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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].
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