To evaluate the limit [tex]\(\lim_{x \rightarrow 0} \frac{x \cos(x) - \sin(x)}{4 x^3}\)[/tex], we follow a step-by-step approach.
Step 1: Analyze the function
We start with the function:
[tex]\[ f(x) = \frac{x \cos(x) - \sin(x)}{4 x^3} \][/tex]
Step 2: Direct substitution
If we substitute [tex]\(x = 0\)[/tex] directly into the function, we get an indeterminate form [tex]\( \frac{0}{0} \)[/tex]. Hence, direct substitution doesn't help, and we need to apply L'Hospital's Rule.
Step 3: Applying L'Hospital's Rule
L'Hospital's Rule states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can compute the limit of the ratio of the derivatives:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \][/tex]
In our function:
[tex]\[ f(x) = x \cos(x) - \sin(x) \][/tex]
[tex]\[ g(x) = 4 x^3 \][/tex]
First, we find the derivatives of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (x \cos(x) - \sin(x)) \][/tex]
[tex]\[ g'(x) = \frac{d}{dx} (4 x^3) \][/tex]
Step 4: Calculating the derivatives
- Derivative of [tex]\(f(x)\)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (x \cos(x) - \sin(x)) \][/tex]
Using the product rule and chain rule:
[tex]\[ f'(x) = \frac{d}{dx} (x \cos(x)) - \frac{d}{dx} (\sin(x)) \][/tex]
[tex]\[ f'(x) = (\cos(x) - x \sin(x)) - \cos(x) \][/tex]
[tex]\[ f'(x) = \cos(x) - x \sin(x) - \cos(x) \][/tex]
[tex]\[ f'(x) = -x \sin(x) \][/tex]
- Derivative of [tex]\(g(x)\)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx} (4 x^3) \][/tex]
[tex]\[ g'(x) = 12 x^2 \][/tex]
Step 5: Applying L'Hospital's Rule
Now, apply L'Hospital's Rule:
[tex]\[ \lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{4 x^3} = \lim_{x \to 0} \frac{-x \sin(x)}{12 x^2} \][/tex]
Simplify the expression:
[tex]\[ \lim_{x \to 0} \frac{-x \sin(x)}{12 x^2} = \lim_{x \to 0} \frac{- \sin(x)}{12 x} \][/tex]
This again gives an indeterminate form [tex]\( \frac{0}{0} \)[/tex], so we apply L'Hospital's Rule a second time:
Step 6: Second application of L'Hospital's Rule
Taking derivatives again, we get:
[tex]\[ f''(x) = \frac{d}{dx} (- \sin(x)) = - \cos(x) \][/tex]
[tex]\[ g''(x) = \frac{d}{dx} (12 x) = 12 \][/tex]
So,
[tex]\[ \lim_{x \to 0} \frac{- \sin(x)}{12 x} = \lim_{x \to 0} \frac{- \cos(x)}{12} \][/tex]
Substituting [tex]\(x = 0\)[/tex]:
[tex]\[ \lim_{x \to 0} \frac{- \cos(x)}{12} = \frac{- \cos(0)}{12} = \frac{- 1}{12} \][/tex]
Therefore, the limit evaluates to:
[tex]\[ \lim_{x \rightarrow 0} \frac{x \cos(x) - \sin(x)}{4 x^3} = -\frac{1}{12} \][/tex]
Thus, the final answer is: [tex]\(\boxed{-\frac{1}{12}}\)[/tex].
