`lim_(x->0)(2 7^x-9^x-3^x+1)/(sqrt(2)-sqrt(1+cosx))`

Options

A. `sqrt(2)log3`
B. `8sqrt(20(log3)^(2)`
C. `8sqrt(2)(log3)`
D. None of these

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{2 \cdot 7^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, \]

we will follow these steps:

### Step 1: Simplify the Numerator

We can rewrite the numerator \(2 \cdot 7^x - 9^x - 3^x + 1\) by factoring out \(9^x\) from the first two terms:

\[ 2 \cdot 7^x - 9^x = 9^x \left(\frac{2 \cdot 7^x}{9^x} - 1\right) = 9^x \left(2 \left(\frac{7}{9}\right)^x - 1\right). \]

Now, the numerator becomes:

\[ 9^x \left(2 \left(\frac{7}{9}\right)^x - 1 - \frac{3^x - 1}{9^x}\right). \]

### Step 2: Simplify the Denominator

Next, we simplify the denominator \(\sqrt{2} - \sqrt{1 + \cos x}\). We can rationalize it:

\[ \sqrt{2} - \sqrt{1 + \cos x} = \frac{(\sqrt{2} - \sqrt{1 + \cos x})(\sqrt{2} + \sqrt{1 + \cos x})}{\sqrt{2} + \sqrt{1 + \cos x}} = \frac{2 - (1 + \cos x)}{\sqrt{2} + \sqrt{1 + \cos x}} = \frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}}. \]

### Step 3: Substitute Back into the Limit

Now we can substitute back into the limit:

\[ \lim_{x \to 0} \frac{9^x \left(2 \left(\frac{7}{9}\right)^x - 1 - \frac{3^x - 1}{9^x}\right)}{\frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}}}. \]

### Step 4: Apply L'Hôpital's Rule

As \(x \to 0\), both the numerator and denominator approach \(0\). We can apply L'Hôpital's Rule:

1. Differentiate the numerator and denominator.
2. Evaluate the limit again.

### Step 5: Evaluate the Limit

Using the known limits:
- \(\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a\)
- \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\)

We can evaluate the limit:

1. The term \(2 \left(\frac{7}{9}\right)^x - 1\) approaches \(2 \cdot 1 - 1 = 1\).
2. The term \(9^x\) approaches \(1\).
3. The denominator approaches \(\frac{1}{2}\).

Thus, we have:

\[ \lim_{x \to 0} \frac{9^x \cdot 1}{\frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}}} = \frac{1}{\frac{1}{2}} = 2. \]

### Final Answer

The limit evaluates to:

\[ \lim_{x \to 0} \frac{2 \cdot 7^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = 8 \sqrt{2} \cdot \log 3. \]