Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Sure! Let's evaluate the limit as \( x \) approaches 0 for the expression \(\frac{3x \sin(4x)}{e^{x^2} - \cos(2x)}\). Here is a step-by-step explanation:
1. Identify the function:
We need to find the limit of the function \( \frac{3x \sin(4x)}{e^{x^2} - \cos(2x)} \) as \( x \) approaches 0.
2. Substitute \( x \) with 0 and analyze:
Substitute \( x = 0 \) directly to see if it yields the indeterminate form \(\frac{0}{0}\):
- Numerator: \( 3 \cdot 0 \cdot \sin(4 \cdot 0) = 0 \)
- Denominator: \( e^{0^2} - \cos(2 \cdot 0) = 1 - 1 = 0 \)
Since both the numerator and the denominator evaluate to 0, we indeed have an indeterminate form \(\frac{0}{0}\).
3. Apply L'Hôpital's Rule:
Given that the limit is in the indeterminate form, we can apply L'Hôpital's Rule, which states that:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is in the indeterminate form.
4. Compute the derivatives of the numerator and the denominator:
First, find the derivative of the numerator \( 3x \sin(4x) \):
[tex]\[ \frac{d}{dx} \left(3x \sin(4x)\right) = 3 \left(\sin(4x) + 4x \cos(4x)\right) = 3 \sin(4x) + 12x \cos(4x) \][/tex]
Next, find the derivative of the denominator \( e^{x^2} - \cos(2x) \):
[tex]\[ \frac{d}{dx} \left( e^{x^2} - \cos(2x) \right) = 2x e^{x^2} + 2\sin(2x) \][/tex]
5. Form the new limit with derivatives:
By applying L'Hôpital's Rule, the limit becomes:
[tex]\[ \lim_{x \to 0} \frac{3 \sin(4x) + 12x \cos(4x)}{2x e^{x^2} + 2 \sin(2x)} \][/tex]
6. Substitute \( x = 0 \) in the new limit:
Evaluate the new expression by substituting \( x = 0 \):
- Numerator: \( 3 \sin(4 \cdot 0) + 12 \cdot 0 \cdot \cos(4 \cdot 0) = 3 \cdot 0 + 12 \cdot 0 = 0 \)
- Denominator: \( 2 \cdot 0 \cdot e^{0^2} + 2 \cdot \sin(2 \cdot 0) = 0 + 2 \cdot 0 = 0 \)
The new fraction still evaluates to \( \frac{0}{0} \), so L'Hôpital's Rule needs to be applied again.
7. Apply L’Hôpital’s Rule again:
Differentiate the numerator and denominator again:
- Second derivative of the numerator \( 3 \sin(4x) + 12x \cos(4x) \):
[tex]\[ \frac{d}{dx}(3 \sin(4x) + 12x \cos(4x)) = 12 \cos(4x) - 48x \sin(4x) + 12 \cos(4x) = 24 \cos(4x) - 48x \sin(4x) \][/tex]
- Second derivative of the denominator \( 2x e^{x^2} + 2 \sin(2x) \):
[tex]\[ \frac{d}{dx}(2x e^{x^2} + 2 \sin(2x)) = 2e^{x^2} + 4x^2 e^{x^2} + 4 \cos(2x) \][/tex]
8. Form the new limit with second derivatives:
The limit becomes:
[tex]\[ \lim_{x \to 0} \frac{24 \cos(4x) - 48x \sin(4x)}{2 e^{x^2} + 4x^2 e^{x^2} + 4 \cos(2x)} \][/tex]
9. Substitute \( x = 0 \) in the new limit:
Evaluate the limit by directly substituting \( x = 0 \):
- Numerator: \( 24 \cos(4 \cdot 0) - 48 \cdot 0 \cdot \sin(4 \cdot 0) = 24 \)
- Denominator: \( 2 e^{0^2} + 4 \cdot 0^2 \cdot e^{0^2} + 4 \cos(2 \cdot 0) = 2 + 0 + 4 = 6 \)
10. Calculate the final limit:
[tex]\[ \lim_{x \to 0} \frac{24}{6} = 4 \][/tex]
Therefore, the limit is [tex]\(\boxed{4}\)[/tex].
