The value of `lim_(xrarr0)(1-cos^(3)x)/(xsinxcosx)`

Options

A. `(3)/(2)`
B. 1
C. 0
D. 2

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{1 - \cos^3 x}{x \sin x \cos x} \), we will follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{x \sin x \cos x} \]
We can use the identity \( \sin x \) in the denominator. We know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{x^2 \cos x} \cdot \frac{\sin x}{x} \]

### Step 2: Simplify the Expression
Now we have:
\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{x^2 \cos x} \]
This limit evaluates to \( \frac{0}{0} \) as \( x \to 0 \) because \( 1 - \cos^3(0) = 0 \) and \( x^2 \cos(0) = 0 \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:
- The derivative of the numerator \( 1 - \cos^3 x \) is:
\[ \frac{d}{dx}(1 - \cos^3 x) = 3 \cos^2 x \sin x \]
- The derivative of the denominator \( x^2 \cos x \) is:
\[ \frac{d}{dx}(x^2 \cos x) = 2x \cos x - x^2 \sin x \]

### Step 4: Substitute Back into the Limit
Now we substitute these derivatives back into the limit:
\[ \lim_{x \to 0} \frac{3 \cos^2 x \sin x}{2x \cos x - x^2 \sin x} \]

### Step 5: Evaluate the Limit
Now we evaluate the limit as \( x \to 0 \):
- The numerator becomes \( 3 \cdot 1^2 \cdot 0 = 0 \).
- The denominator becomes \( 2 \cdot 0 \cdot 1 - 0^2 \cdot 0 = 0 \).

Since we still have \( \frac{0}{0} \), we apply L'Hôpital's Rule again.

### Step 6: Differentiate Again
Differentiate the numerator and denominator again:
- The derivative of \( 3 \cos^2 x \sin x \) is:
\[ 3(2 \cos x (-\sin x) \sin x + \cos^2 x \cos x) = 3(-2 \cos x \sin^2 x + \cos^3 x) \]
- The derivative of \( 2x \cos x - x^2 \sin x \) is:
\[ 2 \cos x - 2x \sin x - (x^2 \cos x + 2x \sin x) \]

### Step 7: Substitute and Simplify
Now we substitute back into the limit and evaluate again. After simplification and substituting \( x = 0 \), we find:
\[ \lim_{x \to 0} \frac{3 \cdot 1^2 \cdot 0}{2 \cdot 0 \cdot 1 - 0^2 \cdot 0} = \frac{3}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{3}{2} \]