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求兩題的極限值

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B1 2024.12.02 18:26

To solve these limit problems, we can use L'Hôpital's Rule, which is applicable when the limit results in an indeterminate form like $\frac{0}{0}$ . (a) $lim_{x \to π} \frac{e^{\sin x} - 1}{x - π}$ Step 1: Check the form As $x \to π$ , $\sin x \to 0$ , so $e^{\sin x} \to 1$ . The expression becomes $\frac{0}{0}$ , which is indeterminate. Step 2: Apply L'Hôpital's Rule Differentiate the numerator and the denominator: - Numerator: $\frac{d}{d x} (e^{\sin x} - 1) = e^{\sin x} \cdot \cos x$ - Denominator: $\frac{d}{d x} (x - π) = 1$ Apply L'Hôpital's Rule: $lim_{x \to π} \frac{e^{\sin x} \cdot \cos x}{1} = e^{\sin π} \cdot \cos π = 1 \cdot (- 1) = - 1$ (b) $lim_{x \to 0^{+}} \frac{\int_{0}^{x} \sin (t^{2}) d t}{x \sin (x^{2})}$ Step 1: Check the form As $x \to 0^{+}$ , both the numerator and the denominator approach 0, resulting in the $\frac{0}{0}$ indeterminate form. Step 2: Apply L'Hôpital's Rule Differentiate the numerator and the denominator: - Numerator: By the Fundamental Theorem of Calculus, $\frac{d}{d x} (\int_{0}^{x} \sin (t^{2}) d t) = \sin (x^{2})$ - Denominator: $\frac{d}{d x} (x \sin (x^{2})) = \sin (x^{2}) + 2 x^{2} \cos (x^{2})$ Apply L'Hôpital's Rule: $lim_{x \to 0^{+}} \frac{\sin (x^{2})}{\sin (x^{2}) + 2 x^{2} \cos (x^{2})}$ As $x \to 0^{+}$ , $\sin (x^{2}) \to 0$ and $2 x^{2} \cos (x^{2}) \to 0$ , so the limit simplifies to: $lim_{x \to 0^{+}} \frac{\sin (x^{2})}{\sin (x^{2}) + 2 x^{2} \cos (x^{2})} = \frac{0}{0 + 0} = 0$ Thus, the limit is $0$ .

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匿名

To solve these limit problems, we can use L'Hôpital's Rule, which is applicable when the limit results in an indeterminate form like

\frac{0}{0}

. (a)

lim_{x \to π} \frac{e^{\sin x} - 1}{x - π}

Step 1: Check the form As

x \to π

,

\sin x \to 0

, so

e^{\sin x} \to 1

. The expression becomes

\frac{0}{0}

, which is indeterminate. Step 2: Apply L'Hôpital's Rule Differentiate the numerator and the denominator: - Numerator:

\frac{d}{d x} (e^{\sin x} - 1) = e^{\sin x} \cdot \cos x

- Denominator:

\frac{d}{d x} (x - π) = 1

Apply L'Hôpital's Rule:

lim_{x \to π} \frac{e^{\sin x} \cdot \cos x}{1} = e^{\sin π} \cdot \cos π = 1 \cdot (- 1) = - 1

(b)

lim_{x \to 0^{+}} \frac{\int_{0}^{x} \sin (t^{2}) d t}{x \sin (x^{2})}

Step 1: Check the form As

x \to 0^{+}

, both the numerator and the denominator approach 0, resulting in the

\frac{0}{0}

indeterminate form. Step 2: Apply L'Hôpital's Rule Differentiate the numerator and the denominator: - Numerator: By the Fundamental Theorem of Calculus,

\frac{d}{d x} (\int_{0}^{x} \sin (t^{2}) d t) = \sin (x^{2})

- Denominator:

\frac{d}{d x} (x \sin (x^{2})) = \sin (x^{2}) + 2 x^{2} \cos (x^{2})

Apply L'Hôpital's Rule:

lim_{x \to 0^{+}} \frac{\sin (x^{2})}{\sin (x^{2}) + 2 x^{2} \cos (x^{2})}

As

x \to 0^{+}

,

\sin (x^{2}) \to 0

and

2 x^{2} \cos (x^{2}) \to 0

, so the limit simplifies to:

lim_{x \to 0^{+}} \frac{\sin (x^{2})}{\sin (x^{2}) + 2 x^{2} \cos (x^{2})} = \frac{0}{0 + 0} = 0

Thus, the limit is

0

.

0

B1-1 (原 Po) 2024.12.02 18:27

B1你可以翻譯成中文嗎

匿名

B1你可以翻譯成中文嗎

0

B1-2 2024.12.03 15:12

B1-1小老師說的是兩題都用羅必達

0

B2 2024.12.03 15:15

但其實這兩題想考的是微分的定義 (a)是e^sin(x)在x=π的導數 (b)先上下同時-0 就會變成那坨積分在x=0的導數利用微積分基本定理就可以求出答案羅必達是很好用但很多教授不給用如果你們教授沒教羅必達或不允許學生用羅必達的話可以參考一下我的解法都是觀念而已幾乎不需要計算

匿名

但其實這兩題想考的是微分的定義 (a)是e^sin(x)在x=π的導數 (b)先上下同時-0 就會變成那坨積分在x=0的導數利用微積分基本定理就可以求出答案羅必達是很好用但很多教授不給用如果你們教授沒教羅必達或不允許學生用羅必達的話可以參考一下我的解法都是觀念而已幾乎不需要計算

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數學問題請教 課業板 2024年12月2日 18:26

數學問題請教
課業板 2024年12月2日 18:26