1. Identify the function:
We need to find the limit of the function \( \frac{3x \sin(4x)}{e^{x^2} - \cos(2x)} \) as \( x \) approaches 0.
2. Substitute \( x \) with 0 and analyze:
Substitute \( x = 0 \) directly to see if it yields the indeterminate form \(\frac{0}{0}\):
- Numerator: \( 3 \cdot 0 \cdot \sin(4 \cdot 0) = 0 \)
- Denominator: \( e^{0^2} - \cos(2 \cdot 0) = 1 - 1 = 0 \)
Since both the numerator and the denominator evaluate to 0, we indeed have an indeterminate form \(\frac{0}{0}\).
3. Apply L'Hôpital's Rule:
Given that the limit is in the indeterminate form, we can apply L'Hôpital's Rule, which states that:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is in the indeterminate form.
4. Compute the derivatives of the numerator and the denominator:
First, find the derivative of the numerator \( 3x \sin(4x) \):
[tex]\[ \frac{d}{dx} \left(3x \sin(4x)\right) = 3 \left(\sin(4x) + 4x \cos(4x)\right) = 3 \sin(4x) + 12x \cos(4x) \][/tex]
Next, find the derivative of the denominator \( e^{x^2} - \cos(2x) \):
[tex]\[ \frac{d}{dx} \left( e^{x^2} - \cos(2x) \right) = 2x e^{x^2} + 2\sin(2x) \][/tex]
5. Form the new limit with derivatives:
By applying L'Hôpital's Rule, the limit becomes:
[tex]\[ \lim_{x \to 0} \frac{3 \sin(4x) + 12x \cos(4x)}{2x e^{x^2} + 2 \sin(2x)} \][/tex]
6. Substitute \( x = 0 \) in the new limit:
Evaluate the new expression by substituting \( x = 0 \):
- Numerator: \( 3 \sin(4 \cdot 0) + 12 \cdot 0 \cdot \cos(4 \cdot 0) = 3 \cdot 0 + 12 \cdot 0 = 0 \)
- Denominator: \( 2 \cdot 0 \cdot e^{0^2} + 2 \cdot \sin(2 \cdot 0) = 0 + 2 \cdot 0 = 0 \)
The new fraction still evaluates to \( \frac{0}{0} \), so L'Hôpital's Rule needs to be applied again.
7. Apply L’Hôpital’s Rule again:
Differentiate the numerator and denominator again:
- Second derivative of the numerator \( 3 \sin(4x) + 12x \cos(4x) \):
[tex]\[ \frac{d}{dx}(3 \sin(4x) + 12x \cos(4x)) = 12 \cos(4x) - 48x \sin(4x) + 12 \cos(4x) = 24 \cos(4x) - 48x \sin(4x) \][/tex]
- Second derivative of the denominator \( 2x e^{x^2} + 2 \sin(2x) \):
[tex]\[ \frac{d}{dx}(2x e^{x^2} + 2 \sin(2x)) = 2e^{x^2} + 4x^2 e^{x^2} + 4 \cos(2x) \][/tex]
8. Form the new limit with second derivatives:
The limit becomes:
[tex]\[ \lim_{x \to 0} \frac{24 \cos(4x) - 48x \sin(4x)}{2 e^{x^2} + 4x^2 e^{x^2} + 4 \cos(2x)} \][/tex]
9. Substitute \( x = 0 \) in the new limit:
Evaluate the limit by directly substituting \( x = 0 \):
- Numerator: \( 24 \cos(4 \cdot 0) - 48 \cdot 0 \cdot \sin(4 \cdot 0) = 24 \)
- Denominator: \( 2 e^{0^2} + 4 \cdot 0^2 \cdot e^{0^2} + 4 \cos(2 \cdot 0) = 2 + 0 + 4 = 6 \)
10. Calculate the final limit:
[tex]\[ \lim_{x \to 0} \frac{24}{6} = 4 \][/tex]
Therefore, the limit is [tex]\(\boxed{4}\)[/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
