`"lt"_(x to 2)(2^(x) -x^(2))/(x^(x) -2^(x))`=

Options

A. `(log 2-1)/(log 2+1)`
B. `(log 2+1)/(log 2-1)`
C. 1
D. `-1`

Answer

To solve the limit

\[ \lim_{x \to 2} \frac{2^x - x^2}{x^x - 2^x} \]

we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \(x = 2\) directly into the expression:

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

This is an indeterminate form \(0/0\), so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

**Numerator:**
\[ \frac{d}{dx}(2^x - x^2) = 2^x \ln(2) - 2x \]

**Denominator:**
\[ \frac{d}{dx}(x^x - 2^x) = x^x(1 + \ln(x)) - 2^x \ln(2) \]

### Step 3: Rewrite the limit with derivatives
Now we rewrite the limit using the derivatives we found:

\[ \lim_{x \to 2} \frac{2^x \ln(2) - 2x}{x^x(1 + \ln(x)) - 2^x \ln(2)} \]

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

**Numerator:**
\[ 2^2 \ln(2) - 2 \cdot 2 = 4 \ln(2) - 4 \]

**Denominator:**
\[ 2^2(1 + \ln(2)) - 2^2 \ln(2) = 4(1 + \ln(2)) - 4 \ln(2) = 4 \]

So, we have:

\[ \lim_{x \to 2} \frac{4 \ln(2) - 4}{4} \]

### Step 5: Simplify the expression
This simplifies to:

\[ \frac{4(\ln(2) - 1)}{4} = \ln(2) - 1 \]

### Final Answer
Thus, the limit is:

\[ \ln(2) - 1 \]