The limit `underset(x to 2)lim (2^(x)+2^(3-x)-6)/(sqrt2^(-x)-2^(1-x))` is equal to

Options

A. 1
B. 2
C. 4
D. 8

Answer

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 \( x = 2 \)

First, we substitute \( x = 2 \) into the expression:

\[ 2^2 + 2^{3-2} - 6 = 4 + 2 - 6 = 0, \]

and for the denominator:

\[ \sqrt{2^{-2}} - 2^{1-2} = \sqrt{\frac{1}{4}} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0. \]

This gives us the 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 exists.

### Step 3: Differentiate the numerator and denominator

1. **Differentiate the numerator**:

The numerator is \( f(x) = 2^x + 2^{3-x} - 6 \).

Using the chain rule:
- The derivative of \( 2^x \) is \( 2^x \ln(2) \).
- The derivative of \( 2^{3-x} \) is \( -2^{3-x} \ln(2) \).

Thus,

\[ f'(x) = 2^x \ln(2) - 2^{3-x} \ln(2). \]

2. **Differentiate the denominator**:

The denominator is \( g(x) = \sqrt{2^{-x}} - 2^{1-x} \).

Rewriting \( \sqrt{2^{-x}} \) as \( (2^{-x})^{1/2} = 2^{-x/2} \):

- The derivative of \( 2^{-x/2} \) is \( -\frac{1}{2} 2^{-x/2} \ln(2) \).
- The derivative of \( 2^{1-x} \) is \( -2^{1-x} \ln(2) \).

Thus,

\[ g'(x) = -\frac{1}{2} 2^{-x/2} \ln(2) + 2^{1-x} \ln(2). \]

### Step 4: Substitute \( x = 2 \) again

Now we substitute \( x = 2 \) into \( f'(x) \) and \( g'(x) \):

1. For \( f'(2) \):

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

2. For \( g'(2) \):

\[ g'(2) = -\frac{1}{2} 2^{-2/2} \ln(2) + 2^{1-2} \ln(2) = -\frac{1}{2} \cdot \frac{1}{2} \ln(2) + \frac{1}{2} \ln(2) = -\frac{1}{4} \ln(2) + \frac{1}{2} \ln(2) = \frac{1}{4} \ln(2). \]

### Step 5: Calculate the limit

Now we can calculate the limit:

\[ \lim_{x \to 2} \frac{f'(x)}{g'(x)} = \frac{2 \ln(2)}{\frac{1}{4} \ln(2)} = 2 \cdot 4 = 8. \]

### Final Answer

Thus, the limit is

\[ \boxed{8}. \]