Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![Title: Calculating the Limit of a Function
Content:
In this example, we are tasked with calculating the limit of the following mathematical expression as \( x \) approaches 0:
\[ \lim_{{x \to 0}} \frac{2x + \sin(x)}{x(1-x)} \]
To solve this, we need to apply limit laws and potentially L'Hopital's Rule, since this limit involves a fraction where the numerator and denominator both approach 0 as \( x \) approaches 0.
Steps:
1. **Substitute x with 0 in the expression**: Initially substitute \( x = 0 \) to check the form of the limit:
\[ \frac{2(0) + \sin(0)}{0(1-0)} = \frac{0 + 0}{0 \cdot 1} = \frac{0}{0} \]
Since we obtain the indeterminate form \( \frac{0}{0} \), we need to apply further methods, such as L'Hopital's Rule. This rule states that for the limits of the indeterminate form \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \), we can take the derivatives of the numerator and the denominator:
2. **Differentiate numerator and denominator separately**:
- Numerator: \( \frac{d}{dx}(2x + \sin(x)) = 2 + \cos(x) \)
- Denominator: \( \frac{d}{dx}[x(1-x)] = \frac{d}{dx}[x - x^2] = 1 - 2x \)
3. **Apply L'Hopital's Rule**: Substitute these derivatives back into the limit expression:
\[ \lim_{{x \to 0}} \frac{2 + \cos(x)}{1 - 2x} \]
4. **Evaluate the limit**: Now substitute \( x = 0 \) to compute the limit:
\[ \frac{2 + \cos(0)}{1 - 2(0)} = \frac{2 + 1}{1 - 0} = \frac{3}{1} = 3 \]
Therefore, the limit of the given function as \( x \) approaches 0 is](https://content.bartleby.com/qna-images/question/aaa612e4-bb6a-4a7c-9bcb-a4980796bdd1/8225c217-98bf-451b-a0c1-09ed19b322e4/3y1tap_reoriented.jpeg)
Transcribed Image Text:Title: Calculating the Limit of a Function
Content:
In this example, we are tasked with calculating the limit of the following mathematical expression as \( x \) approaches 0:
\[ \lim_{{x \to 0}} \frac{2x + \sin(x)}{x(1-x)} \]
To solve this, we need to apply limit laws and potentially L'Hopital's Rule, since this limit involves a fraction where the numerator and denominator both approach 0 as \( x \) approaches 0.
Steps:
1. **Substitute x with 0 in the expression**: Initially substitute \( x = 0 \) to check the form of the limit:
\[ \frac{2(0) + \sin(0)}{0(1-0)} = \frac{0 + 0}{0 \cdot 1} = \frac{0}{0} \]
Since we obtain the indeterminate form \( \frac{0}{0} \), we need to apply further methods, such as L'Hopital's Rule. This rule states that for the limits of the indeterminate form \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \), we can take the derivatives of the numerator and the denominator:
2. **Differentiate numerator and denominator separately**:
- Numerator: \( \frac{d}{dx}(2x + \sin(x)) = 2 + \cos(x) \)
- Denominator: \( \frac{d}{dx}[x(1-x)] = \frac{d}{dx}[x - x^2] = 1 - 2x \)
3. **Apply L'Hopital's Rule**: Substitute these derivatives back into the limit expression:
\[ \lim_{{x \to 0}} \frac{2 + \cos(x)}{1 - 2x} \]
4. **Evaluate the limit**: Now substitute \( x = 0 \) to compute the limit:
\[ \frac{2 + \cos(0)}{1 - 2(0)} = \frac{2 + 1}{1 - 0} = \frac{3}{1} = 3 \]
Therefore, the limit of the given function as \( x \) approaches 0 is
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