The value of `lim_(xto2) (2^(x)+2^(3-x)-6)/(sqrt(2^(-x))-2^(1-x))" is "`

Options

A. `|2x|gtsqrt(3)`
B. `|2x|ltsqrt(3)`
C. `|2x|gesqrt(3)`
D. `|2x|lesqrt(3)`

Explanation

To solve the limit \( \lim_{x \to 2} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} \), we will follow these steps:

### Step 1: Substitute \( x = 2 \) into the expression
First, we substitute \( x = 2 \) into both the numerator and the denominator to check if we get an indeterminate form.

**Numerator:**
\[ 2^2 + 2^{3-2} - 6 = 4 + 2 - 6 = 0 \]

**Denominator:**
\[ \sqrt{2^{-2}} - 2^{1-2} = \sqrt{\frac{1}{4}} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0 \]

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

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists.

### Step 3: Differentiate the numerator and denominator
**Numerator:**
\[ f(x) = 2^x + 2^{3-x} - 6 \]
Differentiating:
\[ f'(x) = 2^x \ln(2) - 2^{3-x} \ln(2) = \ln(2) (2^x - 2^{3-x}) \]

**Denominator:**
\[ g(x) = \sqrt{2^{-x}} - 2^{1-x} \]
Differentiating:
\[ g'(x) = \frac{d}{dx}(2^{-x})^{1/2} - \frac{d}{dx}(2^{1-x}) = \frac{1}{2} (2^{-x})^{-1/2} (-\ln(2) 2^{-x}) - (-\ln(2) 2^{1-x}) \]
This simplifies to:
\[ g'(x) = -\frac{\ln(2)}{2} \cdot \frac{1}{\sqrt{2^{-x}}} \cdot 2^{-x} + \ln(2) \cdot 2^{1-x} \]
\[ = -\frac{\ln(2)}{2} \cdot \frac{1}{\sqrt{2^{-x}}} \cdot 2^{-x} + \ln(2) \cdot 2^{1-x} \]

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the derivatives.

**Numerator:**
\[ f'(2) = \ln(2)(2^2 - 2^{3-2}) = \ln(2)(4 - 2) = 2\ln(2) \]

**Denominator:**
\[ g'(2) = -\frac{\ln(2)}{2} \cdot \frac{1}{\sqrt{2^{-2}}} \cdot 2^{-2} + \ln(2) \cdot 2^{1-2} \]
\[ = -\frac{\ln(2)}{2} \cdot \frac{1}{\frac{1}{2}} \cdot \frac{1}{4} + \ln(2) \cdot \frac{1}{2} \]
\[ = -\frac{\ln(2)}{2} \cdot 2 \cdot \frac{1}{4} + \frac{\ln(2)}{2} = -\frac{\ln(2)}{4} + \frac{\ln(2)}{2} = \frac{\ln(2)}{4} \]

### Step 5: Calculate the limit
Now we can calculate the limit:
\[ \lim_{x \to 2} \frac{f'(x)}{g'(x)} = \frac{2\ln(2)}{\frac{\ln(2)}{4}} = 2\ln(2) \cdot \frac{4}{\ln(2)} = 8 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{8} \]