Evalaute `lim_(xto0) (x2^(x)-x)/(1-cosx)`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{x2^x - x}{1 - \cos x} \]

we will follow these steps:

### Step 1: Identify the Form
First, we substitute \(x = 0\) directly into the expression:

\[ \frac{0 \cdot 2^0 - 0}{1 - \cos(0)} = \frac{0 - 0}{1 - 1} = \frac{0}{0} \]

This is an indeterminate form, 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:**
The numerator is \(x2^x - x\). We differentiate it:

\[ \frac{d}{dx}(x2^x - x) = \frac{d}{dx}(x2^x) - \frac{d}{dx}(x) \]

Using the product rule for the first term:

\[ \frac{d}{dx}(x2^x) = 2^x + x \cdot 2^x \ln(2) \]

So, the derivative of the numerator becomes:

\[ 2^x + x \cdot 2^x \ln(2) - 1 \]

**Denominator:**
The denominator is \(1 - \cos x\). We differentiate it:

\[ \frac{d}{dx}(1 - \cos x) = \sin x \]

### Step 3: Rewrite the Limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{2^x + x \cdot 2^x \ln(2) - 1}{\sin x} \]

### Step 4: Substitute \(x = 0\) Again
Now we substitute \(x = 0\):

\[ \frac{2^0 + 0 \cdot 2^0 \ln(2) - 1}{\sin(0)} = \frac{1 + 0 - 1}{0} = \frac{0}{0} \]

Again, we have an indeterminate form, so we apply L'Hôpital's Rule again.

### Step 5: Differentiate Again
We differentiate the numerator and denominator again.

**Numerator:**
Differentiating \(2^x + x \cdot 2^x \ln(2) - 1\):

\[ \frac{d}{dx}(2^x) + \frac{d}{dx}(x \cdot 2^x \ln(2)) = 2^x \ln(2) + (2^x \ln(2) + x \cdot 2^x \ln^2(2)) \]

This simplifies to:

\[ 2^x \ln(2) + 2^x \ln(2) + x \cdot 2^x \ln^2(2) = 2 \cdot 2^x \ln(2) + x \cdot 2^x \ln^2(2) \]

**Denominator:**
Differentiating \(\sin x\):

\[ \frac{d}{dx}(\sin x) = \cos x \]

### Step 6: Rewrite the Limit Again
Now we have:

\[ \lim_{x \to 0} \frac{2 \cdot 2^x \ln(2) + x \cdot 2^x \ln^2(2)}{\cos x} \]

### Step 7: Substitute \(x = 0\) Again
Substituting \(x = 0\):

\[ \frac{2 \cdot 2^0 \ln(2) + 0 \cdot 2^0 \ln^2(2)}{\cos(0)} = \frac{2 \ln(2)}{1} = 2 \ln(2) \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{x2^x - x}{1 - \cos x} = 2 \ln(2) \]