Evaluate the following limits :
`Lim_(x to 0) (x(2^(x)-1))/(1-cos x)`

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Direct Substitution
First, we try direct substitution by plugging \( x = 0 \) into the expression:

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

This gives us an indeterminate form \( \frac{0}{0} \), 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:**
The numerator is \( x(2^x - 1) \). We will use the product rule for differentiation:

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

So, the derivative of the numerator is:

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

**Denominator:**
The denominator is \( 1 - \cos x \). The derivative is:

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

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

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

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

- For the numerator:
\[ 2^0 - 1 + 0 \cdot 2^0 \ln(2) = 1 - 1 + 0 = 0 \]

- For the denominator:
\[ \sin(0) = 0 \]

We again have the indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 5: Differentiate Again
**Numerator:**
Now we differentiate \( 2^x - 1 + x \cdot 2^x \ln(2) \):

1. The derivative of \( 2^x \) is \( 2^x \ln(2) \).
2. The derivative of \( x \cdot 2^x \ln(2) \) using the product rule is:

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

Thus, the derivative of the numerator becomes:

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

**Denominator:**
The derivative of \( \sin x \) is \( \cos x \).

### 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 Again
Substituting \( x = 0 \):

- For the numerator:
\[ 2 \cdot 2^0 \ln(2) + 0 \cdot 2^0 (\ln(2))^2 = 2 \ln(2) \]

- For the denominator:
\[ \cos(0) = 1 \]

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

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

So, the final answer is:

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