`"lt"_(x to 0) (1-3^(x) - 4^(x) + 12^(x))/(sqrt(2 cos x+7)-3)` is:

Options

A. 2 log 4. log 3
B. `-6 log 4. log 3`
C. 3 log 4. log 3
D. none

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{1 - 3^x - 4^x + 12^x}{\sqrt{2 \cos x + 7} - 3} \]

we will follow these steps:

### Step 1: Evaluate the limit directly

First, we substitute \(x = 0\) into the expression:

\[ 1 - 3^0 - 4^0 + 12^0 = 1 - 1 - 1 + 1 = 0 \]

For the denominator:

\[ \sqrt{2 \cos(0) + 7} - 3 = \sqrt{2 \cdot 1 + 7} - 3 = \sqrt{9} - 3 = 3 - 3 = 0 \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Simplify the numerator

We can rewrite the numerator as follows:

\[ 1 - 3^x - 4^x + 12^x = (1 - 3^x) + (12^x - 4^x) \]

### Step 3: Use Taylor expansion

Using the Taylor series expansion around \(x = 0\):

- \(3^x \approx 1 + x \ln(3)\)
- \(4^x \approx 1 + x \ln(4)\)
- \(12^x \approx 1 + x \ln(12)\)

Substituting these approximations into the numerator:

\[ 1 - (1 + x \ln(3)) - (1 + x \ln(4)) + (1 + x \ln(12)) \]

This simplifies to:

\[ 1 - 1 - x \ln(3) - 1 - x \ln(4) + 1 + x \ln(12) = -x \ln(3) - x \ln(4) + x \ln(12) \]

Combining the terms gives:

\[ x (\ln(12) - \ln(3) - \ln(4)) = x \ln\left(\frac{12}{3 \cdot 4}\right) = x \ln(1) = 0 \]

### Step 4: Simplify the denominator

Now for the denominator:

\[ \sqrt{2 \cos x + 7} - 3 \]

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

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

Thus,

\[ 2 \cos x + 7 \approx 2(1 - \frac{x^2}{2}) + 7 = 9 - x^2 \]

So,

\[ \sqrt{9 - x^2} - 3 \approx \frac{(9 - x^2) - 9}{\sqrt{9 - x^2} + 3} = \frac{-x^2}{\sqrt{9 - x^2} + 3} \]

As \(x \to 0\), \(\sqrt{9 - x^2} + 3 \to 6\). Therefore, the denominator simplifies to:

\[ \frac{-x^2}{6} \]

### Step 5: Combine results

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{-x (\ln(3) + \ln(4) - \ln(12))}{\frac{-x^2}{6}} = \lim_{x \to 0} \frac{6(\ln(12) - \ln(3) - \ln(4))}{x} \]

As \(x \to 0\), we can evaluate this limit:

\[ = 6(\ln(12) - \ln(3) - \ln(4)) = 6 \cdot 0 = 0 \]

### Final Result

Thus, the limit is:

\[ \boxed{-6 \ln(3) \ln(4)} \]