The value of `lim_(xrarr0)(1-cos^(3)(sinx))/(sinxsin(sinx)cos(sinx))`

Options

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

Answer

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

### Step 1: Substitute \( \sin x \) with \( t \)
Let \( t = \sin x \). As \( x \to 0 \), \( t \to 0 \) as well. The limit can be rewritten as:
\[ \lim_{t \to 0} \frac{1 - \cos^3(t)}{t \sin(t) \cos(t)} \]

### Step 2: Use the identity for \( 1 - \cos^3(t) \)
Recall the identity for the difference of cubes:
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
Here, let \( a = 1 \) and \( b = \cos(t) \):
\[ 1 - \cos^3(t) = (1 - \cos(t))(1 + \cos^2(t) + \cos(t)) \]
Thus, we can rewrite the limit as:
\[ \lim_{t \to 0} \frac{(1 - \cos(t))(1 + \cos^2(t) + \cos(t))}{t \sin(t) \cos(t)} \]

### Step 3: Simplify the expression
Now, we know that \( 1 - \cos(t) \) can be approximated using the Taylor series expansion:
\[ 1 - \cos(t) \approx \frac{t^2}{2} \quad \text{as } t \to 0 \]
Substituting this into our limit gives:
\[ \lim_{t \to 0} \frac{\frac{t^2}{2}(1 + \cos^2(t) + \cos(t))}{t \sin(t) \cos(t)} \]

### Step 4: Substitute \( \sin(t) \) and simplify
Using the approximation \( \sin(t) \approx t \) as \( t \to 0 \), we have:
\[ \lim_{t \to 0} \frac{\frac{t^2}{2}(1 + \cos^2(t) + \cos(t))}{t \cdot t \cdot \cos(t)} = \lim_{t \to 0} \frac{\frac{t^2}{2}(1 + \cos^2(t) + \cos(t))}{t^2 \cos(t)} \]
This simplifies to:
\[ \lim_{t \to 0} \frac{1 + \cos^2(t) + \cos(t)}{2 \cos(t)} \]

### Step 5: Evaluate the limit
As \( t \to 0 \), \( \cos(t) \to 1 \):
\[ \lim_{t \to 0} \frac{1 + 1 + 1}{2 \cdot 1} = \frac{3}{2} \]

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