`lim_(xrarr0)(((1-cos^(2)(3x))/(cos^(3)(4x)))((sin^(3)(4x))/(log_(e)(2x+1)^(5)))` is equal to ____.

Options

A. `18`
B. `15`
C. `9`
D. `24`

Answer

To solve the limit

\[ \lim_{x \to 0} \left( \frac{(1 - \cos^2(3x))}{\cos^3(4x)} \cdot \frac{\sin^3(4x)}{\log_e((2x + 1)^5)} \right), \]

we will break it down step by step.

### Step 1: Simplify \(1 - \cos^2(3x)\)

Using the identity \(1 - \cos^2(\theta) = \sin^2(\theta)\), we can rewrite \(1 - \cos^2(3x)\) as:

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

Thus, the limit becomes:

\[ \lim_{x \to 0} \left( \frac{\sin^2(3x)}{\cos^3(4x)} \cdot \frac{\sin^3(4x)}{\log_e((2x + 1)^5)} \right). \]

### Step 2: Analyze \(\sin^2(3x)\) and \(\sin^3(4x)\)

Using the small angle approximation \(\sin(kx) \approx kx\) as \(x \to 0\):

\[ \sin(3x) \approx 3x \quad \text{and} \quad \sin(4x) \approx 4x. \]

Thus,

\[ \sin^2(3x) \approx (3x)^2 = 9x^2, \]
\[ \sin^3(4x) \approx (4x)^3 = 64x^3. \]

### Step 3: Analyze \(\cos^3(4x)\)

Using the small angle approximation for cosine, \(\cos(kx) \approx 1\) as \(x \to 0\):

\[ \cos(4x) \approx 1 \quad \Rightarrow \quad \cos^3(4x) \approx 1. \]

### Step 4: Analyze \(\log_e((2x + 1)^5)\)

Using the logarithmic approximation \(\log_e(1 + u) \approx u\) for small \(u\):

\[ \log_e(2x + 1) \approx 2x \quad \Rightarrow \quad \log_e((2x + 1)^5) \approx 5 \cdot 2x = 10x. \]

### Step 5: Substitute back into the limit

Now substituting back into our limit:

\[ \lim_{x \to 0} \left( \frac{9x^2}{1} \cdot \frac{64x^3}{10x} \right). \]

This simplifies to:

\[ \lim_{x \to 0} \left( \frac{9 \cdot 64}{10} x^4 \right) = \frac{576}{10} x^4. \]

### Step 6: Evaluate the limit

As \(x \to 0\), \(x^4 \to 0\):

\[ \lim_{x \to 0} \frac{576}{10} x^4 = 0. \]

### Final Answer

The limit is:

\[ \boxed{0}. \]