$

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(8) $\displaystyle{\lim_{x \to \frac{\pi}{2}} (1+\cos x)^{2\sec x}}$

令 $\displaystyle t = \frac{\pi}{2} - x$，则 $\displaystyle x \to \frac{\pi}{2}$ 时 $t \to 0$，且 $\cos x = \sin t$，$\displaystyle \sec x = \frac{1}{\cos x} = \frac{1}{\sin t}$，于是
$$
\lim_{x \to \frac{\pi}{2}} (1+\cos x)^{2\sec x} = \lim_{t \to 0} (1+\sin t)^{2/\sin t}.
$$
令 $u = \sin t$，则 $t \to 0$ 时 $u \to 0$，原式 $= \lim_{u \to 0} (1+u)^{2/u} = e^2$。

答案： $e^2$

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(9) $\displaystyle{\lim_{x \to \infty} \left(1-\frac{1}{x}\right)^{\frac{1}{\sin(1/x)}}}$

令 $\displaystyle t = \frac{1}{x}$，则 $x \to \infty$ 时 $t \to 0$，原式 $= \lim_{t \to 0} (1-t)^{1/\sin t}$。
取对数：$\displaystyle \lim_{t \to 0} \frac{\ln(1-t)}{\sin t} = \lim_{t \to 0} \frac{-t + o(t)}{t + o(t)} = -1$，因此原极限 $= e^{-1}$。

答案： $e^{-1}$

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(10) $\displaystyle{\lim_{n \to \infty} \left(1+\frac{2}{3^n}\right)^{3^n}}$

令 $\displaystyle u = \frac{2}{3^n}$，则 $n \to \infty$ 时 $u \to 0$，且 $\displaystyle 3^n = \frac{2}{u}$，
$$
\lim_{n \to \infty} \left(1+\frac{2}{3^n}\right)^{3^n} = \lim_{u \to 0} (1+u)^{2/u} = e^2.
$$

答案： $e^2$

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(11) $\displaystyle{\lim_{x \to \infty} \left(\frac{3x-1}{3x+1}\right)^{3x-1}}$

改写：$\displaystyle \frac{3x-1}{3x+1} = 1 - \frac{2}{3x+1}$，令 $\displaystyle u = -\frac{2}{3x+1}$，则 $\displaystyle 3x-1 = -\frac{2}{u} - 2$，当 $x \to \infty$ 时 $u \to 0$，
$$
\lim_{x \to \infty} \left(1-\frac{2}{3x+1}\right)^{3x-1} = \lim_{u \to 0} (1+u)^{-\frac{2}{u} - 2} = e^{-2} \cdot 1 = e^{-2}.
$$

答案： $e^{-2}$

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(12) $\displaystyle{\lim_{x \to 1} (2-x)^{\sec \frac{\pi x}{2}}}$

令 $t = 1-x$，则 $x \to 1$ 时 $t \to 0$，且 $2-x = 1+t$，$\displaystyle \sec \frac{\pi x}{2} = \frac{1}{\cos(\frac{\pi}{2}(1-t))} = \frac{1}{\sin(\frac{\pi t}{2})}$，
$$
\lim_{x \to 1} (2-x)^{\sec \frac{\pi x}{2}} = \lim_{t \to 0} (1+t)^{1/\sin(\pi t/