value of `lim_(x->0)(1-cos^3x)/(xsinx*cosx)` is

Options

A. `2//5`
B. `3//5`
C. `3//2`
D. `3//4`

Answer

To find the limit

\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{x \sin x \cos x}, \]

we start by substituting \(x = 0\):

\[ \frac{1 - \cos^3(0)}{0 \cdot \sin(0) \cdot \cos(0)} = \frac{1 - 1}{0} = \frac{0}{0}. \]

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule or manipulate the expression.

### Step 1: Use the identity for \(1 - \cos^3 x\)

We can use the identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) with \(a = 1\) and \(b = \cos x\):

\[ 1 - \cos^3 x = (1 - \cos x)(1 + \cos^2 x + \cos x). \]

### Step 2: Substitute into the limit

Now substitute this back into the limit:

\[ \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos^2 x + \cos x)}{x \sin x \cos x}. \]

### Step 3: Split the limit

We can split the limit into two parts:

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \lim_{x \to 0} \frac{1 + \cos^2 x + \cos x}{\sin x \cos x}. \]

### Step 4: Evaluate the first limit

Using the known limit:

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \]

### Step 5: Evaluate the second limit

For the second limit, we know:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{and} \quad \cos(0) = 1. \]

Thus,

\[ \lim_{x \to 0} \frac{1 + \cos^2 x + \cos x}{\sin x \cos x} = \frac{1 + 1 + 1}{1 \cdot 1} = 3. \]

### Step 6: Combine the results

Now we can combine the results of both limits:

\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{x \sin x \cos x} = \frac{1}{2} \cdot 3 = \frac{3}{2}. \]

Thus, the final answer is:

\[ \boxed{\frac{3}{2}}. \]