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[Homework] 1.6

Posted by haifeng on 2020-10-12 12:12:27 last update 2020-10-12 15:18:00 | Edit | Answers (2)

P. 44--45

4. 求下列极限

(4) $lim_{x \to \frac{π}{2}} \frac{\cos x}{2 x - π}$

(5) $lim_{x \to π} \frac{\sin x}{x - π}$

(7) $lim_{x \to 0^{-}} \frac{x}{\sqrt{1 - \cos x}}$

5. 求下列极限

(4) $lim_{x \to 1} (3 - 2 x)^{\frac{3}{x - 1}}$

(6) $lim_{x \to \infty} (\frac{x + 2}{x - 2})^{x}$

(8) $lim_{x \to \infty} (\frac{n^{2} + 3}{n^{2} + 1})^{n^{2}}$

Posted by haifeng on 2020-10-12 15:14:59

4. 求下列极限

(4) $lim_{x \to \frac{π}{2}} \frac{\cos x}{2 x - π}$

解:

$原式 = lim_{x \to \frac{π}{2}} \frac{\sin (\frac{π}{2} - x)}{2 (x - \frac{π}{2})} = - \frac{1}{2}$

(5) $lim_{x \to π} \frac{\sin x}{x - π}$

解:

$原式 = lim_{x \to π} \frac{\sin (π - x)}{x - π} = - 1$

(7) $lim_{x \to 0^{-}} \frac{x}{\sqrt{1 - \cos x}}$

解:

$原式 = lim_{x \to 0^{-}} \frac{x}{\sqrt{2 \sin^{2} \frac{x}{2}}} = lim_{x \to 0^{-}} \frac{2 \cdot \frac{x}{2}}{\sqrt{2} \cdot \sin \frac{| x |}{2}} = \sqrt{2} \cdot lim_{x \to 0^{-}} \frac{- \frac{| x |}{2}}{\sin \frac{| x |}{2}} = - \sqrt{2}$

Posted by haifeng on 2020-10-12 20:17:34

5. 求下列极限

(4) $lim_{x \to 1} (3 - 2 x)^{\frac{3}{x - 1}}$

解:

$\begin{aligned} 原式 & = lim_{x \to 1} (1 + 2 (1 - x))^{\frac{3}{x - 1}} \\ = lim_{x \to 1} [(1 + 2 (1 - x))^{\frac{1}{2 (1 - x)}}]^{- 6} \\ = [lim_{x \to 1} (1 + 2 (1 - x))^{\frac{1}{2 (1 - x)}}]^{- 6} \\ = e^{- 6} \end{aligned}$

(6) $lim_{x \to \infty} (\frac{x + 2}{x - 2})^{x}$

解:

$\begin{aligned} 原式 & = lim_{x \to \infty} (1 + \frac{4}{x - 2})^{x} \\ = lim_{x \to \infty} {[(1 + \frac{4}{x - 2})^{\frac{x - 2}{4}}]^{4} \cdot (1 + \frac{4}{x - 2})^{2}} \\ = e^{4} \cdot 1 \\ = e^{4} \end{aligned}$

(8) $lim_{x \to \infty} (\frac{n^{2} + 3}{n^{2} + 1})^{n^{2}}$

解:

$\begin{aligned} 原式 & = lim_{x \to \infty} (1 + \frac{2}{n^{2} + 1})^{n^{2}} \\ = lim_{x \to \infty} {[(1 + \frac{2}{n^{2} + 1})^{\frac{n^{2} + 1}{2}}]^{2} \cdot (1 + \frac{2}{n^{2} + 1})^{- 1}} \\ = e^{2} \cdot 1 \\ = e^{2} \end{aligned}$

问题及解答

[Homework] 1.6