Evaluate the following limits :
`lim_(x to 0)(x(2^(x)-1))/(1-cosx)`.

Answer

To evaluate the limit

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

we can use some known limits and properties of logarithms and trigonometric functions. Let's go through the steps:

### Step 1: Rewrite the limit
We can rewrite the limit as follows:

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

### Step 2: Apply known limits
We know from calculus that:

1. \(\lim_{x \to 0} \frac{2^x - 1}{x} = \log(2)\).
2. \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\).

### Step 3: Substitute the limits
Using these known limits, we can substitute them into our rewritten limit:

\[ \lim_{x \to 0} \frac{2^x - 1}{x} = \log(2), \]
and for the denominator:

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

### Step 4: Rewrite the limit using the known results
Now, we can express our limit as:

\[ \lim_{x \to 0} \frac{2^x - 1}{\frac{1 - \cos x}{x}} = \lim_{x \to 0} \frac{2^x - 1}{x} \cdot \frac{x}{1 - \cos x}. \]

### Step 5: Substitute the limits into the expression
Now we substitute the limits we found:

\[ \lim_{x \to 0} \frac{2^x - 1}{x} = \log(2), \]
and we know that:

\[ \lim_{x \to 0} \frac{x}{1 - \cos x} = \lim_{x \to 0} \frac{x}{\frac{1 - \cos x}{x^2} \cdot x^2} = \lim_{x \to 0} \frac{x^2}{\frac{1 - \cos x}{x^2}} = \frac{x^2}{\frac{1}{2}} = 2x^2. \]

### Step 6: Combine the results
Now we can combine our results:

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

### Step 7: Final result
Thus, the limit evaluates to:

\[ 2 \log(2). \]

Alternatively, we can express this as:

\[ \log(4). \]

### Final Answer:
\[ \lim_{x \to 0} \frac{x(2^x - 1)}{1 - \cos x} = 2 \log(2) = \log(4). \]