`"lt"_(x to 0) (8^(x) -4^(x) -2^(x)+1)/(sqrt(2)- sqrt(1 + cos x))`=

Options

A. `4sqrt(2) log 2`
B. `8sqrt(2) log 2`
C. `8sqrt(2) (log 2)^(2)`
D. none

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, \]

we will follow these steps:

### Step 1: Analyze the limit
First, we substitute \(x = 0\) into the expression.

\[ 8^0 - 4^0 - 2^0 + 1 = 1 - 1 - 1 + 1 = 0, \]
and for the denominator:

\[ \sqrt{2} - \sqrt{1 + \cos(0)} = \sqrt{2} - \sqrt{2} = 0. \]

This gives us the indeterminate form \(\frac{0}{0}\).

### Step 2: Simplify the numerator
We can rewrite the numerator \(8^x - 4^x - 2^x + 1\) using the fact that \(8^x = (2^3)^x = (2^x)^3\), \(4^x = (2^2)^x = (2^x)^2\), and \(2^x = (2^x)^1\):

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

Let \(y = 2^x\). Then the expression becomes:

\[ y^3 - y^2 - y + 1. \]

### Step 3: Factor the numerator
We can factor \(y^3 - y^2 - y + 1\):

\[ y^3 - y^2 - y + 1 = (y - 1)(y^2 + 1). \]

Thus, we have:

\[ 8^x - 4^x - 2^x + 1 = (2^x - 1)(2^{2x} + 1). \]

### Step 4: Rewrite the limit
Now, substituting back into the limit gives:

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

### Step 5: Simplify the denominator
For the denominator, we can use the identity \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\):

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

Using the identity \(1 - \cos\left(\frac{x}{2}\right) = 2\sin^2\left(\frac{x}{4}\right)\):

\[ \sqrt{2}(1 - \cos\left(\frac{x}{2}\right)) = \sqrt{2}(2\sin^2\left(\frac{x}{4}\right)). \]

### Step 6: Substitute back into the limit
Now we have:

\[ \lim_{x \to 0} \frac{(2^x - 1)(2^{2x} + 1)}{\sqrt{2} \cdot 2\sin^2\left(\frac{x}{4}\right)}. \]

### Step 7: Evaluate the limit
As \(x \to 0\), \(2^x - 1 \sim \ln(2) \cdot x\) and \(2^{2x} + 1 \to 2 + 1 = 3\). Thus:

\[ \lim_{x \to 0} \frac{(\ln(2) \cdot x)(3)}{\sqrt{2} \cdot 2\sin^2\left(\frac{x}{4}\right)}. \]

Using \(\sin\left(\frac{x}{4}\right) \sim \frac{x}{4}\):

\[ \sin^2\left(\frac{x}{4}\right) \sim \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}. \]

Thus, the limit becomes:

\[ \lim_{x \to 0} \frac{3\ln(2) \cdot x}{\sqrt{2} \cdot 2 \cdot \frac{x^2}{16}} = \lim_{x \to 0} \frac{3\ln(2) \cdot 16}{2\sqrt{2} \cdot x} = \frac{48\ln(2)}{2\sqrt{2}} = 24\frac{\ln(2)}{\sqrt{2}}. \]

### Final Answer
The limit evaluates to:

\[ \lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = 8\sqrt{2} \ln(2). \]