`"lt"_(x to 0)(64^(x) -32^(x) - 16^(x) + 4^(x) + 2^(x) -1)/([sqrt(15 + cos x) - 4]sin x)=`

Options

A. `-96 (log 2)^(2)`
B. `48 (log 2)^(2)`
C. `(2 log 2)^(3)`
D. none

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{64^x - 32^x - 16^x + 4^x + 2^x - 1}{\left(\sqrt{15 + \cos x} - 4\right) \sin x} \]

we will follow these steps:

### Step 1: Rewrite the Exponential Terms
We can express the exponential terms using the property \(a^x = e^{x \ln a}\):

\[ 64^x = e^{x \ln 64}, \quad 32^x = e^{x \ln 32}, \quad 16^x = e^{x \ln 16}, \quad 4^x = e^{x \ln 4}, \quad 2^x = e^{x \ln 2} \]

### Step 2: Simplify the Numerator
As \(x \to 0\), we can use the Taylor expansion \(e^u \approx 1 + u\) for small \(u\):

\[ 64^x \approx 1 + x \ln 64, \quad 32^x \approx 1 + x \ln 32, \quad 16^x \approx 1 + x \ln 16, \quad 4^x \approx 1 + x \ln 4, \quad 2^x \approx 1 + x \ln 2 \]

Substituting these approximations into the numerator:

\[ \begin{align*} 64^x - 32^x - 16^x + 4^x + 2^x - 1 &\approx (1 + x \ln 64) - (1 + x \ln 32) - (1 + x \ln 16) + (1 + x \ln 4) + (1 + x \ln 2) - 1 \\ &= x \ln 64 - x \ln 32 - x \ln 16 + x \ln 4 + x \ln 2 \\ &= x (\ln 64 - \ln 32 - \ln 16 + \ln 4 + \ln 2) \end{align*} \]

### Step 3: Combine the Logarithms
Using the properties of logarithms:

\[ \ln 64 - \ln 32 - \ln 16 + \ln 4 + \ln 2 = \ln \left(\frac{64 \cdot 4 \cdot 2}{32 \cdot 16}\right) = \ln \left(\frac{512}{512}\right) = \ln(1) = 0 \]

### Step 4: Evaluate the Denominator
Next, we simplify the denominator:

\[ \sqrt{15 + \cos x} - 4 \]

Using the Taylor expansion for \(\cos x\):

\[ \cos x \approx 1 - \frac{x^2}{2} \]

Thus,

\[ \sqrt{15 + \cos x} \approx \sqrt{15 + 1 - \frac{x^2}{2}} = \sqrt{16 - \frac{x^2}{2}} \approx 4 - \frac{x^2}{16} \]

So,

\[ \sqrt{15 + \cos x} - 4 \approx -\frac{x^2}{16} \]

### Step 5: Combine the Results
Now substituting back into the limit:

\[ \lim_{x \to 0} \frac{x (\ln 64 - \ln 32 - \ln 16 + \ln 4 + \ln 2)}{\left(-\frac{x^2}{16}\right) \sin x} \]

Since \(\sin x \approx x\) as \(x \to 0\):

\[ \lim_{x \to 0} \frac{x \cdot 0}{-\frac{x^2}{16} \cdot x} = \lim_{x \to 0} \frac{0}{-\frac{x^3}{16}} = 0 \]

### Final Answer
Thus, the limit evaluates to:

\[ \boxed{0} \]